blocks|key|2945472|text|使用正则表达式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2945473|类总和|2945474|pattern+=+/%5E\d{1,14)(\.{1,4}\)?$/;

if+(patten.test(yourNumber))+{
//+Hunky+dory
}
else
{
//+have+another+bash
}|code-block|syntax|javascript|2945475|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|$]]

Use a regular expression

summat like

<pre><code>pattern = /^\d{1,14)(\.{1,4}\)?$/;

if (patten.test(yourNumber)) {
// Hunky dory
}
else
{
// have another bash
}
</code></pre>

blocks|key|2945480|text|为什么不使用split()方法(下面未经测试的代码)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2945481|function+Comma(num)+{
++var+s+=+num.split('.');
++if+(s[0].length+>+14)+{
++++//+Too+many+numbers+before+decimal.
++}
++if+(s[1].length+>+4)+{
++++//+Too+many+numbers+after+decimal.
++}
}|code-block|syntax|javascript|2945482|编辑|style|BOLD|2945483|下面的代码将接受任何数字，并返回一个小数点前至多14位，小数点后至多4位的数字(它实际上并不验证输入的是一个数字，但您得到了图像)：|2945484|function+Comma(num)+{
++var+s+=+num.split('.');
++var+beforeDecimal+=+s[0];+++++++++//+This+is+the+number+BEFORE+the+decimal.
++var+afterDecimal+=+'0000';++++++++//+Default+value+for+digits+after+decimal
++if+(s.length+>+1)+++++++++++++++++//+Check+that+there+indeed+is+a+decimal+separator.
++++afterDecimal+=+s[1];++++++++++++//+This+is+the+number+AFTER+the+decimal.
++if+(beforeDecimal.length+>+14)+{
++++//+Too+many+numbers+before+decimal.
++++//+Get+the+first+14+digits+and+discard+the+rest.
++++beforeDecimal+=+beforeDecimal.substring(0,+14);
++}
++if+(afterDecimal.length+>+4)+{
++++//+Too+many+numbers+after+decimal.
++++//+Get+the+first+4+digits+and+discard+the+rest.
++++afterDecimal+=+afterDecimal.substring(0,+4);
++}

++//+Return+the+new+number+with+at+most+14+digits+before+the+decimal
++//+and+at+most+4+after.
++return+beforeDecimal+%2B+"."+%2B+afterDecimal;
}|2945485|(和往常一样，代码是未经测试的。)|2945486|entityMap|0|LINK|mutability|MUTABLE|url|http://www.w3schools.com/jsref/jsref_split.asp^0|6|7|0|0|0|0|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|15|8|@$A|16|B|17|K|L]]|9|@]|C|$]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|1A|8|@]|9|@]|C|$]]|$1|S|3|-4|5|6|7|1B|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]]]

Why not use the <a href="http://www.w3schools.com/jsref/jsref_split.asp" rel="noreferrer">split()</a> method (untested code below):

<pre><code>function Comma(num) {
 var s = num.split('.');
 if (s[0].length &gt; 14) {
 // Too many numbers before decimal.
 }
 if (s[1].length &gt; 4) {
 // Too many numbers after decimal.
 }
}
</code></pre>

Edit 
The following will take any number and return a number with at most 14 digits before the decimal point and at most 4 digits after (well it doesn't actually verify that the input is a number but you get the picture):

<pre><code>function Comma(num) {
 var s = num.split('.');
 var beforeDecimal = s[0]; // This is the number BEFORE the decimal.
 var afterDecimal = '0000'; // Default value for digits after decimal
 if (s.length &gt; 1) // Check that there indeed is a decimal separator.
 afterDecimal = s[1]; // This is the number AFTER the decimal.
 if (beforeDecimal.length &gt; 14) {
 // Too many numbers before decimal.
 // Get the first 14 digits and discard the rest.
 beforeDecimal = beforeDecimal.substring(0, 14);
 }
 if (afterDecimal.length &gt; 4) {
 // Too many numbers after decimal.
 // Get the first 4 digits and discard the rest.
 afterDecimal = afterDecimal.substring(0, 4);
 }

