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).
Thanks,
Jim
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