Question

任何人都有任何见解为什么这段代码没有显示剩下的字符，只显示最大字符数，我已经经历了好几次，我似乎无法找到错误。

HTML

<form class="comment" method="post" action="postComment.php">
    <input type="text" placeholder="Name" name="Cname" onKeyUp="charLeft(this.value,30,'n')"><br/><span id="nCharLeft"></span><br/>
    <input type="text" placeholder="Email" name="Cemail" onKeyUp="charLeft(this.value,50,'e')"><br/><span id="eCharLeft"></span><br/>
    <textarea rows="4" placeholder="Please leave a comment." name="Ccomment" onKeyUp="charLeft(this.value,300,'c')"></textarea><br/><span id="cCharLeft"></span><br>
    <input type="submit" value="Post Comment"><br/>
</form>

的JavaScript

function charLeft(val,len,indi) {
    var output = indi + "CharLeft";
    if (val.length==0) { 
        return;
    }
var xmlhttp=new XMLHttpRequest();
xmlhttp.onreadystatechange=function() {
    if (xmlhttp.readyState==4 && xmlhttp.status==200) {
       document.getElementById(output).innerHTML=xmlhttp.responseText;
    }
}
xmlhttp.open("GET","charLeft.php?q="+val+"&m="+len,true);
xmlhttp.send();
}

PHP（charLeft.php）

<?php
$v = $_REQUEST['v'];
$m = $_REQUEST['m'];
$len = strlen($v);
$charleft = $m - $len;

echo $charleft==="1" ? "$charleft character left." : "$charleft characters left.";
?>

Answer 1

在您的AJAX调用中，您正在q中发送值但是将其作为 $ v = $ _REQUEST [＆＃39; v＆＃39;];

因此，要么将AJAX调用更改为 xmlhttp.open（＆＃34; GET＆＃34;＆＃34; charLeft.php V =＆＃34; + VAL +＆＃34;＆安培; M =＆＃34 + LEN，TRUE）;

或将PHP更改为 $ v = $ _REQUEST [＆＃39; q＆＃39;];

Answer 2

您发送了“charLeft.php？q =”+ val +“＆amp; m =”+ len，但是通过$ v = $ _REQUEST ['v']获取值;

尝试

$v = $_REQUEST['q'];

表格输入的字符计数器 - 不计数

2 个答案: