blocks|key|1494342|text|正则表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1494343|var+numberPattern+=+/\d%2B/g;

'something102asdfkj1948948'.match(+numberPattern+)|code-block|syntax|javascript|1494344|这将返回一个数组，里面有两个元素，'102‘和'1948948’。按您希望的方式操作。如果不匹配，它将返回null。|1494345|要连接它们，请执行以下操作：|1494346|'something102asdfkj1948948'.match(+numberPattern+).join('')|1494347|假设您不是在处理复杂的小数，我想这应该足够了。|1494348|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Regular expressions:

<pre class="lang-js prettyprint-override"><code>var numberPattern = /\d+/g;

'something102asdfkj1948948'.match( numberPattern )
</code></pre>

This would return an Array with two elements inside, '102' and '1948948'. Operate as you wish. If it doesn't match any it will return null.

To concatenate them:

<pre class="lang-js prettyprint-override"><code>'something102asdfkj1948948'.match( numberPattern ).join('')
</code></pre>

Assuming you're not dealing with complex decimals, this should suffice I suppose.

blocks|key|1279628|text|您还可以去掉所有非数字字符(\D或[%5E0-9])：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1279629|​|1279630|let+word_With_Numbers+=+'abc123c+def4567hij89'
let+word_Without_Numbers+=+word_With_Numbers.replace(/\D/g,+'');

console.log(word_Without_Numbers)|code-block|syntax|javascript|1279631|1279632|entityMap^0|E|2|H|6|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|$]]

You could also strip all the non-digit characters (<code>\D</code> or <code>[^0-9]</code>):

<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>let word_With_Numbers = 'abc123c def4567hij89'
let word_Without_Numbers = word_With_Numbers.replace(/\D/g, '');

console.log(word_Without_Numbers)</code></pre>
</div>
</div>

blocks|key|1279697|text|如果您只想要数字：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279698|var+value+=+'675-805-714';
var+numberPattern+=+/\d%2B/g;
value+=+value.match(+numberPattern+).join([]);
alert(value);
//Show:+675805714|code-block|syntax|javascript|1279699|现在你得到了连接的数字|1279700|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|$]]

If you want only digits:

<pre><code>var value = '675-805-714';
var numberPattern = /\d+/g;
value = value.match( numberPattern ).join([]);
alert(value);
//Show: 675805714
</code></pre>

Now you get the digits joined

blocks|key|1494372|text|我猜您想要从字符串中获取数字。在这种情况下，您可以使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1494373|//+Returns+an+array+of+numbers+located+in+the+string
function+get_numbers(input)+{
++++return+input.match(/[0-9]%2B/g);
}

var+first_test+=+get_numbers('something102');
var+second_test+=+get_numbers('something102or12');
var+third_test+=+get_numbers('no+numbers+here!');

alert(first_test);+//+[102]
alert(second_test);+//+[102,12]
alert(third_test);+//+null|code-block|syntax|javascript|1494374|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|$]]

I guess you want to get number(s) from the string. In which case, you can use the following:

<pre><code>// Returns an array of numbers located in the string
function get_numbers(input) {
 return input.match(/[0-9]+/g);
}

var first_test = get_numbers('something102');
var second_test = get_numbers('something102or12');
var third_test = get_numbers('no numbers here!');

alert(first_test); // [102]
alert(second_test); // [102,12]
alert(third_test); // null
</code></pre>

