blocks|key|1111870|text|我最终选择了这个解决方案，它适用于正整数和负整数的求和(对正则表达式稍作修改也适用于小数)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1111871|function+sum(string)+{
++return+(string.match(/%5E(-?\d%2B)(\%2B-?\d%2B)*$/))+?+string.split('%2B').stringSum()+:+NaN;
}+++

Array.prototype.stringSum+=+function()+{
++++var+sum+=+0;
++++for(var+k=0,+kl=this.length;k<kl;k%2B%2B)
++++{
++++++++sum+%2B=+%2Bthis[k];
++++}
++++return+sum;
}|code-block|syntax|javascript|1111872|我不确定它是否比eval()快，但由于我必须执行该操作很多次，所以运行这个脚本比创建大量javascript编译器实例要舒服得多|1111873|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

<p>I've eventually gone for this solution, which works for summing positive and negative integers (and with a little modification to the regex will work for decimals too):</p>

<pre><code>function sum(string) {
  return (string.match(/^(-?\d+)(\+-?\d+)*$/)) ? string.split('+').stringSum() : NaN;
}   

Array.prototype.stringSum = function() {
    var sum = 0;
    for(var k=0, kl=this.length;k&lt;kl;k++)
    {
        sum += +this[k];
    }
    return sum;
}
</code></pre>

<p>I'm not sure if it's faster than eval(), but as I have to carry out the operation lots of times I'm far more comfortable runing this script than creating loads of instances of the javascript compiler</p>


blocks|key|1327180|text|您可以使用JavaScript+Expression+Evaluator+library，它允许您执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1327181|Parser.evaluate("2+%5E+x",+{+x:+3+});|code-block|syntax|javascript|1327182|或者mathjs，它允许这样的东西：|1327183|math.eval('sin(45+deg)+%5E+2');|1327184|我最终为我的一个项目选择了数学。|1327185|entityMap|0|LINK|mutability|MUTABLE|url|http://silentmatt.com/javascript-expression-evaluator/|1|http://mathjs.org/^0|5|13|0|0|0|2|6|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@$A|14|B|15|1|16]]|C|$]]|$1|K|3|L|5|F|7|17|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<p>You can use the <a href="http://silentmatt.com/javascript-expression-evaluator/" rel="noreferrer">JavaScript Expression Evaluator library</a>, which allows you to do stuff like: </p>

<pre><code>Parser.evaluate("2 ^ x", { x: 3 });
</code></pre>

<p>Or <a href="http://mathjs.org/" rel="noreferrer">mathjs</a>, which allows stuff like:</p>

<pre><code>math.eval('sin(45 deg) ^ 2');
</code></pre>

<p>I ended up choosing mathjs for one of my projects.</p>


blocks|key|1111836|text|您可以轻松地执行%2B或-操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1111837|​|1111838|function+addbits(s)+{
++var+total+=+0,
++++++s+=+s.match(/[%2B\-]*(\.\d%2B%7C\d%2B(\.\d%2B)?)/g)+%7C%7C+[];
++++++
++while+(s.length)+{
++++total+%2B=+parseFloat(s.shift());
++}
++return+total;
}

var+string+=+'1%2B23%2B4%2B5-30';
console.log(
++addbits(string)
)|code-block|syntax|javascript|1111839|1111840|更复杂的数学使得eval更有吸引力--当然也更容易编写。|1111841|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|R|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|M|$]]

<p>You can do + or - easily:</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function addbits(s) {
  var total = 0,
      s = s.match(/[+\-]*(\.\d+|\d+(\.\d+)?)/g) || [];
      
  while (s.length) {
    total += parseFloat(s.shift());
  }
  return total;
}

var string = '1+23+4+5-30';
console.log(
  addbits(string)
)</code></pre>
</div>
</div>
</p>

