blocks|key|2136530|text|您可以使用<input+type="number">，它具有min和max属性：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2136531|​|2136532|<input+type="number"+min="1"+max="999"+value="1">|code-block|syntax|javascript|2136533|2136534|浏览器支持|2136535|entityMap|0|LINK|mutability|MUTABLE|url|http://caniuse.com/#feat=input-number^0|5|L|U|3|Y|3|0|0|0|0|0|5|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|17|8|@]|D|@$9|18|A|19|1|1A]]|E|$]]|$1|P|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|Q|$R|$5|S|T|U|E|$V|W]]]]

<p>You can use <code>&lt;input type="number"&gt;</code> which has a <code>min</code> and <code>max</code> attribute:</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true">
<div class="snippet-code">
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;input type="number" min="1" max="999" value="1"&gt;</code></pre>
</div>
</div>
</p>

<p><a href="http://caniuse.com/#feat=input-number" rel="nofollow">Browser support</a></p>


blocks|key|1383779|text|您可以使用<input+type="number">元素、带有参数0,+-1的input事件String.prototype.slice()删除input值的最后一个字符，如果在第一个if条件下使用valueAsNumber属性代替parseInt()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1383780|​|1383781|$(document).ready(function()+{
++$(".numbers").on("input",+function(e)+{
++++var+value+=+this.valueAsNumber;
++++if+(value+<=+0+%7C%7C+value+>+999)+{
++++++this.value+=+this.value.slice(0,+-1);
++++}
++++//+if+(e.which+!=+8+&&+e.which+!=+0+&&+(e.which+<+48+%7C%7C+e.which+>+57))+{
++++//+++++return+false;
++++//+}
++});
});|code-block|syntax|javascript|1383782|<script+src="//ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js">
</script>
<input+name="quantity"+type="number"+class="numbers"+value=""+/>|1383783|1383784|jsfiddle+https://jsfiddle.net/kcet86s8/5/|1383785|entityMap|0|LINK|mutability|MUTABLE|url|https://jsfiddle.net/kcet86s8/5/^0|5|L|X|5|13|5|1A|O|20|5|2K|2|2R|D|38|A|0|0|0|0|0|9|W|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1G|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1H|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|J|7|1I|8|@]|D|@]|E|$K|L]]|$1|O|3|G|5|6|7|1J|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1K|8|@]|D|@$9|1L|A|1M|1|1N]]|E|$]]|$1|R|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|S|$T|$5|U|V|W|E|$X|Y]]]]

<p>You can use <code>&lt;input type="number"&gt;</code> element, <code>input</code> event <code>String.prototype.slice()</code> with parameters <code>0, -1</code> to remove last character of <code>input</code> value substituting <code>valueAsNumber</code> property for <code>parseInt()</code> if <code>false</code> at first <code>if</code> condition</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>$(document).ready(function() {
  $(".numbers").on("input", function(e) {
    var value = this.valueAsNumber;
    if (value &lt;= 0 || value &gt; 999) {
      this.value = this.value.slice(0, -1);
    }
    // if (e.which != 8 &amp;&amp; e.which != 0 &amp;&amp; (e.which &lt; 48 || e.which &gt; 57)) {
    //     return false;
    // }
  });
});</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="//ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"&gt;
&lt;/script&gt;
&lt;input name="quantity" type="number" class="numbers" value="" /&gt;</code></pre>
</div>
</div>
</p>

<p>jsfiddle <a href="https://jsfiddle.net/kcet86s8/5/" rel="nofollow">https://jsfiddle.net/kcet86s8/5/</a></p>


