Table is not showing a valid JSON values

Table is not showing a valid JSON values

DenonthDenonth Posts: 76Questions: 0Answers: 0
edited June 2012 in General
Hi all,
I have validated my output in JSON format and it is passing it normally.
Here is for example my JSON output:
[code]{"sEcho":1,"iTotalRecords":"1","iTotalDisplayRecords":"1","aaData":[["1","1","8808.348207","10000.00","12000.00","1322224306","99999999.00","5000.00"]]}[/code]

When I am loading my page I am getting JSON parse error. Any idea what can be wrong?

Replies

  • allanallan Posts: 63,542Questions: 1Answers: 10,476 Site admin
    Can you run your table through the DataTables debugger and give me the unique code please?

    Allan
  • DenonthDenonth Posts: 76Questions: 0Answers: 0
    http://debug.datatables.net/oletop
  • allanallan Posts: 63,542Questions: 1Answers: 10,476 Site admin
    edited June 2012
    The return from the server is:

    [code]
    < br / > < b > Warning < /b>: include(C:/Program
    Files(x86) / EasyPHP - 5.4.0RC4 / www / datatables / mysql.php): failed to open stream: No such file or directory in < b > C: \Program Files(x86)\EasyPHP - 5.4.0RC4\www\jackpot\live_table.php < /b> on line 32 < br / > < br / > < b > Warning < /b>: include(): Failed opening 'C:/Program
    Files(x86) / EasyPHP - 5.4.0RC4 / www / datatables / mysql.php ' for inclusion (include_path='.;
    C: \php\pear ') in C:\Program Files (x86)\EasyPHP-5.4.0RC4\www\jackpot\live_table.php on line 32

    You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' = !ORDER BY`PROGRESSIVE_ID`asc
    LIMIT 0,
    10 ' at line 3
    [/code]

    so not valid JSON :-)

    Allan
This discussion has been closed.