<p>More complicated math makes eval more attractive- and certainly simpler to write.</p>


blocks|key|1326994|text|总得有人来解析这个字符串。如果不是解释器(通过eval)，那么就需要您编写解析例程来提取数字、运算符以及您希望在数学表达式中支持的任何其他内容。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1326995|所以，不，没有eval是没有任何(简单)方法的。如果您担心安全性(因为您解析的输入不是来自您控制的源)，那么在将其传递给eval之前，您可以检查输入的格式(通过白名单正则表达式过滤器|1326996|entityMap^0|N|4|0|7|4|1O|4|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

<p>Somebody has to parse that string.  If it's not the interpreter (via <code>eval</code>) then it'll need to be you, writing a parsing routine to extract numbers, operators, and anything else you want to support in a mathematical expression.</p>

<p>So, no, there isn't any (simple) way without <code>eval</code>.  If you're concerned about security (because the input you're parsing isn't from a source you control), maybe you can check the input's format (via a whitelist regex filter) before passing it to <code>eval</code>?</p>


blocks|key|1112198|text|出于同样的目的，我创建了BigEval。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1112199|在求解表达式时，它的性能与Eval()完全相同，并支持%25、%5E、&、**+(幂)和！等运算符。(阶乘)。您还可以在表达式中使用函数和常量(或者说变量)。表达式是在包括JavaScript在内的编程语言中常见的PEMDAS+order中求解的。|style|CODE|1112200|var+Obj+=+new+BigEval();
var+result+=+Obj.exec("5!+%2B+6.6e3+*+(PI+%2B+E)");+//+38795.17158152233
var+result2+=+Obj.exec("sin(45+*+deg)**2+%2B+cos(pi+/+4)**2");+//+1
var+result3+=+Obj.exec("0+&+-7+%5E+-7+-+0%251+%2B+6%252");+//-7|code-block|syntax|javascript|1112201|在处理任意精度的数字时，也可以使用这些Big+Number库进行算术运算。|1112202|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/aviaryan/BigEval.js|1|https://en.wikipedia.org/wiki/Order_of_operations#Programming_languages^0|C|7|0|0|D|6|2V|C|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|6|7|12|8|@$A|13|B|14|F|G]]|9|@$A|15|B|16|1|17]]|C|$]]|$1|H|3|I|5|J|7|18|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<p>I created <a href="https://github.com/aviaryan/BigEval.js" rel="noreferrer">BigEval</a> for the same purpose.<br>
In solving expressions, it performs exactly same as <code>Eval()</code> and supports operators like %, ^, &amp;, ** (power) and ! (factorial).
You are also allowed to use functions and constants (or say variables) inside the expression. The expression is solved in <a href="https://en.wikipedia.org/wiki/Order_of_operations#Programming_languages" rel="noreferrer">PEMDAS order</a> which is common in programming languages including JavaScript.</p>

<pre><code>var Obj = new BigEval();
var result = Obj.exec("5! + 6.6e3 * (PI + E)"); // 38795.17158152233
var result2 = Obj.exec("sin(45 * deg)**2 + cos(pi / 4)**2"); // 1
var result3 = Obj.exec("0 &amp; -7 ^ -7 - 0%1 + 6%2"); //-7
</code></pre>

<p>It can also be made to use those Big Number libraries for arithmetic in case you are dealing with numbers with arbitrary precision.</p>


blocks|key|1112532|text|简单而优雅地使用Function()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1112533|​|1112534|function+parse(str)+{
++return+Function(`'use+strict';+return+(${str})`)()
}

parse("1%2B2%2B3");+|code-block|syntax|javascript|1112535|1112536|entityMap^0|8|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|T|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|O|$]]

<p>Simple and elegant with <code>Function()</code></p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="true">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function parse(str) {
  return Function(`'use strict'; return (${str})`)()
}

parse("1+2+3"); </code></pre>
</div>
</div>
</p>


blocks|key|1327147|text|我最近在C#中做到了这一点(我们没有Eval()+...)通过计算Reverse+Polish+Notation中的表达式(这是比较容易的部分)。困难的部分实际上是解析字符串，并将其转换为反向波兰语表示法。我使用了Shunting+Yard+algorithm，因为在维基百科和伪代码上有一个很好的例子。我发现实现两者都非常简单，如果你还没有找到解决方案或正在寻找替代方案，我建议你这样做。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1327148|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Reverse_Polish_notation|1|https://en.wikipedia.org/wiki/Shunting-yard_algorithm^0|I|6|X|N|0|2Z|N|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]|$9|V|A|W|1|X]]|E|$]]|$1|F|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]|N|$5|I|J|K|E|$L|O]]]]

<p>I've recently done this in C# (no <code>Eval()</code> for us...) by evaluating the expression in <a href="https://en.wikipedia.org/wiki/Reverse_Polish_notation" rel="nofollow noreferrer">Reverse Polish Notation</a> (that's the easy bit). The hard part is actually parsing the string and turning it into Reverse Polish Notation. I used the <a href="https://en.wikipedia.org/wiki/Shunting-yard_algorithm" rel="nofollow noreferrer">Shunting Yard algorithm</a>, as there's a great example on Wikipedia and pseudocode. I found it really simple to implement both and I'd recommend this if you haven't already found a solution or are looking for alternatives.</p>


blocks|key|1112066|text|这是我刚刚拼凑的一个小函数来解决这个问题-它通过一次分析字符串中的一个字符来构建表达式(虽然它实际上相当快)。这将接受任何数学表达式(仅限于%2B、-、*、/运算符)并返回结果。它还可以处理负值和无限的数字操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112067|剩下的“要做的”就是确保它在%2B&+-之前计算*&/。稍后我将添加该功能，但现在这是我需要的……|1112068|/**
*+Evaluate+a+mathematical+expression+(as+a+string)+and+return+the+result
*+@param+{String}+expr+A+mathematical+expression
*+@returns+{Decimal}+Result+of+the+mathematical+expression
*+@example
*++++//+Returns+-81.4600
*++++expr("10.04%2B9.5-1%2B-100");
*/+
function+expr+(expr)+{

++++var+chars+=+expr.split("");
++++var+n+=+[],+op+=+[],+index+=+0,+oplast+=+true;

++++n[index]+=+"";

++++//+Parse+the+expression
++++for+(var+c+=+0;+c+<+chars.length;+c%2B%2B)+{

++++++++if+(isNaN(parseInt(chars[c]))+&&+chars[c]+!==+"."+&&+!oplast)+{
++++++++++++op[index]+=+chars[c];
++++++++++++index%2B%2B;
++++++++++++n[index]+=+"";
++++++++++++oplast+=+true;
++++++++}+else+{
++++++++++++n[index]+%2B=+chars[c];
++++++++++++oplast+=+false;
++++++++}
++++}

++++//+Calculate+the+expression
++++expr+=+parseFloat(n[0]);
++++for+(var+o+=+0;+o+<+op.length;+o%2B%2B)+{
++++++++var+num+=+parseFloat(n[o+%2B+1]);
++++++++switch+(op[o])+{
++++++++++++case+"%2B":
++++++++++++++++expr+=+expr+%2B+num;
++++++++++++++++break;
++++++++++++case+"-":
++++++++++++++++expr+=+expr+-+num;
++++++++++++++++break;
++++++++++++case+"*":
++++++++++++++++expr+=+expr+*+num;
++++++++++++++++break;
++++++++++++case+"/":
++++++++++++++++expr+=+expr+/+num;
++++++++++++++++break;
++++++++}
++++}

++++return+expr;
}|code-block|syntax|javascript|1112069|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

<p>This is a little function I threw together just now to solve this issue - it builds the expression by analyzing the string one character at a time (it's actually pretty quick though). This will take any mathematical expression (limited to +,-,*,/ operators only) and return the result. It can handle negative values and unlimited number operations as well.</p>

