By outputting mysql_error() to the page you will receive exactly the same
error string that the client will give you
Just a little tip to save the time wasting of firing up your client ;-)
----- Original Message -----
From: "Shahmat Dahlan" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>; <php-db@lists.php.net>
Sent: Tuesday, December 13, 2005 11:15 AM
Subject: Re: [PHP-DB] Minor Change
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.
running it through the mysql client would definitely verify whether your
query is correct or not.
[EMAIL PROTECTED] wrote:
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 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 ($row =
mysql_fetch_assoc
# ($data_set)) That is what always come up when there is an error in the
# query.
**************
On Mon, 12 Dec 2005 14:13:10 -0500, Micah Stevens
<[EMAIL PROTECTED]> wrote:
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 109fh5 to 109fh6.
In the table, I changed 6 cells, leaving a couple blank. Then I changed
only the digit "5" to make it a "6" (109fh6) in the following:
$get_data_query = "select rep, party, state, cd, minority, afr_am,
asian,
am_indian, hispanic, med_hsehold_income, poverty from 109fh6 order by
$sort_field $sort_order";
Now I get "Warning: mysql_fetch_assoc(): supplied argument is not a
valid
MySQL result resource in" etc.
I've done this many times without a problem (this is the 6th time in
this
sequence). What could be wrong after such a minor change?
Ken
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