Ajax的成功返回空字符串

Question

我想用ajax進行重置密碼模塊，但是從PHP文件返回值（console.log()）始終是空白。在我的php文件中也沒有錯誤，我甚至不會回顯空字符串。請幫助，這是我的Ajax請求

$.ajax({ 
    type:'POST', 
    url:'verify-change-pass', 
    data: {_oldpass : op, _pass1 : p1, _pass2 : p2}, 
    success:function(data) { 
     data = data.trim(); 

     switch(data){ 
      case "er_no_acc" : 
       $('#grp_oldpass').addClass('has-error'); 
       $('#feedback_oldpass').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_oldpass').innerHTML = "*No account found. Please refresh the page to generate new session."; 
       //alert('Please enter your username'); 
       break; 
      case "er_oldpass_empty" : 
       $('#grp_oldpass').addClass('has-error'); 
       $('#feedback_oldpass').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_oldpass').innerHTML = "*Please enter old password."; 
       // alert('Please enter your password'); 
       break; 
      case "er_oldpass_invalid" : 
       $('#grp_oldpass').addClass('has-error'); 
       $('#feedback_oldpass').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_oldpass').innerHTML = "*Old password does not match.";     
       // alert('Please enter your password'); 
       break; 
      case "er_pass1_empty" : 
       $('#grp_pass1').addClass('has-error'); 
       $('#feedback_pass1').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_pass1').innerHTML = "*Please enter password."; 
       // alert('Please enter your password'); 
       break; 
      case "er_pass2_empty" : 
       $('#grp_pass2').addClass('has-error'); 
       $('#feedback_pass2').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_pass2').innerHTML = "*Please enter password.";     
       // alert('Please enter your password'); 
       break; 
      case "er_pass1_tooshort" : 
       $('#grp_pass1').addClass('has-error'); 
       $('#feedback_pass1').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_pass1').innerHTML = "*Password must be more than 5 characters."; 
       // alert('Password too short.'); 
       break; 
      case "er_pass2_tooshort" : 
       $('#grp_pass2').addClass('has-error'); 
       $('#feedback_pass2').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_pass2').innerHTML = "*Password must be more than 5 characters."; 
       // alert('Password too short.'); 
       break; 
      case "er_pass1_toolong" : 
       $('#grp_pass1').addClass('has-error'); 
       $('#feedback_pass1').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_pass1').innerHTML = "*Password must be less than 20 characters."; 
       // alert('Password too long.'); 
       break; 
      case "er_pass2_toolong" : 
       $('#grp_pass2').addClass('has-error'); 
       $('#feedback_pass2').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_pass2').innerHTML = "*Password must be less than 20 characters."; 
       // alert('Password too long.'); 
       break; 
      case "er_pass_notsame" : 
       $('#grp_pass1,#grp_pass2').addClass('has-error'); 
       $('#feedback_pass1,#feedback_pass2').addClass('glyphicon glyphicon-remove form-control-feedback'); 
       document.getElementById('msg_pass1').innerHTML = "*Password does not match."; 
       document.getElementById('msg_pass2').innerHTML = "*Password does not match."; 
       // alert('Password does not match'); 
       break;     
      case "y" : 
        ResetMyPassword(p1);  
       break; 
      default: 
       alert('Error occured : '+data); 
     } 


     /* Always put console.log for debugging */ 
     console.log(data); 
    } 
}); 

這是我verify-change-pass.php

session_start(); 
if(!isset($_SESSION['username'])){ 
    echo "Error 404 : Page not found."; 
    exit(); 
} 
include("connectdb.php"); 

$par_uname = $_SESSION['username']; 
$par_oldpass = isset($_POST['_oldpass']) ? $_POST['_oldpass'] : ''; 
$par_pass1 = isset($_POST['_pass1']) ? $_POST['_pass1'] : ''; 
$par_pass2 = isset($_POST['_pass2']) ? $_POST['_pass2'] : ''; 

trim($par_oldpass); 
trim($par_pass1); 
trim($par_pass2); 
htmlspecialchars($par_oldpass); 
htmlspecialchars($par_pass1); 
htmlspecialchars($par_pass2); 

$count=0; 
$stmt = $mysqli->prepare("SELECT acc_hashedpass FROM admin_acc WHERE acc_uname=?"); 
$stmt->bind_param('s',$par_uname); 
$stmt->execute(); 
$stmt->store_result(); 
$count = $stmt->num_rows; 
$stmt->bind_result($hashed); 
$stmt->fetch(); 
$stmt->close(); 

if($count != 1){ 
    echo "er_no_acc"; //no username found. check the session if fetched properly 
    exit(); 
} 

if ($par_oldpass == ""){ 
    echo "er_oldpass_empty"; //oldpass is empty 
    exit(); 
} 

if(!password_verify($par_oldpass,$hashed)){ 
    echo "er_oldpass_invalid" . password_verify($par_oldpass,$hashed); //invalid old password 
    exit(); 
} 

if ($par_pass1 == ""){ 
    echo "er_pass1_empty"; //password 1 is empty 
    exit(); 
} 

if($par_pass2 == ""){ 
    echo "er_pass2_empty"; //password 2 is empty 
    exit(); 
} 

if(mb_strlen($par_pass1) < 6){ 
    echo "er_pass1_tooshort"; //password is too short. Must be more than 8 characters 
    exit(); 
} 
if(mb_strlen($par_pass2) < 6){ 
    echo "er_pass2_tooshort"; //password is too short. Must be more than 8 characters 
    exit(); 
} 

if(mb_strlen($par_pass1) > 20){ 
    echo "er_pass1_toolong"; //password is too long. Must be less than 20 characters 
    exit(); 
} 
if(mb_strlen($par_pass2) > 20){ 
    echo "er_pass2_toolong"; //password is too long. Must be less than 20 characters 
    exit(); 
} 

if ($par_pass1 != $par_pass2){ 
    echo "er_pass_notsame"; //password doesnt match 
    exit(); 
} 

echo "y"; 

從verify-change-pass.php的第一個高達6日if回聲工作，之後，返回值始終是一個空字符串。請幫助..

順便說一句，這個片段適用於localhost，但我在網上上傳後，該代碼不起作用。

Answer 1

顯然，我需要+50的聲譽做一個評論，所以讓我通過一個答案評論（笑）

error_reporting(E_ALL); 
ini_set('display_errors', 1); 

嘗試添加到您的代碼的頂部？

此外，檢查是否

include("connectdb.php"); 

內部有任何錯誤/可能退出。

您是否嘗試過在最高層呼應的東西嗎？

Answer 2

嘗試從由ajax 調用的php文件的代碼中刪除退出函數，並打印出您想要的消息並嘗試提醒您在ajax成功開始的響應。 - >你可以試試jQuery.trim（data）;而不是data.trim（）; 希望它可以工作。 - >嘗試否則，如果而不是個人，如果在PHP