<p>The only "to do" left is to make sure it calculates * &amp; / before + &amp; -. Will add that functionality later, but for now this does what I need...</p>

<pre><code>/**
* Evaluate a mathematical expression (as a string) and return the result
* @param {String} expr A mathematical expression
* @returns {Decimal} Result of the mathematical expression
* @example
*    // Returns -81.4600
*    expr("10.04+9.5-1+-100");
*/ 
function expr (expr) {

    var chars = expr.split("");
    var n = [], op = [], index = 0, oplast = true;

    n[index] = "";

    // Parse the expression
    for (var c = 0; c &lt; chars.length; c++) {

        if (isNaN(parseInt(chars[c])) &amp;&amp; chars[c] !== "." &amp;&amp; !oplast) {
            op[index] = chars[c];
            index++;
            n[index] = "";
            oplast = true;
        } else {
            n[index] += chars[c];
            oplast = false;
        }
    }

    // Calculate the expression
    expr = parseFloat(n[0]);
    for (var o = 0; o &lt; op.length; o++) {
        var num = parseFloat(n[o + 1]);
        switch (op[o]) {
            case "+":
                expr = expr + num;
                break;
            case "-":
                expr = expr - num;
                break;
            case "*":
                expr = expr * num;
                break;
            case "/":
                expr = expr / num;
                break;
        }
    }

    return expr;
}
</code></pre>


blocks|key|1112222|text|试试nerdamer|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1112223|​|1112224|var+result+=+nerdamer('12%2B2%2BPI').evaluate();
document.getElementById('text').innerHTML+=+result.text();|code-block|syntax|javascript|1112225|<script+src="http://nerdamer.com/js/nerdamer.core.js"></script>
<div+id="text"></div>|1112226|1112227|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/jiggzson/nerdamer^0|2|8|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|6|7|Z|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|10|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|H|7|11|8|@]|9|@]|C|$I|J]]|$1|M|3|E|5|6|7|12|8|@]|9|@]|C|$]]|$1|N|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|O|$P|$5|Q|R|S|C|$T|U]]]]

<p>Try <a href="https://github.com/jiggzson/nerdamer" rel="nofollow">nerdamer</a></p>

<p><div class="snippet" data-lang="js" data-hide="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var result = nerdamer('12+2+PI').evaluate();
document.getElementById('text').innerHTML = result.text();</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="http://nerdamer.com/js/nerdamer.core.js"&gt;&lt;/script&gt;
&lt;div id="text"&gt;&lt;/div&gt;</code></pre>
</div>
</div>
</p>


blocks|key|1327570|text|您可以使用for循环来检查字符串是否包含任何无效字符，然后使用带有eval的try...catch来检查计算是否会像eval("2%2B%2B")那样抛出错误。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1327571|​|1327572|function+evaluateMath(str)+{
++for+(var+i+=+0;+i+<+str.length;+i%2B%2B)+{
++++if+(isNaN(str[i])+&&+!['%2B',+'-',+'/',+'*',+'%25',+'**'].includes(str[i]))+{
++++++return+NaN;
++++}
++}
++
++
++try+{
++++return+eval(str)
++}+catch+(e)+{
++++if+(e.name+!==+'SyntaxError')+throw+e
++++return+NaN;
++}
}

console.log(evaluateMath('2+%2B+6'))|code-block|syntax|javascript|1327573|1327574|或者，您可以设置Math.eval，而不是函数|1327575|1327576|Math.eval+=+function(str)+{
++for+(var+i+=+0;+i+<+str.length;+i%2B%2B)+{
++++if+(isNaN(str[i])+&&+!['%2B',+'-',+'/',+'*',+'%25',+'**'].includes(str[i]))+{
++++++return+NaN;
++++}
++}
++
++
++try+{
++++return+eval(str)
++}+catch+(e)+{
++++if+(e.name+!==+'SyntaxError')+throw+e
++++return+NaN;
++}
}

console.log(Math.eval('2+%2B+6'))|1327577|1327578|entityMap^0|1M|B|0|0|0|0|8|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Z|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|10|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|P|3|G|5|6|7|14|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|S|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|U|$]]