 // Return the new number with at most 14 digits before the decimal
 // and at most 4 after.
 return beforeDecimal + "." + afterDecimal;
}
</code></pre>

(And as always the code is untested.)

blocks|key|255948|text|我认为我们可以同意，将数字转换为字符串并计算小数点左边的数字是非常粗俗的，特别是考虑到非常大的数字可以转换为科学记数法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|255949|我不是计算机科学家，但这里有一些我在JS中拼凑起来的东西，它将在数学上做到这一点。支持负数、科学记数法，可以解析10%5E308到10%5E323的值。|255950|function+countDigits(value)+{
++if+(value+===+0)+return+{+wholePlaces:+0,+decimalPlaces:+0+};

++var+absValue+=+Math.abs(value);+//+-15.555+becomes+15.555
++var+wholePlaces+=+0;
++for+(;+wholePlaces+<=+308;+%2B%2BwholePlaces)+{+//+Number.MAX_VALUE+is+1.798e%2B308
++++if+(absValue+<+Math.pow(10,+wholePlaces))
++++++break;
++}

++var+decimalValue+=+absValue+-+Math.floor(absValue);+//+15.555+-+15+=+0.555
++var+decimalPlaces+=+0;
++for+(;+decimalPlaces+>=+-323;+--decimalPlaces)+{+//+Number.MIN_VALUE+is+5e-324
++++var+temp+=+(decimalValue+/+Math.pow(10,+decimalPlaces))+%2B+0.09;+//+Adding+0.09+to+counter+float+errors
++++if+(temp+-+Math.floor(temp)+<+0.1)++//+If+the+decimal+remaining+is+smaller+that+0.1,+we've+reached+the+end
++++++break;
++}
++decimalPlaces+=+Math.abs(decimalPlaces);
++return+{
++++wholePlaces,
++++decimalPlaces,
++}
}

countDigits(0);+++++++++//+{+wholePlaces:+0,+decimalPlaces:+0+}
countDigits(0.10);++++++//+{+wholePlaces:+0,+decimalPlaces:+1+}
countDigits(-0.10);+++++//+{+wholePlaces:+0,+decimalPlaces:+1+}
countDigits(0.10000);+++//+{+wholePlaces:+0,+decimalPlaces:+1+}
countDigits(-0.10000);++//+{+wholePlaces:+0,+decimalPlaces:+1+}
countDigits(5);+++++++++//+{+wholePlaces:+1,+decimalPlaces:+0+}
countDigits(-5);++++++++//+{+wholePlaces:+1,+decimalPlaces:+0+}
countDigits(15.555);++++//+{+wholePlaces:+2,+decimalPlaces:+3+}
countDigits(-15.555);+++//+{+wholePlaces:+2,+decimalPlaces:+3+}
countDigits(215.555);+++//+{+wholePlaces:+3,+decimalPlaces:+3+}
countDigits(-215.555);++//+{+wholePlaces:+3,+decimalPlaces:+3+}
countDigits(1.55555e%2B4)+//+{+wholePlaces:+5,+decimalPlaces:+1+}+(15555.5)|code-block|syntax|javascript|255951|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|$]]

I think we can agree that converting numbers into strings and counting the digits left of the decimal is gross, especially considering that very large numbers can get converted to scientific notation.

I am not a computer scientist but here is something I cobbled together in JS that will do this mathematically. Supports negative numbers, scientific notation, and can parse a value as large as 10^308 and as small as 10^-323.

