>任何人都知道为什么这段代码不显示剩余的字符而只显示最大数量的字符,我已经经历了几次,我似乎找不到错误。
.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>
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 (字符左.php)
<?php
$v = $_REQUEST['v'];
$m = $_REQUEST['m'];
$len = strlen($v);
$charleft = $m - $len;
echo $charleft==="1" ? "$charleft character left." : "$charleft characters left.";
?>
在 AJAX 调用中,您将 q 中的值发送,但访问它为$v = $_REQUEST['v'];
因此,要么将 AJAX 调用更改为xmlhttp.open("GET","charLeft.php?v="+val+"&m="+len,true);
或在 PHP 中更改为$v = $_REQUEST['q'];
你发送了"charLeft.php?q="+val+"&m="+len,但通过$v = $_REQUEST['v']获得值;
尝试使用
$v = $_REQUEST['q'];