<p>You could use a for loop to check if the string contains any invalid characters and then use a try...catch with eval to check if the calculation throws an error like <code>eval(&quot;2++&quot;)</code> would.</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function evaluateMath(str) {
  for (var i = 0; i &lt; str.length; i++) {
    if (isNaN(str[i]) &amp;&amp; !['+', '-', '/', '*', '%', '**'].includes(str[i])) {
      return NaN;
    }
  }
  
  
  try {
    return eval(str)
  } catch (e) {
    if (e.name !== 'SyntaxError') throw e
    return NaN;
  }
}

console.log(evaluateMath('2 + 6'))</code></pre>
</div>
</div>
</p>
<p>or instead of a function, you could set <code>Math.eval</code></p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>Math.eval = function(str) {
  for (var i = 0; i &lt; str.length; i++) {
    if (isNaN(str[i]) &amp;&amp; !['+', '-', '/', '*', '%', '**'].includes(str[i])) {
      return NaN;
    }
  }
  
  
  try {
    return eval(str)
  } catch (e) {
    if (e.name !== 'SyntaxError') throw e
    return NaN;
  }
}

console.log(Math.eval('2 + 6'))</code></pre>
</div>
</div>
</p>


blocks|key|1112287|text|尝试使用选择器表达式计算AutoCalculator+https://github.com/JavscriptLab/autocalculate的输入值和输出|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1112288|只需为输出输入添加一个属性，如data-ac="(#firstinput%2B#secondinput)“|1112289|不需要任何初始化，只需添加data-ac属性即可。它会自动找出动态添加的元素|1112290|FOr添加'Rs‘，输出只需添加大括号内的数据-ac=“{Rs}(#firstinput%2B#secondinput)”|1112291|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/JavscriptLab/autocalculate^0|R|19|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]]|C|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|C|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|C|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|K|$L|$5|M|N|O|C|$P|Q]]]]

<p>Try AutoCalculator
<a href="https://github.com/JavscriptLab/autocalculate" rel="nofollow noreferrer">https://github.com/JavscriptLab/autocalculate</a>
Calculate Inputs value and Output By using selector expressions</p>

<p>Just add an attribute for your output input like
data-ac="(#firstinput+#secondinput)"</p>

<p>No Need of any initialization just add data-ac attribute only.
It will find out dynamically added elements automatically</p>

<p>FOr add 'Rs' with Output just add inside curly bracket
data-ac="{Rs}(#firstinput+#secondinput)"</p>


blocks|key|1112378|text|const+operatorToFunction+=+{
++++"%2B":+(num1,+num2)+=>+%2Bnum1+%2B+%2Bnum2,
++++"-":+(num1,+num2)+=>+%2Bnum1+-+%2Bnum2,
++++"*":+(num1,+num2)+=>+%2Bnum1+*+%2Bnum2,
++++"/":+(num1,+num2)+=>+%2Bnum1+/+%2Bnum2
}

const+findOperator+=+(str)+=>+{
++++const+[operator]+=+str.split("").filter((ch)+=>+["%2B",+"-",+"*",+"/"].includes(ch))
++++return+operator;
}

const+executeOperation+=+(str)+=>+{
++++const+operationStr+=+str.replace(/[+]/g,+"");
++++const+operator+=+findOperator(operationStr);
++++const+[num1,+num2]+=+operationStr.split(operator)
++++return+operatorToFunction[operator](num1,+num2);
};

