$.getJSON获取不了回调函数,没有返回值,请高手指点指点,具体代码如下
php代码:
<?php
include 'global.php';$user = new User();if(!empty($_POST["pname"]))
{
$pname=$_POST["pname"];
$where = "pname = $pname";
$res = $user->selectAll("pname='$pname'");
$arr = array();
if(!empty($res))
{
$arr['res'] = "可以注册";
}
else
{
$arr['res'] = "不可以注册";
}
return json_encode($arr);
}
?>HTML代码:
<script src="jquery-1.7.2.min.js"></script>
<script>$(function(){ $("input[name='sub']").click(function(){ var pname = $("input[name='pname']").val();
$.getJSON("ajax.php",{'pname':pname},function(data){
alert(data);
});
});
});
</script><form method="post">
姓名:<input type="text" name="pname" ><span id="sp"></span><br>
<input type="submit" name="sub" value="提交">
</form>
php代码:
<?php
include 'global.php';$user = new User();if(!empty($_POST["pname"]))
{
$pname=$_POST["pname"];
$where = "pname = $pname";
$res = $user->selectAll("pname='$pname'");
$arr = array();
if(!empty($res))
{
$arr['res'] = "可以注册";
}
else
{
$arr['res'] = "不可以注册";
}
return json_encode($arr);
}
?>HTML代码:
<script src="jquery-1.7.2.min.js"></script>
<script>$(function(){ $("input[name='sub']").click(function(){ var pname = $("input[name='pname']").val();
$.getJSON("ajax.php",{'pname':pname},function(data){
alert(data);
});
});
});
</script><form method="post">
姓名:<input type="text" name="pname" ><span id="sp"></span><br>
<input type="submit" name="sub" value="提交">
</form>
echo json_encode($arr);最简单的方式。浏览器输入////url/xxxx/ajax.php?pname=123看看输出结果是啥