Table is not showing a valid JSON values
Table is not showing a valid JSON values
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?
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?
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Replies
Allan
[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