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: <
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.
r
Couple of things you can do:
1. Drop the song ID and only get the artist information "SELECT
distinct(Artist) from songtable". It doesn't look like your SELECT statement
needs a song, but you include the song ID as $id anyway. Any reason for that or
can you drop it so you only get artist?
2.
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")
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 tab
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_ass
I have table and insert data using song_id, that means one artist can have
many song_ids. my question is how can I query distinct artist with his/her
song_id while I will not get duplicate data like duplicate artist?
my code is like this
$content .="
Select Artist";
$result= $db->sql_
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 10
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_da
Eternity Records Webmaster wrote:
I need to figure out how to format the date format: year-month-day
(2005-12-06). It is a date(8) field in a mysql table that i pulled out of
the table with php 5.0.5. I need it in the format: Tue December 6, 2005 (or
in mysql's formatting: 6-12-2005). I am using
farily straight forward to withinn the query itself, check out
http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html
or for php
http://uk.php.net/date
RTFM
Regards
Adrian
Eternity Records Webmaster wrote:
I need to figure out how to format the date format: year-month-day
(200
I need to figure out how to format the date format: year-month-day
(2005-12-06). It is a date(8) field in a mysql table that i pulled out of
the table with php 5.0.5. I need it in the format: Tue December 6, 2005 (or
in mysql's formatting: 6-12-2005). I am using the reformatting for display
only. S
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