blocks|key|1279776|text|如果您想捕获任何类型的数字，那么陈大超的#3+answer+at+this+time是正确的方法，但正则表达式可以缩写为：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1279777|/[-]{0,1}[\d]*[\.]{0,1}[\d]%2B/g|code-block|syntax|javascript|1279778|至：|1279779|/-?\d*\.?\d%2B/g|1279780|例如，下面的代码：|1279781|"lin-grad.ient(217deg,rgba(255,+0,+0,+-0.8),+rgba(-255,0,0,0)+70.71%25)".match(/-?\d*\.?\d%2B/g)|1279782|生成此数组：|1279783|["217","255","0","0","-0.8","-255","0","0","0","70.71"]|1279784|我去掉了一个MDN线性渐变示例，这样它就可以完全测试regexp，而不需要在这里滚动。我想我已经在文本"lin-grad.ient“中包含了负数、小数、单位后缀(如deg和%25+)、不一致的逗号和空格用法以及额外的点/句点和连字符/破折号的所有可能性。如果我遗漏了什么，请告诉我。我能看到的唯一不能处理的是一个格式错误的十进制数，比如"0..8“。|1279785|如果您确实需要一个数字数组，您可以在同一行代码中转换整个数组：|1279786|array+=+whatever.match(/-?\d*\.?\d%2B/g).map(Number);|1279787|我的特殊代码是解析CSS函数，它不需要担心点/句点字符的非数字用法，因此正则表达式可以更简单：|1279788|/-?[\d\.]%2B/g|1279789|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/42264780/4941356^0|K|M|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@$A|1D|B|1E|1|1F]]|C|$]]|$1|D|3|E|5|F|7|1G|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1H|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|1I|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1J|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1K|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|1L|8|@]|9|@]|C|$]]|$1|S|3|T|5|F|7|1M|8|@]|9|@]|C|$G|H]]|$1|U|3|V|5|6|7|1N|8|@]|9|@]|C|$]]|$1|W|3|X|5|6|7|1O|8|@]|9|@]|C|$]]|$1|Y|3|Z|5|F|7|1P|8|@]|9|@]|C|$G|H]]|$1|10|3|11|5|6|7|1Q|8|@]|9|@]|C|$]]|$1|12|3|13|5|F|7|1R|8|@]|9|@]|C|$G|H]]|$1|14|3|-4|5|6|7|1S|8|@]|9|@]|C|$]]]|15|$16|$5|17|18|19|C|$1A|1B]]]]

IMO the <a href="https://stackoverflow.com/a/42264780/4941356">#3 answer at this time</a> by Chen Dachao is the right way to go if you want to capture any kind of number, but the regular expression can be shortened from:

<pre><code>/[-]{0,1}[\d]*[\.]{0,1}[\d]+/g
</code></pre>

to:

<pre><code>/-?\d*\.?\d+/g
</code></pre>

For example, this code:

<pre><code>"lin-grad.ient(217deg,rgba(255, 0, 0, -0.8), rgba(-255,0,0,0) 70.71%)".match(/-?\d*\.?\d+/g)
</code></pre>

generates this array:

<pre><code>["217","255","0","0","-0.8","-255","0","0","0","70.71"]
</code></pre>

I've butchered an MDN linear gradient example so that it fully tests the regexp and doesn't need to scroll here. I think I've included all the possibilities in terms of negative numbers, decimals, unit suffixes like deg and %, inconsistent comma and space usage, and the extra dot/period and hyphen/dash characters within the text "lin-grad.ient". Please let me know if I'm missing something. The only thing I can see that it does not handle is a badly formed decimal number like "0..8".

If you really want an array of numbers, you can convert the entire array in the same line of code:

<pre><code>array = whatever.match(/-?\d*\.?\d+/g).map(Number);
</code></pre>

My particular code, which is parsing CSS functions, doesn't need to worry about the non-numeric use of the dot/period character, so the regular expression can be even simpler:

<pre><code>/-?[\d\.]+/g
</code></pre>

blocks|key|1279668|text|var+result+=+input.match(/\d%2B/g).join([])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1279669|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>var result = input.match(/\d+/g).join([])
</code></pre>

blocks|key|1494734|text|使用拆分和正则表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1494735|++++var+str+=+"fooBar0123".split(/(\d%2B)/);
++++console.log(str[0]);+//+fooBar
++++console.log(str[1]);+//+0123|code-block|syntax|javascript|1494736|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|$]]

Using split and regex :