<pre><code>function countDigits(value) {
 if (value === 0) return { wholePlaces: 0, decimalPlaces: 0 };

 var absValue = Math.abs(value); // -15.555 becomes 15.555
 var wholePlaces = 0;
 for (; wholePlaces &lt;= 308; ++wholePlaces) { // Number.MAX_VALUE is 1.798e+308
 if (absValue &lt; Math.pow(10, wholePlaces))
 break;
 }

 var decimalValue = absValue - Math.floor(absValue); // 15.555 - 15 = 0.555
 var decimalPlaces = 0;
 for (; decimalPlaces &gt;= -323; --decimalPlaces) { // Number.MIN_VALUE is 5e-324
 var temp = (decimalValue / Math.pow(10, decimalPlaces)) + 0.09; // Adding 0.09 to counter float errors
 if (temp - Math.floor(temp) &lt; 0.1) // If the decimal remaining is smaller that 0.1, we've reached the end
 break;
 }
 decimalPlaces = Math.abs(decimalPlaces);
 return {
 wholePlaces,
 decimalPlaces,
 }
}

countDigits(0); // { wholePlaces: 0, decimalPlaces: 0 }
countDigits(0.10); // { wholePlaces: 0, decimalPlaces: 1 }
countDigits(-0.10); // { wholePlaces: 0, decimalPlaces: 1 }
countDigits(0.10000); // { wholePlaces: 0, decimalPlaces: 1 }
countDigits(-0.10000); // { wholePlaces: 0, decimalPlaces: 1 }
countDigits(5); // { wholePlaces: 1, decimalPlaces: 0 }
countDigits(-5); // { wholePlaces: 1, decimalPlaces: 0 }
countDigits(15.555); // { wholePlaces: 2, decimalPlaces: 3 }
countDigits(-15.555); // { wholePlaces: 2, decimalPlaces: 3 }
countDigits(215.555); // { wholePlaces: 3, decimalPlaces: 3 }
countDigits(-215.555); // { wholePlaces: 3, decimalPlaces: 3 }
countDigits(1.55555e+4) // { wholePlaces: 5, decimalPlaces: 1 } (15555.5)
</code></pre>

blocks|key|261203|text|通过数学的另一种选择|type|unstyled|depth|inlineStyleRanges|entityRanges|data|261204|261205|您可以通过以下行获取逗号前的数字部分的数字|ordered-list-item|261206|261207|let+digitsBefore+=+Math.ceil(Math.log10(Math.floor(Math.abs(x))%2B1))，其中x是|offset|length|style|CODE|261208|的数字，字符串总长度不应超过digitsBefore%2B1%2B4。%2B1表示小数点，4表示允许的小数位数。|unordered-list-item|261209|261210|entityMap^0|0|0|0|0|0|1V|0|E|G|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|T|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|U|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|V|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|O|7|Z|8|@$I|10|J|11|K|L]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

Another option via math
<ol>
<li>You can get the digits of the number-part before the comma via the following line 
<code>let digitsBefore = Math.ceil(Math.log10(Math.floor(Math.abs(x))+1))</code>
where x is the number</li>
<li>The total string length should not exceed <code>digitsBefore+1+4</code>. The +1 is for the decimal point, the four is the number of decimals you allow.</li>
</ol>

I have a requirement where I should allow a maximum of 14 digits before decimal place and a maximum of 4 digits after decimal place.
Is there a way that I can let the user know if he is entering 222222222222222.222 -- 15 digits before decimal is invalid once he is out of that textbox using Javascript.
I tried this but it did not help me:
<pre><code> MynewTextBox.Attributes.Add(&quot;onkeyup&quot;, &quot;javascript:this.value=Comma(this.value);&quot;);

function Comma( Num ) {

 var period = Num.indexOf('.'); 
 if ( Num.length &gt; (period + 4)) 
 alert(&quot;too many after decimal point&quot;);
 if ( period != -1 ) 
 {
 Num += '00000'; 
 Num = Num.substr( 0, (period + 4));
 } 
</code></pre>
Also, the above function is giving me the error:
<blockquote>
Object Expected.
</blockquote>
Can anyone help me with this.

Count the number of digits before and after decimal places using Javascript

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我有一个要求，我应该允许小数位前最多14位，小数位后最多4位。有没有一种方法可以让用户知道，一旦他使用Javascript离开文本框，他输入的是222222222222222.222 --小数之前的15位数字是无效的。我试过了，但它对我没有帮助：  MynewTextBox.Attributes.Add("onkeyu...

问使用Javascript计算小数位前后的位数
EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问使用Javascript计算小数位前后的位数EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问使用Javascript计算小数位前后的位数
EN