const+addition+=+executeOperation('1+%2B+1');+//+ans+is+2
const+subtraction+=+executeOperation('4+-+1');+//+ans+is+3
const+multiplication+=+executeOperation('2+*+5');+//+ans+is+10
const+division+=+executeOperation('16+/+4');+//+ans+is+4|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1112379|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>const operatorToFunction = {
    "+": (num1, num2) =&gt; +num1 + +num2,
    "-": (num1, num2) =&gt; +num1 - +num2,
    "*": (num1, num2) =&gt; +num1 * +num2,
    "/": (num1, num2) =&gt; +num1 / +num2
}

const findOperator = (str) =&gt; {
    const [operator] = str.split("").filter((ch) =&gt; ["+", "-", "*", "/"].includes(ch))
    return operator;
}

const executeOperation = (str) =&gt; {
    const operationStr = str.replace(/[ ]/g, "");
    const operator = findOperator(operationStr);
    const [num1, num2] = operationStr.split(operator)
    return operatorToFunction[operator](num1, num2);
};

const addition = executeOperation('1 + 1'); // ans is 2
const subtraction = executeOperation('4 - 1'); // ans is 3
const multiplication = executeOperation('2 * 5'); // ans is 10
const division = executeOperation('16 / 4'); // ans is 4
</code></pre>


blocks|key|1112653|text|最好也是最简单的方法就是使用math.js库。下面是一些示例代码，演示如何使用该库。单击here来摆弄。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1112654|//+functions+and+constants
math.round(math.e,+3)++++++++++++++++//+2.718
math.atan2(3,+-3)+/+math.pi++++++++++//+0.75
math.log(10000,+10)++++++++++++++++++//+4
math.sqrt(-4)++++++++++++++++++++++++//+2i
math.derivative('x%5E2+%2B+x',+'x')++++++//+2*x%2B1
math.pow([[-1,+2],+[3,+1]],+2)
+++++//+[[7,+0],+[0,+7]]

//+expressions
math.evaluate('1.2+*+(2+%2B+4.5)')+++++//+7.8
math.evaluate('12.7+cm+to+inch')+++++//+5+inch
math.evaluate('sin(45+deg)+%5E+2')+++++//+0.5
math.evaluate('9+/+3+%2B+2i')++++++++++//+3+%2B+2i
math.evaluate('det([-1,+2;+3,+1])')++//+-7

//+chaining
math.chain(3)
++++.add(4)
++++.multiply(2)
++++.done()+//+14|code-block|syntax|javascript|1112655|entityMap|0|LINK|mutability|MUTABLE|url|https://mathjs.org/|1|https://jsbin.com/devacu/edit^0|E|7|0|18|4|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

<p>The best way and easiest way is to use <a href="https://mathjs.org/" rel="nofollow noreferrer">math.js</a> library.
Here some example code demonstrating how to use the library. Click <a href="https://jsbin.com/devacu/edit" rel="nofollow noreferrer">here</a> to fiddle around.</p>
<pre><code>// functions and constants
math.round(math.e, 3)                // 2.718
math.atan2(3, -3) / math.pi          // 0.75
math.log(10000, 10)                  // 4
math.sqrt(-4)                        // 2i
math.derivative('x^2 + x', 'x')      // 2*x+1
math.pow([[-1, 2], [3, 1]], 2)
     // [[7, 0], [0, 7]]

// expressions
math.evaluate('1.2 * (2 + 4.5)')     // 7.8
math.evaluate('12.7 cm to inch')     // 5 inch
math.evaluate('sin(45 deg) ^ 2')     // 0.5
math.evaluate('9 / 3 + 2i')          // 3 + 2i
math.evaluate('det([-1, 2; 3, 1])')  // -7

