从php返回的json无法解析为jQuery dataTables

Question

I have a simple mysql database table with library books. I am using a php page to retrieve the list of books. This is what it returns:

php get_books.php

{"iTotalRecords":"1","aaData":[{"author":"Tim Powers","title":"The Anubis Gates","genre":"Fiction","publisher":null,"year":null,"location":"Bookshelf","notes":null}]}

In jQuery dataTables, I have:

<script >
    $(document).ready(function() {
    $('#books').DataTable({
        "bServerSide": true,
        "sAjaxSource": "./get_books.php"
    });
});
</script>

When I run the web page with that script I get an alert:

DataTables warning (table id = 'books'): DataTables warning: JSON data from server could not be parsed. This is caused by a JSON formatting error.

I can't find what the formatting error is. How should the data be formatted.

Here is the php page that does the return of the JSON data:

<?php
    $page = isset($_POST['page']) ? intval($_POST['page']) : 1;
    $rows = isset($_POST['rows']) ? intval($_POST['rows']) : 10;
    $offset = ($page-1)*$rows;
    $result = array();

    include 'conn.php';

    $rs = mysql_query("select count(*) from books");
    $row = mysql_fetch_row($rs);
    $result["iTotalRecords"] = $row[0];
    $rs = mysql_query("select * from books limit $offset,$rows");

    $items = array();
    while($row = mysql_fetch_object($rs)){
        array_push($items, $row);
    }
    $result["aaData"] = $items;

    echo json_encode($result);

?>

What should the return look like and how do I produce it?

Answer 1

There are many problems both with your JS and PHP code.

If you have less than a few thousands records in books table I would recommend disabling server-side processing mode by removing "bServerSide": true to enable client-side processing mode.

JavaScript :

$(document).ready(function() {
    $('#books').DataTable({
        "ajax": "get_books.php",
        "columns": [
           { "data": "author" },
           { "data": "title" },
           { "data": "genre" },          
           { "data": "location" }         
        ]
    });
});

PHP:

<?php   
    include 'conn.php';

    $rs = mysql_query("select author, title, genre, location from books");

    $result = array();
    $items = array();
    while($row = mysql_fetch_object($rs)){
        array_push($items, $row);
    }
    $result["data"] = $items;

    header("Content-type: application/json");
    header("Cache-Control: no-cache, must-revalidate");

    echo json_encode($result);
?>

HTML

<table id="books" class="display tablesorter">
   <thead>
      <tr>
         <th>Author</th>
         <th>Title</th>
         <th>Genre</th>
         <th>Location</th>
      </tr>
   </thead>
</table> 

See this jsFiddle for code and demonstration.

If you have more than a few thousands records, you will gain more performance by using server-side processing mode. But in this case I recommend using ssp.class.php helper library that comes in jQuery DataTables distribution (see examples/server_side/scripts folder).

Answer 2

Found the problem and it was stupid on my part! On OS X El Capitan ( 10.11.2 ), php files were NOT getting recognized by Safari or any other browser because they were in my home directory and not in the /Library/WebServer/Documents root for apache!

I moved my project into that directory as I could not get things set up to use my user directory ~/Sites as you could before El Capitan ( will work on that some more in the future ).

With that change, the php documents get executed as they are included in the ajax argument and all works fine!

Thanks every one for your help. Sorry to have wasted time, but my test.php worked, but I hadn't noticed that it was in the web server root directory!