doujiao1981 2015-02-23 05:21
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通过AJAX请求PHP脚本。 AJAX请求返回实际的PHP脚本

Requesting JSON from php script :

var channelList;

$(document).ready(function() {
        url: 'channellookup.php',    
        dataType: 'json',    
        error: function(){console.log(arguments)},     
        success: function(data) {
            channelList = data;

Now to the interesting part: The error message in the console reads like this:

Arguments { 0: XMLHttpRequest, 1: "parsererror", 2: "Invalid JSON: <?php
   header('Content-type: application/json');     // To ensure output json type.
   class MyDB extends SQLite3

And so on. My whole PHP code is in that message. Something must go completely wrong here.

Here is my PHP in full


   header('Content-type: application/json');     // To ensure output json type.
   class MyDB extends SQLite3

      function __construct()


   $db = new MyDB();

      echo $db->lastErrorMsg();    
   } else {    
      echo "Opened database successfully

   $sql =<<<EOF

      SELECT * from channels;


   $ret = $db->query($sql);

   $channelList = array();

   while($row = $ret->fetchArray(SQLITE3_ASSOC) ){    
      $channelList[] = $row;    
   echo json_encode($channelList);

as you can see I do encode as json. As I said this works in the terminal. Please keep in mind that I do not want to use the json in my html yet so the page generating before the asynchronous request is completed isn't an issue yet.

Is the problem maybe that I am not doing this on a remote server but on local files? As I understand the Browser should be able to handle this case.

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2条回答 默认 最新

  • dooo61733 2015-02-23 05:32

    I'm quite sure that your PHP didn't compile at all. This is because the result that was returned, is in fact the body of your php file, indicating that no translation took place.

    You can check if your PHP instance is running by creating a file with the code:

    <?php phpinfo() ?>

    And use your browser to navigate to that page and see if there are any output. You want to check your PHP/Apache installation until the output is correct.

    As to why your webtool work, I cannot phantom any reason. It should fail in the same way.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?



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