William Curry wrote:
$qry1 = "SELECT *,CONVERT(Char(24),CALL_ENTRY_DATE,101) as MYDATE from PcarsCallComplete
where Location_address = " .$Location2. " order by CALL_NO";
This method usually works well for me for debugging purposes:

immediately prior to query, echo it to see exactly what you're telling mysql and then exit the script so it's the only output. Log in to mysql as the user that your script is logging in as (just in case it's a permission setting)
Enter sql via copy & paste
Check results.

So, for you:
echo "\$qry1 = \"SELECT *,CONVERT(Char(24),CALL_ENTRY_DATE,101) as MYDATE from PcarsCallComplete where Location_address = $Location2 order by CALL_NO\";";

Enter output into mysql (obviously only the code between the quotes) and run query. Be sure you're logged in as the same user that the script logs in with, else you may have different privileges!

If you don't receive an error or an empty set in mysql, then it could be something simple yet hard to diagnose. Perhaps you're inserting (and logging directly in) to a different database than your script is reading from (such as if you have multiple comps on your network acting as servers, perhaps you're trying to select from the wrong machine's database).



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