--
$query1 = "SELECT distinct link_id, url, title, description, $fulltxt, size
FROM ".$mysql_table_prefix."links WHERE link_id in ($inlist)";
$result = mysql_query($query1);
echo mysql_error();
and what does $fulltxt contain when you get the error
Note: forwarded message attached.
-
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house payment--- Begin Message ---
I have this code already, have a look at this few lines of code, which is
already existing.
-
Asif Qureshi wrote:
Here I narrow down the problem.
I meant narrow down the problem in your code.
Add
echo "mysql error? " . mysql_error() . "";
after your mysql_query calls for example.
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