<pre><code> var str = "fooBar0123".split(/(\d+)/);
 console.log(str[0]); // fooBar
 console.log(str[1]); // 0123
</code></pre>

blocks|key|1279963|text|根据@Syntle的回答，如果只有非数字字符，则会得到一个Uncaught+TypeError:+Cannot+read+property+'join'+of+null。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1279964|这将防止在未找到匹配项并返回空字符串时出现错误：|1279965|('something'.match(+/\d%2B/g+)%7C%7C[]).join('')|1279966|entityMap^0|T|1J|0|0|0|16|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|K|$]]

As per @Syntle's answer, if you have only non numeric characters you'll get an <code>Uncaught TypeError: Cannot read property 'join' of null</code>.
This will prevent errors if no matches are found and return an empty string:
<code>('something'.match( /\d+/g )||[]).join('')</code>

blocks|key|1279833|text|如果您还想要点/逗号分隔的数字，那么：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279834|\d*\.?\d*|code-block|syntax|javascript|1279835|或|1279836|[0-9]*\.?[0-9]*|1279837|您可以使用https://regex101.com/测试您的正则表达式。|offset|length|1279838|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/^0|0|0|0|0|5|L|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

If you want dot/comma separated numbers also, then:

<pre><code>\d*\.?\d*
</code></pre>

or

<pre><code>[0-9]*\.?[0-9]*
</code></pre>

You can use <a href="https://regex101.com/" rel="nofollow noreferrer">https://regex101.com/</a> to test your regexes.

blocks|key|1279887|text|其他解决方案拥有的所有功能，但需要进行一些验证|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279888|//+value+=+'675-805-714'
const+validateNumberInput+=+(value)+=>+{+
++++let+numberPattern+=+/\d%2B/g+
++++let+numbers+=+value.match(numberPattern)

++++if+(numbers+===+null)+{
++++++++return+0
++++}++

++++return+parseInt(numbers.join([]))
}
//+675805714|code-block|syntax|javascript|1279889|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|$]]

Everything that other solutions have, but with a little validation

<pre><code>// value = '675-805-714'
const validateNumberInput = (value) =&gt; { 
 let numberPattern = /\d+/g 
 let numbers = value.match(numberPattern)

 if (numbers === null) {
 return 0
 } 

 return parseInt(numbers.join([]))
}
// 675805714
</code></pre>

blocks|key|1494810|text|下面是使用Regex将字符串转换为有效的普通或十进制数字的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1494811|//something123.777.321something+to+123.777321
const+str+=+'something123.777.321something';
let+initialValue+=+str.replace(/[%5E0-9.]%2B/,+'');
//initialValue+=+'123.777.321';

//characterCount+just+count+the+characters+in+a+given+string
if+(characterCount(intitialValue,+'.')+>+1)+{
+const+splitedValue+=+intitialValue.split('.');
+//splittedValue+=+['123','777','321'];
+intitialValue+=+splitedValue.shift()+%2B+'.'+%2B+splitedValue.join('');
+//result+i.e.+initialValue+=+'123.777321'
}|code-block|syntax|javascript|1494812|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|$]]

Here is the solution to convert the string to valid plain or decimal numbers using Regex:
<pre><code>//something123.777.321something to 123.777321
const str = 'something123.777.321something';
let initialValue = str.replace(/[^0-9.]+/, '');
//initialValue = '123.777.321';

//characterCount just count the characters in a given string
if (characterCount(intitialValue, '.') &gt; 1) {
 const splitedValue = intitialValue.split('.');
 //splittedValue = ['123','777','321'];
 intitialValue = splitedValue.shift() + '.' + splitedValue.join('');
 //result i.e. initialValue = '123.777321'
}
</code></pre>

If I have a string like "something12" or "something102", how would I use a regex in javascript to return just the number parts?

Regex using javascript to return just numbers

如果我有一个像"something12“或"something102”这样的字符串，我如何在javascript中使用正则表达式来只返回数字部分？

问正则表达式使用javascript只返回数字
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回答 11

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问正则表达式使用javascript只返回数字EN

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问正则表达式使用javascript只返回数字
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