php json返回在響應中創建語法錯誤

我看過很多帖子在這裏和這個錯誤的其他地方沒有成功解決我的錯誤。在控制檯中，表單數據按照預期通過json發送到我的php處理頁面，php過程沒有返回任何錯誤，所以我相信這個過程正在完成。我已經檢查的結果被格式化的方式並不能看到什麼不對勁的地方，雖然我在阿賈克斯/ JSON沒有專家，所以我可能是錯的，這是我的代碼php json返回在響應中創建語法錯誤

jQuery代碼

<script type="text/javascript"> 

$(document).ready(function() { 

    $("#admin_login_but").click(function() { 

    if($("#admin_name").val()=="" || $("#admin_password").val()=="" || $("#admin_email").val()=="") { 

    $(".adminloginError").html("Please enter Admin Name, Admin Email and Admin Password"); 

     return false; 

    }else{ 


    $(".adminloginError").html('<img src="image/ajax-loader.gif" width="16" height="16" alt=""/>'); 
    var adminName=$("#admin_name").val(); 
    var adminPassword=$("#admin_password").val(); 
    var adminEmail=$("#admin_email").val(); 

$.post("includes/admin_login.inc.php",{admin_name:adminName,admin_password:adminPassword, admin_email:adminEmail},function(json) { 
    if(json.result === "success") { 
     $(".adminloginError").html("Welcome "+adminName+"!"); 

     setTimeout(function(){ 
       window.location= "admin_secure.php"; 
     },2000);       

    }else{ 
     $(".adminloginError").html(json.message); 
    } 
}); 
    }

這裏是我的PHP處理代碼

<?php 
include_once 'functions.php'; 
include_once 'db_connect.php'; 

header("Content-Type: application/json"); //this will tell the browser to send a json object back to client not text/html (as default) 

//convert variable (array) into a JSON object 

function result($var){ 
    echo json_encode($var); 
    exit(); 
} 

sec_session_start(); 

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


//check if surname is empty 
    if (isset($_POST["admin_login_but"])) { 

//check if first_name is empty 
    if (empty($_POST["admin_name"])) { 

     $response = array('result'=>'fail', 'message' => 'Missing Admin Name'); 
     result($response); 

    }else{ 

// if not empty sanitize first_name input 
    $admin_name = filter_input(INPUT_POST, 'admin_name', FILTER_SANITIZE_STRING); 

    } 


//Check if email is empty and santize and validate email 
     if (empty($_POST['admin_email'])) { 

     $response = array('result'=>'fail', 'message' => 'Missing Admin Email'); 
     result($response); 

     }else{ 

     $admin_email = filter_var($_POST['admin_email'], FILTER_SANITIZE_EMAIL); 
} 

     if (!filter_var($admin_email, FILTER_VALIDATE_EMAIL)) { 

     $response = array('result'=>'fail', 'message' => 'The Email is not in a Valid Email Format!'); 
     result($response); 

     } 



//check if register password input is empty 
    if (empty($_POST["admin_password"])) { 

     $response = array('result'=>'fail', 'message' => 'Missing Admin Password'); 
     result($response); 

    } else { 
//Sanitize the data passed in 'password' 
    $admin_password = filter_input(INPUT_POST, 'admin_password', FILTER_SANITIZE_STRING); 
} 

    //validate the data passed in 'password' 
    if (!preg_match("/^.*(?=.{8,})(?=.*\d)(?=.*[a-z])(?=.*[A-Z]).*$/", $admin_password)) { 

     $response = array('result'=>'fail', 'message' => 'Password is in the Wrong Format!'); 
     result($response); 

     } 


//query database 
$results = mysqli_query($mysqli, "SELECT * FROM admin WHERE name = '$admin_name' AND email = '$admin_email' AND hash = '$admin_password'"); 

// Check if SQL query had erors 
if(!$results){ 

     $response = array('result'=>'fail', 'message' => 'sql error: ' . mysqli_error($mysqli)); 
     result($response); 
} 

// If query was successfull and there are rows do this: 
if (mysqli_num_rows($results)>0){ 

    $_GET['name'] = $admin_name; 

    $response = array('result'=>'success', 'message' => 'User is Authenticated'); 
    result($response); 

} else { 

    $response = array('result'=>'fail', 'message' => 'User Authentication Failed'); 
    result($response); 

} 
} 

?>

我無法弄清楚如何解決這個問題。任何人都可以幫助請

來源

2015-05-04 Tiny

可能重複：意外結束數據在JSON數據的第1行第1列]（http://stackoverflow.com/questions/29291222/json-parse-error-unexpected-end-of-data-at-line-1-column-1-of- json-data） –

我已經看過這個，它側重於jQuery腳本，我不相信我的問題是與jQuery的，但如何PHP返回JSON和你提到的問題不涉及，就像我說過的那樣搜索全部解決方案，我只是在這裏作爲最後的手段發佈 – Tiny

是否檢查過腳本開始或結尾沒有空行/換行符？（如果是的話，它會創建httpheaders而不是你的json） – FeedTheWeb

我想通了，我還沒有把其他的選擇在我的PHP結束了初始後檢查

if(!isset($_POST)) { 

//all the processing code 


}else{ 
    $response = array('result'=>'success', 'message' => 'Post is Empty'); 
    result($response); 
}

[JSON解析錯誤的

來源

2015-05-04 16:23:03 Tiny

php json返回在響應中創建語法錯誤

回答

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