Question

我正在使用ajax运行测试ID，但是它不适用于代码：200错误。

由于ajax没有返回值，因此它不断出错并打印“失败”

add_user.html的ID通过id_check.js的ajax进行实时检查。但是，从id_check.js发送数据的memid_check.php似乎没有运行。确认

memid_check.php

echo "<script> alert('test!!!'); </script>";
.....

但没有运行。

所有文件都在一个文件夹中，因此路径似乎很好

id_check.js

$(function(){

var id = $('.id_tbox');
var pwd =$('.pwd_tbox');
var name =$('.name_tbox');
var email =$('.email_tbox');
var idCheck = $('.idCheck');

$(".memcheck_button").click(function(){
    console.log(id.val());
    $.ajax({
        type: 'post',
        dataType: 'json',
        url: "memid_check.php",
        data:{id:id.val()},

        success: function(json){
            if(json.res == 'good'){
                console.log(json.res);
                alert("사용가능한 아이디");
                idCheck.val('1');
            }
            else{
                alert("다른 아이디 입력");
                id.focus();
            }
        },

        error:function(request,status,error){
alert("code:"+request.status+"\n"+"message:"+request.responseText+"\n"+"error:"+error);
            console.log("failed");
        }




    });
});
});

memid_check.php

<?php

echo "<script> alert('test!!!'); </script>";


include "db_c.php";

$id = $_POST['id'];

$sql = "SELECT * FROM add_user WHERE id = '{$id}'";

$res = $database -> query($sql);

if( res->num_rows >= 1){
echo json_encode(array('res'=>'bad'));
}
else{

echo json_encode(array('res'=>'good'));

} 

?>

add_user.html

<?php  
include "database.php";
?>

<!DOCTYPE html>
<html>
<head>
    <meta name="viewport" content="width=device-width, initial-scale=1.0,         target-densitydpi=medium-dpi">
    <link rel="stylesheet" href="add_user_style.css">
    <script src="https://code.jquery.com/jquery-3.4.1.min.js"></script>
    <script src="js/default.js"></script>
    <link href="https://fonts.googleapis.com/css?family=Nothing+You+Could+Do&display=swap" rel="stylesheet">
     <link href="https://fonts.googleapis.com/css?family=Cinzel|Permanent+Marker|Rajdhani&display=swap" rel="stylesheet">
    <script type="text/javascript" src="id_check.js"></script>


</head>


<body>
    <div class="aus_portrait"></div>
    <div class ="aus_form" align ="center">
        <div class="aus_box">Sign Up</div>
    <form action="loginP.php" method="post">
        <p class = "id">ID</p><input type="text" name="id" class="id_tbox">
        <p class = "pwd">PASSWORD</p><input type="text" name="pwd" class="pwd_tbox">
        <p class = "name">MAIL</p><input type="text" name="email" class="email_tbox">
        <p class = "email">NAME</p><input type="text" name="name" class="name_tbox">
        <input type="submit" class="sub_button" value="Submit">
        <input type="button" class="exit_button" value="Cancel">
        <input type="hidden" name="idCheck" class="idCheck">
        <div class="memcheck_button">중복확인</div>
    </form>
    </div>


    </body>
</html>

<script>
$(document).ready(function() { $(".exit_button").on("click", function(){ location.href="login.html"});}); </script>

ajax code200错误无法连接到URL

0 个答案: