You're getting an error, after the query, put:
echo mysql_error();
to find out what's happening.
On Monday 12 December 2005 11:05 am, [EMAIL PROTECTED] wrote:
I made tiny changes to my php file and sql table and the table won't come
up. I updated the table name (and php file name) from
After adding echo mysql_error(); I get the same result. I tried changing
the query to include 109fh7 (a table which doesn't exist) and got the same
result as with 109fh6. Changing to 109fh5 does pull up that table. The
line to which the error message refers is while ($row =
Ken:
Echo out the query. Check it. Run it from the MySQL client.
What does the error message say? Does the table 109fh6 actually exist in
the database?
David
# Ken responded:
#
# After adding echo mysql_error(); I get the same result. I tried changing
# the query to include 109fh7 (a
This isn't really a MySQL error (sorta), it's a PHP error. You probably forgot
to update a variable name when you updated everything else.
Here's an example sequence for querying using PHP/MySQL:
$TestQY = SELECT * from SomeTable;
$TestRS = mysql_query($TestQY) or die(Error executing query);
did it say which line does the error
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result
resource in
is referring to. if it is referring to the line which contains the
function call mysql_query (),
dpirago is right, there could be some mistake in the sql statement.