blocks|key|2136601|text|为了清楚起见，你应该在你的问题上加上这个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2136602|不允许用户输入除1到999以外的任何值，包括小数。|blockquote|2136603|您的问题是:您希望限制用户输入，这取决于输入的值。但是"keypress“事件是在更新值之前触发的，因此您得到的值是旧值，而不是新值(这很容易检查)|2136604|您的需求中的问题是，在更新输入之前无法获得新值，至少jquery不能这样做。因此，您不能限制用户输入在“按键”。|2136605|以另一种方式，你可以使用“键”。如果用户输入了错误的值，则恢复最后的值。.i.e：https://jsfiddle.net/kcet86s8/7/|offset|length|2136606|$(document).ready(function+()+{
++$.myval+=+$('.numbers').val();
++++$('.numbers').keyup(function+(e)+{
++++++++var+value+=+parseInt($(this).val());
++++++++if+(value+<=+0+%7C%7C+value+>+999)+{
++++++++$(this).val($.myval);
++++++++++++return+false;
++++++++}
++++++++if+((e.which+!=+8+&&+e.which+!=+0+&&+(e.which+<+48+%7C%7C+e.which+>+57)))+{
++++++++$(this).val($.myval);
++++++++++++return+false;
++++++++}
++++++$.myval+=+$(this).val();
++++});
});|code-block|syntax|javascript|2136607|用户在恢复输入值之前仍然可以看到它。我认为这并不完美，但这是唯一的办法。|2136608|更新键代码条件以启用numpad输入，如@Midas建议的那样：https://jsfiddle.net/kcet86s8/10/|2136609|$(document).ready(function+()+{
++$.myval+=+$('.numbers').val();
++++$('.numbers').keyup(function+(e)+{
++++++++var+value+=+parseInt($(this).val());
++++++++if+(value+<=+0+%7C%7C+value+>+999)+{
++++++++$(this).val($.myval);
++++++++++++return+false;
++++++++}
++++++++if+(e.which+!=+8+&&+e.which+!=+0+&&+!(e.which+>=+48+&&+e.which+<=+57)+&&+!(e.which+>=+96+&&+e.which+<=+105))+{
++++++++$(this).val($.myval);
++++++++++++return+false;
++++++++}
++++++$.myval+=+$(this).val();
++++});
});|2136610|entityMap|0|LINK|mutability|MUTABLE|url|https://jsfiddle.net/kcet86s8/7/|1|https://jsfiddle.net/kcet86s8/10/^0|0|0|0|0|15|W|0|0|0|0|W|X|1|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|18|8|@]|9|@]|A|$]]|$1|E|3|F|5|6|7|19|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|1A|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|1B|8|@]|9|@$K|1C|L|1D|1|1E]]|A|$]]|$1|M|3|N|5|O|7|1F|8|@]|9|@]|A|$P|Q]]|$1|R|3|S|5|6|7|1G|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1H|8|@]|9|@$K|1I|L|1J|1|1K]]|A|$]]|$1|V|3|W|5|O|7|1L|8|@]|9|@]|A|$P|Q]]|$1|X|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|Y|$Z|$5|10|11|12|A|$13|14]]|15|$5|10|11|12|A|$13|16]]]]

<p>You should add this to your question, for clear.</p>

<blockquote>
  <p>disallow user to enter any value apart from 1 to 999 including decimals</p>
</blockquote>

<p>Your problem is: you want to restrict user input, which depend to the inputed value. But the "keypress" event is fire before the value updated, therefore the value you get is the old value, not the new one(that's easy to check this out)</p>

<p>And the problem in your requirement is you can't get the new value before the input updated, atleast jquery can't do that. So you can't restrict user input at "keypress".</p>

<p>In other way, you can use "keyup". Then revert last value if user input wrong value.
.i.e: <a href="https://jsfiddle.net/kcet86s8/7/" rel="nofollow">https://jsfiddle.net/kcet86s8/7/</a></p>

<pre><code>$(document).ready(function () {
  $.myval = $('.numbers').val();
    $('.numbers').keyup(function (e) {
        var value = parseInt($(this).val());
        if (value &lt;= 0 || value &gt; 999) {
        $(this).val($.myval);
            return false;
        }
        if ((e.which != 8 &amp;&amp; e.which != 0 &amp;&amp; (e.which &lt; 48 || e.which &gt; 57))) {
        $(this).val($.myval);
            return false;
        }
      $.myval = $(this).val();
    });
});
</code></pre>

<p>User still can sees the input value before it have been reverted. I think it's not perfect but the only way.</p>

<hr>

<p>Update keycode condition to enable numpad input, as @Midas suggested: <a href="https://jsfiddle.net/kcet86s8/10/" rel="nofollow">https://jsfiddle.net/kcet86s8/10/</a></p>

<pre><code>$(document).ready(function () {
  $.myval = $('.numbers').val();
    $('.numbers').keyup(function (e) {
        var value = parseInt($(this).val());
        if (value &lt;= 0 || value &gt; 999) {
        $(this).val($.myval);
            return false;
        }
        if (e.which != 8 &amp;&amp; e.which != 0 &amp;&amp; !(e.which &gt;= 48 &amp;&amp; e.which &lt;= 57) &amp;&amp; !(e.which &gt;= 96 &amp;&amp; e.which &lt;= 105)) {
        $(this).val($.myval);
            return false;
        }
      $.myval = $(this).val();
    });
});
</code></pre>


<p>I want to allow user to enter value from 1 to 999 in the input box</p>

<p>I tried the following</p>

<pre><code>&lt;input name="quantity" class="numbers" value=""/&gt;

$('.numbers').keypress(function (e) {
    var value = parseInt($(this).val());
    if (value &lt;= 0 || value &gt; 999) {
        return false;
    }
    if (e.which != 8 &amp;&amp; e.which != 0 &amp;&amp; (e.which &lt; 48 || e.which &gt; 57)) {
        return false;
    }
});
</code></pre>

<p>This does not work, when i enter zero value, i am stuck, here is the link to fiddle <a href="https://jsfiddle.net/kcet86s8/4/" rel="nofollow">https://jsfiddle.net/kcet86s8/4/</a></p>

<p>Where am i going wrong ?</p>


Disallow user to enter any value apart from 1 to 999 including decimals

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

我希望允许用户在输入框中输入1到999之间的值。我尝试了以下方法<input name="quantity" class="numbers" value=""/>$('.numbers').keypress(function (e) {    var value = parseInt($(this).val());  ...

问不允许用户输入除1到999以外的任何值，包括小数。
EN

Stack Overflow用户

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

问不允许用户输入除1到999以外的任何值，包括小数。EN