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Problem: How to test/check if a value is an Unsigned Integer (UINT) in JavaScript.

Alternative Answers: Thanks goes to the following for improving this answer!

@jbabey:if Number(val) > 0

@SLaks:if (/^\d+$/.test(someString))

About: This shows how to test/check if a value is an Unsigned Integer (UINT) in JavaScript.

Usage: JS: isUINT( value );

Returns: True OR False

Expanded Version:

<script type="text/javascript">
    function isUINT(v)
    {
        var r = RegExp(/(^[^\-]{0,1})?(^[\d]*)$/);
        return r.test(v) && v.length > 0;
    }
</script>

Minified Version:

function isUINT(v){var r=RegExp(/(^[^\-]{0,1})?(^[\d]*)$/);return r.test(v)&&v.length>0}

Comments / Alternatives are Weled!

Problem: How to test/check if a value is an Unsigned Integer (UINT) in JavaScript.

Alternative Answers: Thanks goes to the following for improving this answer!

@jbabey:if Number(val) > 0

@SLaks:if (/^\d+$/.test(someString))

About: This shows how to test/check if a value is an Unsigned Integer (UINT) in JavaScript.

Usage: JS: isUINT( value );

Returns: True OR False

Expanded Version:

<script type="text/javascript">
    function isUINT(v)
    {
        var r = RegExp(/(^[^\-]{0,1})?(^[\d]*)$/);
        return r.test(v) && v.length > 0;
    }
</script>

Minified Version:

function isUINT(v){var r=RegExp(/(^[^\-]{0,1})?(^[\d]*)$/);return r.test(v)&&v.length>0}

Comments / Alternatives are Weled!

  • javascript
  • regex
Share Improve this question edited Nov 15, 2012 at 20:07 Trenton Bost asked Nov 15, 2012 at 18:09 Trenton BostTrenton Bost 4015 silver badges7 bronze badges 2
  • 2 I don't think this question makes sense. There's no such thing as an "unsigned" type in JavaScript, so no value will ever be an "unsigned integer". If you just mean "non-negative integer", then there are better ways to do it than a regular expression. – Pointy Commented Nov 15, 2012 at 18:12
  • @Pointy: Correct. There is no such thing as an 'Unsigned' type in JavaScript. I wanted to make one. – Trenton Bost Commented Nov 15, 2012 at 19:06
Add a ment  | 

3 Answers 3

Reset to default 10 
if (/^\d+$/.test(someString))

​

Old question but now you can use

const isUINT = (val) => Number.isInteger(val) && val >= 0

console.log('"a":', isUINT('a'))
console.log('"1":', isUINT('1'))
console.log('null:', isUINT(null))
console.log('-1:', isUINT(-1))
console.log('0:', isUINT(0))
console.log('1:', isUINT(1))
console.log('1.25:', isUINT(1.25))
console.log('0.25:', isUINT(0.25))

Alternative Answers: Thanks goes to the following for improving this answer!

@jbabey:if Number(val) > 0

@SLaks:if (/^\d+$/.test(someString))

About: This shows how to test/check if a value is an Unsigned Integer (UINT) in JavaScript.

Usage: JS: isUINT( value );

Returns: True OR False

Expanded Version:

<script type="text/javascript">
    function isUINT(v)
    {
        var r = RegExp(/(^[^\-]{0,1})?(^[\d]*)$/);
        return r.test(v) && v.length > 0;
    }
</script>

Minified Version:

function isUINT(v){var r=RegExp(/(^[^\-]{0,1})?(^[\d]*)$/);return r.test(v)&&v.length>0}

Comments / Alternatives are Weled!

本文标签： regex(Howto) JavaScriptTest for Unsigned Integer (UINT) using Regular Expressions (RegExp)Stack Overflow

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