php - 实时用户名搜索不起作用。提交也停止工作

Question

我有用于实时用户名搜索的代码，但无法正常工作。对ajax来说是非常新的东西，并且对PHP掌握得很好。

这是我使用的脚本

<script>
    $(document).ready(function(){
        $("#uname").keyup(function() {
            var name = $('#uname').val();
            if(uname=="")
            {
                $("#disp").html("");
            }
            else
            {
                $.ajax({
                    type: "POST",
                    url: "getuser.php",
                    data: "name="+ uname ,
                    success: function(html){
                    $("#disp").html(html);
                    }
                });
            return false;
            }
        });
    });
</script>

随后是表格

form name="registerationform" id="registerationform" method="post" autocomplete="off">
    <input type="hidden" name="action" value="signup">
    <div class="editProfileCont">
            <div class="formFieldRow">
             <div class="fields">

        First Name<sup><font color="#FF0000">*</sup></font>: <input type="text" name="fname"  maxlength="30" id="fname" class="required" minlength="1" value="<?php if(isset($_POST[fname])){ echo $_POST[fname];}?>"></div></div><br><br>
         <div class="formFieldRow">
             <div class="fields">
        Last Name<sup><font color="#FF0000">*</sup></font>: <input type="text" name="lname"  maxlength="30" id="lname" class="required" minlength="1" value="<?php if(isset($_POST[lname])){ echo $_POST[lname];}?>"></div></div> <br><br>
         <div class="formFieldRow">
             <div class="fields">
        Age<sup><font color="#FF0000">*</sup></font>:       <input type="text" name="age" maxlength="2" id="age" class="required" minlength="1" value="<?php if(isset($_POST[age])){ echo $_POST[age];}?>"></div></div><br>
        <b><?php echo $errorEmail?><br></b>
         <div class="formFieldRow">
             <div class="fields">
        E-Mail Id<sup><font color="#FF0000">*</sup></font>: <input type="text" name="email" maxlength="40" id="email" class="required email" minlength="1" value="<?php if(isset($_POST[email])){ echo $_POST[email];}?>"></div></div><br>
        <b><?php echo $errorUname?><br></b>
         <div class="formFieldRow">
             <div class="fields">
        User Name<sup><font color="#FF0000">*</sup></font>: <input type="text" name="uname"  maxlength="10" id="uname" class="required" minlength="1" value="<?php if(isset($_POST[uname])){ echo $_POST[uname];}?>" onchange="showUser(this.value)"></div></div> <br>
        <div id="disp"></div><br /><br>
         <div class="formFieldRow">
             <div class="fields">
        Password<sup><font color="#FF0000">*</sup></font>:  <input type="password" name="pword" id="pword" class="required" minlength="5"></div></div><br><br><br>


<input type="submit" value="Submit" class="commenButton submitResetBTN">&nbsp;&nbsp;<input name="reset" type="reset" value="Reset" class="commenButton submitResetBTN">


now this is my getuser.php

<?php
session_start();
include "connection.php";
$q=$_POST['name'];
$sql="SELECT * FROM persons WHERE UserName = '".$q."'";

$result = mysqli_query($con,$sql);
$row=mysqli_num_rows($result);
if($row==0)
{
echo "<span style='color:green;'>Available</span>";
}
else
{
echo "<span style='color:red;'>Already exist</span>";
}

?>

它只是停留在原地。什么都没发生。该怎么办？？救命。

Answer 1

$(document).ready(function(){
    $("#uname").keyup(function() {
        var uname = $('#uname').val();
        if(uname=="")
        {
            $("#disp").html("");
        }
        else
        {
            $.ajax({
                type: "POST",
                url: "getuser.php",
                data: "name="+ uname ,
                success: function(html){
                $("#disp").html(html);
                }
            });
        return false;
        }
    });
});

您的脚本说var name = ...，然后一行检查uname。上面的代码已更正。您的其余代码对我来说似乎是正确的。