// chaining
math.chain(3)
    .add(4)
    .multiply(2)
    .done() // 14
</code></pre>


blocks|key|1112716|text|我做了一个小函数来解析一个包含%2B，/，-，*的数学表达式。我使用了if语句，我认为switch+cases会更好。首先，我将字符串分离为运算符，然后将其数字从字符串转换为浮点数，然后在执行操作时遍历。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1112717|​|1112718|+const+evaluate=(mathExpStr)+=>+{
++++mathExpStr.replace(/[%2B-\/*]$/,+"");
++++let+regExp+=+/\d%2B/g;
++++let+valueArr+=+(mathExpStr.match(regExp)+%7C%7C+[]).map((val)+=>
++++++Number.parseFloat(val)
++++);
++++let+operatorArr+=+mathExpStr.match(/[%2B-\/*]/g)+%7C%7C+[];
++++return+converter(valueArr,+operatorArr)
++}

const+converter+=+(arr,operators)=>{
++let+arr2=[...arr]
++for(let+i=0;i<arr.length;i%2B%2B){
++++let+o;
++++if(arr2.length<2){return+arr2[0]}
++++if(operators[i]=="%2B"){
++++++o=arr2[0]%2Barr2[1]
++++++arr2.splice(0,+2,+o)
++++++console.log(o,arr2,+operators[i])
++++}
++++if(operators[i]=="-"){
++++++o=arr2[0]-arr2[1]
++++++arr2.splice(0,2,+o)
++++++console.log(o,arr2,+operators[i])
++++}
++++if(operators[i]=="*"){
++++++o=arr2[0]*arr2[1]
++++++arr2.splice(0,2,o)
++++++console.log(o,arr2,+operators[i])
++++}
++++if(operators[i]=="/"){
++++++o=arr2[0]/arr2[1]
++++++arr2.splice(0,2,+o)
++++++console.log(o,arr2,+operators[i])
++++}
++}
}
//+console.log(converter(valueArr,+operatorArr))
console.log(evaluate("1%2B3%2B5%2B6-4*2/4"))|code-block|syntax|javascript|1112719|1112720|entityMap^0|X|2|15|C|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|V|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|W|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|O|$]]

<p>I made a small function to parse a math expression, containing +,/,-,*. I used <code>if</code> statements I think <code>switch cases</code> will be better.
Firstly I separated the string into the operator and its numbers convert then from string to float then iterate through while performing the operation.</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code> const evaluate=(mathExpStr) =&gt; {
    mathExpStr.replace(/[+-\/*]$/, "");
    let regExp = /\d+/g;
    let valueArr = (mathExpStr.match(regExp) || []).map((val) =&gt;
      Number.parseFloat(val)
    );
    let operatorArr = mathExpStr.match(/[+-\/*]/g) || [];
    return converter(valueArr, operatorArr)
  }

const converter = (arr,operators)=&gt;{
  let arr2=[...arr]
  for(let i=0;i&lt;arr.length;i++){
    let o;
    if(arr2.length&lt;2){return arr2[0]}
    if(operators[i]=="+"){
      o=arr2[0]+arr2[1]
      arr2.splice(0, 2, o)
      console.log(o,arr2, operators[i])
    }
    if(operators[i]=="-"){
      o=arr2[0]-arr2[1]
      arr2.splice(0,2, o)
      console.log(o,arr2, operators[i])
    }
    if(operators[i]=="*"){
      o=arr2[0]*arr2[1]
      arr2.splice(0,2,o)
      console.log(o,arr2, operators[i])
    }
    if(operators[i]=="/"){
      o=arr2[0]/arr2[1]
      arr2.splice(0,2, o)
      console.log(o,arr2, operators[i])
    }
  }
}
// console.log(converter(valueArr, operatorArr))
console.log(evaluate("1+3+5+6-4*2/4"))</code></pre>
</div>
</div>
</p>


blocks|key|1112364|text|这是一个类似于jMichael的算法解决方案，它逐个字符地遍历表达式，并逐步跟踪left/operator/right。该函数在找到一个运算符之后，在每一轮之后累加结果。此版本仅支持'%2B‘和'-’运算符，但编写为使用其他运算符进行扩展。注意:我们在循环前将'currOp‘设置为'%2B’，因为我们假设表达式以正浮点数开始。实际上，总的来说，我假设输入类似于计算器的输入。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112365|function+calculate(exp)+{
++const+opMap+=+{
++++'%2B':+(a,+b)+=>+{+return+parseFloat(a)+%2B+parseFloat(b)+},
++++'-':+(a,+b)+=>+{+return+parseFloat(a)+-+parseFloat(b)+},
++};
++const+opList+=+Object.keys(opMap);

