Question

我有这个脚本

<script>
    function trigger(){
    var x = document.getElementById('xcoord');
    var y = document.getElementById('ycoord');
    var box = document.getElementById('touch');

    if (x.value >= 325 || x.value <= 300 && y.value >= 55 || y.value <= 25) {
    $('#touch').value('passed');

    }

    }  
</script>

这是html

 <input onchange="trigger()" required type="number" id="xcoord" name="xcoord">
 <input onchange="trigger()" required type="number" id="ycoord" name="ycoord">
 <input required type="text" id="touch" name="touch">

每当xcoord和ycoord的值适合条件时，我都需要显示一个值，但即使条件为真，也不会打印任何值。

Answer 1

您将JavaScript与JQuery混合在一起。

使用JavaScript

将值放入输入框

box.value = 'passed';

CodeSnippet

＆＃13;

function trigger(){
    var x = document.getElementById('xcoord');
    var y = document.getElementById('ycoord');
    var box = document.getElementById('touch');

    if (x.value >= 325 || x.value <= 300 && y.value >= 55 || y.value <= 25) {
         box.value = 'passed';
    }
}

＆＃13;

<input onchange="trigger()" required type="number" id="xcoord" name="xcoord">
 <input onchange="trigger()" required type="number" id="ycoord" name="ycoord">
 <input required type="text" id="touch" name="touch">

＆＃13;

使用JQuery将值放到输入框

$('#touch').val('passed'); //Although you already have the box variable

CodeSnippet

＆＃13;

 function trigger(){
    var x = document.getElementById('xcoord');
    var y = document.getElementById('ycoord');
    var box = document.getElementById('touch');

    if (x.value >= 325 || x.value <= 300 && y.value >= 55 || y.value <= 25) {
        $('#touch').val('passed');
    }

}

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input onchange="trigger()" required type="number" id="xcoord" name="xcoord">
 <input onchange="trigger()" required type="number" id="ycoord" name="ycoord">
 <input required type="text" id="touch" name="touch">

＆＃13;

<强> TL;博士;

由于您使用普通JavaScript启动了代码，因此您混淆了如何使用jQuery语法更改值。

JavaScript box.value = 'passed';

jQuery $('#touch').val('passed');

Answer 2

您正在使用jQuery库。使用优于 jQuery 方法。

$("#xcoord , #ycoord").on("keyup",  function(){
    var x = $('#xcoord').val();
    var y = $('#ycoord').val();
   $('#touch').val((x >= 325 || x <= 300) && (y >= 55 || y <= 25)?'passed':'');
} );

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input required type="number" id="xcoord" name="xcoord" value="0">
 <input  required type="number" id="ycoord" name="ycoord" value="0">
 <input required type="text" id="touch" name="touch">

Answer 3

应该是

 $('#touch').val('passed');

jquery将值放到输入框中

3 个答案: