varifying JSON響應

Question

我這裏測試我的jsno例如：

data.php

<?php 

ob_start(); 
echo "string1"; 
$div1 = ob_get_clean(); 
ob_start(); 
echo "string2"; 
$div2 = ob_get_clean(); 

$resultArray = array("resultOne" => $div1,"resultTwo" => $div2); 
echo json_encode($resultArray); 

?> 

start.php

xmlhttp.onreadystatechange=function() 
       { 
        var div1=document.getElementById("myDiv1"); 
        var div2=document.getElementById("myDiv2"); 
        if (xmlhttp.readyState==4 && xmlhttp.status==200) 
        { 
         var data=JSON.parse(xmlhttp.responseText); 
         //document.getElementById("myDiv1").innerHTML=xmlhttp.responseText; 
         div1.innerHTML=data['resultOne']; 
            div2.innerHTML=data['resultTwo']; 
        } 
       } 

這給正確的結果，並顯示數據。

現在我試着實施我的實際應用程序。

login.php中（驗證用戶登錄，如果成功） - > start.php（裝入通過Ajax的login.php） - > data.php（返回JSON數組值如在上述情況下）

只是差是data.php以下列方式：

<?php 
    $url = $_POST['url']; 
    $user_id = $_POST ['userid']; 
    if(isset($_POST['rate'])) 
    { 
     $rate =$_POST['rate']; 
    } 
    else 
     $rate = 0; 
    $data = file_get_contents($url); 
    function get_title($html) 
    { 
     return preg_match('!<title>(.*?)</title>!i', $html, $matches) ? $matches[1] : ''; 
    } 

    function get_logo($html) 
    { 
     preg_match_all('/\bhttps?:\/\/\S+(?:png|jpg)\b/', $html, $matches); 
     //echo "mactch : $matches[0][0]"; 
     return $matches[0][0]; 
    } 

    function plaintext($html) 
    { 
     // remove comments and any content found in the the comment area (strip_tags only removes the actual tags). 
     $plaintext = preg_replace('#<!--.*?-->#s', '', $html); 

     // put a space between list items (strip_tags just removes the tags). 
      $plaintext = preg_replace('#</li>#', ' </li>', $plaintext); 

      // remove all script and style tags 
     $plaintext = preg_replace('#<(script|style)\b[^>]*>(.*?)</(script|style)>#is', "", $plaintext); 

     // remove br tags (missed by strip_tags) 
      $plaintext = preg_replace("#<br[^>]*?>#", " ", $plaintext); 

      // remove all remaining html 
      $plaintext = strip_tags($plaintext); 

     return $plaintext; 
    } 

    function trim_display($size,$string) 
    { 
     $trim_string = substr($string, 0, $size); 

     $trim_string = $trim_string . "..."; 
     return $trim_string; 
    } 

    $title = get_title($data); 
    $logo = get_logo($data); 
    $title_display = trim_display(30,$title); 
    $content = plaintext($data); 
    $Preview = trim_display(100,$content); //to Show first 100 char of the web page as preview 

    function MakeTinyUrl($url) 
    { 
     return md5($url); 
    } 

    $hash = MakeTinyUrl($url); 
    ob_start(); 
    $con = mysqli_connect('127.0.0.1', 'root', '', 'mysql'); 
    if (mysqli_connect_errno()) 
    { 
     echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
     return; 
    } 

    $content=mysqli_real_escape_string($con,$content); 
    $Preview=mysqli_real_escape_string($con,$Preview); 
    $title_display=mysqli_real_escape_string($con,$title_display); 

    $result = mysqli_query($con,"SELECT COUNT(*) as val FROM post_data WHERE url ='".$url."' and userid='".$user_id."'"); 
    $bool= mysqli_fetch_assoc($result); 
    if($bool['val']) 
    { 
     echo '<div style="clear:both;"><i>You have already worked on this url..</i> </div>'; 
    } 
    else 
    { 
     $insertQuery = "INSERT INTO post_data(`userid`, `url`, `desc`, `preview`, `img_url`, `title` ,`hash`) VALUES ('".$user_id."','".$url."','".$content."','".$Preview."','".$logo."','".$title_display."','".$hash."')"; 
     if (!mysqli_query($con,$insertQuery)) 
     { 
      die('Error: ' . mysqli_error($con)); 
     } 
    } 

    $result = mysqli_query($con,"SELECT * FROM post_data WHERE userid ='".$user_id."' and url='".$url."'"); 
    //This will fetch only one row from db 
    while ($row = @mysqli_fetch_array($result)) 
    { 
     $title = $row['title']; 
     $url = $row['url']; 
     $preview = $row['preview']; 
     $image = $row['img_url'];  
    } 
    //Update Rate value in table 
    $update = "update post_data set rate='".$rate."' where url='".$url."'"; 
    if (mysqli_query($con, $update)) 
    { 
     //echo "updated"; 
    } 
    else 
    { 
     //echo "Not updated"; 
    } 

?> 
<html> 
<body> 
    <a class="fragment" href="google.com"> 
     <div> 
      <span id='close' onclick='this.parentNode.parentNode.parentNode.removeChild(this.parentNode.parentNode); return false;'>x</span> 
      <img src ="<?php echo $image ?>" height="116" width="116" alt="some description"/> 
      <h3><?php echo $title ?></h3> 
      <h4><?php echo $url ?> </h4> 
      <p class="text"> <?php echo $preview ?> 
       </p> 

     </div> 
    </a> 

    <?php 
     $div1 = ob_get_clean(); 
     ob_start(); 
     $result = mysqli_query($con,"SELECT * FROM post_data WHERE userid ='".$user_id."'"); 
     echo "Records for user : $user_id"; echo "<br/>"; 
     while ($row = @mysqli_fetch_array($result)) 
     { 
      $title = $row['title']; 
      $url = $row['url']; 
      $preview = $row['preview']; 
      $image = $row['img_url'];  
      echo "Title: $title"; echo '<br/>'; 
      echo "URL: $url"; echo '<br/>'; 
      echo "Preview : $preview"; echo '<br/>'; 
      echo "Image url: $image"; echo '<br/>'; 
      //echo '</br>'; 
     }  
     $div2 = ob_get_clean(); 
     $resultArray = array("resultOne" => $div1,"resultTwo" => $div2); 
     echo json_encode($resultArray); 

    ?> 


</body> 
</html> 

我收到錯誤消息：

語法錯誤：JSON.parse：意外的非空格字符後JSON數據

var data = JSON.parse（xmlhttp.responseText）;

Answer 1

您可以通過以下網址驗證您的JSON響應，這樣就可以斷定你的結果一些點，

$resultArray = array("resultOne" => $div1,"resultTwo" => $div2); 
    echo json_encode($resultArray); 

編號：http://jsonformatter.curiousconcept.com/#jsonformatter