++let+acc+=+0;
++let+next+=+'';
++let+currOp+=+'%2B';

++for+(let+char+of+exp)+{
++++if+(opList.includes(char))+{
++++++acc+=+opMap[currOp](acc,+next);
++++++currOp+=+char;
++++++next+=+'';
++++}+else+{
++++++next+%2B=+char;
++++}+
++}

++return+currOp+===+'%2B'+?+acc+%2B+parseFloat(next)+:+acc+-+parseFloat(next);
}|code-block|syntax|javascript|1112366|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>Here is an algorithmic solution similar to jMichael's that loops through the expression character by character and progressively tracks left/operator/right. The function accumulates the result after each turn it finds an operator character. This version only supports '+' and '-' operators but is written to be extended with other operators. Note: we set 'currOp' to '+' before looping because we assume the expression starts with a positive float. In fact, overall I'm making the assumption that input is similar to what would come from a calculator. </p>

<pre><code>function calculate(exp) {
  const opMap = {
    '+': (a, b) =&gt; { return parseFloat(a) + parseFloat(b) },
    '-': (a, b) =&gt; { return parseFloat(a) - parseFloat(b) },
  };
  const opList = Object.keys(opMap);

  let acc = 0;
  let next = '';
  let currOp = '+';

  for (let char of exp) {
    if (opList.includes(char)) {
      acc = opMap[currOp](acc, next);
      currOp = char;
      next = '';
    } else {
      next += char;
    } 
  }

  return currOp === '+' ? acc + parseFloat(next) : acc - parseFloat(next);
}
</code></pre>


blocks|key|1327701|text|基于Aniket+Kudale的parse|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1327702|将上下文变量添加到表达式|1327703|function+parseExpr(str:+string,+params:+any)+{
++const+names+=+Object.keys(params);
++const+vals+=+Object.values(params);
++return+Function(...names,+`'use+strict';+return+(${str})`)(...vals);
}|code-block|syntax|javascript|1327704|示例|1327705|>+parseExpr('age+>+50?+x+:+x/2',+{x:+40,+age:+46})
20

>+parseExpr('age+>+50?+x+:+x/2',+{x:+40,+age:+60})
40|1327706|entityMap^0|G|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

<p>Based on Aniket Kudale's <code>parse</code></p>
<p>To add context variables to the expression</p>
<pre><code>function parseExpr(str: string, params: any) {
  const names = Object.keys(params);
  const vals = Object.values(params);
  return Function(...names, `'use strict'; return (${str})`)(...vals);
}
</code></pre>
<p>example</p>
<pre><code>&gt; parseExpr('age &gt; 50? x : x/2', {x: 40, age: 46})
20

&gt; parseExpr('age &gt; 50? x : x/2', {x: 40, age: 60})
40

</code></pre>


<p>How do I parse and evaluate a mathematical expression in a string (e.g. <code>'1+1'</code>) without invoking <code>eval(string)</code> to yield its numerical value?</p>

<p>With that example, I want the function to accept <code>'1+1'</code> and return <code>2</code>.</p>


Evaluating a string as a mathematical expression in JavaScript

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如何在不调用eval(string)的情况下解析和计算字符串中的数学表达式(例如'1+1')并生成其数值？在这个例子中，我希望函数接受'1+1'并返回2。

问在JavaScript中将字符串作为数学表达式求值
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回答 16

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问在JavaScript中将字符串作为数学表达式求值
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