我有以下html5输入元素:

<input type="number">

为什么这个输入允许在输入字段中输入字符“e”?无法输入其他字母字符(如预期的那样)

使用chrome v. 44.0.2403.107

下面的例子来看看我的意思。

= >“当家”型输入<


当前回答

<input type="number" onkeydown="return FilterInput(event)" onpaste="handlePaste(event)"  >

function FilterInput(event) {
    var keyCode = ('which' in event) ? event.which : event.keyCode;

    isNotWanted = (keyCode == 69 || keyCode == 101);
    return !isNotWanted;
};
function handlePaste (e) {
    var clipboardData, pastedData;

    // Get pasted data via clipboard API
    clipboardData = e.clipboardData || window.clipboardData;
    pastedData = clipboardData.getData('Text').toUpperCase();

    if(pastedData.indexOf('E')>-1) {
        //alert('found an E');
        e.stopPropagation();
        e.preventDefault();
    }
};

其他回答

上述解决方案只适用于普通html。 对于reactJS,我建议你这样做

  <input type="number" onKeyDown={(e) =>["e", "E", "+", "-"].includes(e.key) && e.preventDefault()} >

最简单的解决方案是parseInt()。

<input type="number" onkeyup="this.value = parseInt(this.value); this.paste(this.onkeyup);"/>

ParseInt()对于E, E, +, -和任何不是数字的东西返回空白(NaN)。

使用onkeyup和keydown根据您的方便与粘贴事件。对我来说,onkeyup工作最好。这也适用于复制粘贴

我们可以像下面这样简单

<input type="number" onkeydown="javascript:返回事件。keyCode == 69 ?/> . False: true

更新后的答案

我们可以让它更简单,就像@88 MPG建议的那样

<输入类型=“编号”=“回归事件”。键盘代码!

因为这正是规范所规定的。数字输入可以接受浮点数,包括负号和e或e字符(其中指数是e或e后面的数字):

A floating-point number consists of the following parts, in exactly the following order: Optionally, the first character may be a "-" character. One or more characters in the range "0—9". Optionally, the following parts, in exactly the following order: a "." character one or more characters in the range "0—9" Optionally, the following parts, in exactly the following order: a "e" character or "E" character optionally, a "-" character or "+" character One or more characters in the range "0—9".

使用angular,你可以限制输入e,+,-, e

 <input type="number"  (keypress)="numericOnly($event)"/>


  numericOnly(event): boolean { // restrict e,+,-,E characters in  input type number
    debugger
    const charCode = (event.which) ? event.which : event.keyCode;
    if (charCode == 101 || charCode == 69 || charCode == 45 || charCode == 43) {
      return false;
    }
    return true;

  }