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Javascript 未返回php ajax jquery responseText_Javascript_Php_Jquery_Ajax - Fatal编程技术网

Javascript 未返回php ajax jquery responseText

Javascript 未返回php ajax jquery responseText,javascript,php,jquery,ajax,Javascript,Php,Jquery,Ajax,我正在尝试用jquery、ajax和php编写一个插入查询。正在插入记录,但返回状态错误。首先,我尝试在php中echo该消息,因为它不起作用,我尝试使用print json_encode进行响应,但都返回了错误状态。为什么它不返回responseText {readyState:0,responseText:,状态:0,状态文本:“错误”} 这是addmember.php文件 <?php require '../database.php'; function rand

我正在尝试用jquery、ajax和php编写一个插入查询。正在插入记录,但返回状态错误。首先,我尝试在php中
echo
该消息,因为它不起作用,我尝试使用
print json_encode
进行响应,但都返回了错误状态。为什么它不返回responseText

{readyState:0,responseText:,状态:0,状态文本:“错误”}

这是addmember.php文件

<?php
     require '../database.php';

    function random_password( $length = 8 ) {
            $chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*()_-=+;:,.?";
            $password = substr( str_shuffle( $chars ), 0, $length );
            return $password;
             }


    $password = random_password(8);
    //$regno=$_POST['regNo'];
    $adminno=$_POST['adminNo'];
    $batch=$_POST['batchText'];
    $type=$_POST["memberType"];
    $initials=$_POST["initialName"];
    $fullname=$_POST["fullName"];
    $address=$_POST["address"];
    $telephone=$_POST["contact"];
    $email=$_POST["email"];
    $nic=$_POST["nic"];
    $dob=$_POST["birthDate"];
    $priv=$_POST["memberType"];
    $userid="";

    $sql="select username from memberinfo where username='$adminno'";
    $result=mysqli_query($con,$sql); 
    if(mysqli_num_rows($result)==0){
        $sql="insert into memberinfo(username,nic_no,class,name_initial,full_name,address,telephone,email,date_of_birth) VALUES ('$adminno','$nic','$batch','$initials', '$fullname', '$address', '$telephone','$email','$dob')";
        $result1=mysqli_query($con,$sql);   
        $sql = "select * from memberinfo where username='$adminno'";
                                $result = $con->query($sql);
                                if ($result->num_rows > 0) {
                                    // output data of each row
                                    while($row = $result->fetch_assoc()) {

                                        $userid = $row['user_id'];
                                    }
                                }

        $sql="insert into userlogin(user_id,username,privilege,password) VALUES ('$userid','$adminno','$priv','$password')";
        $result2=mysqli_query($con,$sql);   

        if ($result1 && $result2) {

               $message = "<p>New record created successfully</p>";

         } else {
                $message = "<p>Error: " . $sql . "<br>" . $con->error.".</p>";
        }
    } else{
        $message = "<p>Admission no already exists.</p>";
    }

    print json_encode($message); 


     $con->close()                


?>

您没有从服务器返回有效的JSON。您正在对字符串进行json编码,但有效的json需要一个对象或数组来封装返回的日期

因此,至少:

echo json_encode(array($message));

状态为零通常意味着页面正在离开。阻止它发生

$('#addmember').click(function(evt){   //<--add the evt
    evt.preventDefault(); //cancel the click

$('#addmember')。单击(函数(evt){/不需要JSON响应。只需从PHP脚本返回消息,如下所示(注意
echo
的使用以及
close()
后面的分号):

PHP 另外,从AJAX调用中删除JSON文件类型,而是附加
response.responseText
而不是
response

JS
@AsmazinnerAjabir您可以打开web检查器,转到“网络”选项卡,检查网络请求以查看是否可以获得更多信息。一切似乎正常。猜测它不会显示,因为成功消息不是来自服务器。是否有方法显示自定义消息?完成,感谢您指出。修复了一个错误此外,ajax中的nge。必须附加response.responseText而不是response
$('#addmember').click(function(evt){   //<--add the evt
    evt.preventDefault(); //cancel the click
$con->close();
echo $message;
$.ajax({
       type:"POST",
       url:"../ajax/addmember.php",
       data:{'adminNo':adminno,'nic':nic,'batchText':batch,'initialName':initials, 'fullName':fullname, 'address':address, 'contact':telephone,'email':email,'birthDate':dob,'memberType':priv},
       success:function(response){
                   console.log(response);
                   $("#result").append(response.responseText);
                 },
       error:function(response){
                   console.log(response);
                 }
       });