mysql_insert_id error

Anon

Is there a problem with mysql_insert_id with
4.0.1pl2 ?

Note: Yes i need to upgrade, but on another machine im still running PHP3 and it works fine.

Warning: Supplied argument is not a valid MySQL-Link resource

I would be guessing this is a PHP bug in the version im running?

Derek

Anon

"Supplied argument is not a valid MySQL-Link resource"

means that you are trying to work on a mysql_connection that does not exist.

Most likely you need to check to make sure your database connection is established properly.

Error-checking is very important.
Check the manual about mysql_error and mysql_errno

Anon

Well It has to exist because I just ran a insert query which worked.

Code looks like this (also very odd this worked fine in PHP3)

$query = mysql_query("INSERT INTO table (subject, message, date, username, email) VALUES ('$subject', '$message', '" . mktime() . "', '$username', '$email')") or die ("Error blah blah");
$this->newsID = mysql_insert_id($query);

Nnow the actual insert works, so there is not connection issue here. Any ideas?

Derek

Anon

This is eiteher a typo or a DUH alert :-)

mysql_insert_id takes a database_link_identifier as parameter, NOT a resultset

int mysql_insert_id ([int link_identifier])

Anon

Alright im confused, because what i showed you was exactly what i used to do and oddly it worked. But time to do it right. So what exactly is my link_identifier?

If i run my insert query

mysql_query("INSERT yadd yadda blah blah");

then i use:

mysql_insert_id();

what do i put as my link identifier?

Anon

Well if you allways used mysq_insert_id($resulset) then
there must have been a bug up until 4.0 :-)

The proper procedure is (possible variants aside):

$db_link_identifier=mysql_connect();
mysql_select_db("databasename",$db_link_identifier);
$resultset=mysql_query("select bla",$db_link_identifier);
$insert_id=mysql_insert_id($db_link_identifier);
mysql_close($db_link_identifier);

Note: this should also contain about 45 IF statements to make sure that every step actually worked.
For ex: if the mysql_connect failed, the rest of the program should be skipped because it won't work anyway.

Anon

I have exactly the same problem:-

ERROR :
Supplied argument is not a valid MySQL-Link resource

My programs ran on PHP3.0.12 before and as soon as I migrated to 4.03, I am getting the above problem, which didn't ever used to crop up.

Any thoughts!!?

Jason.

[deleted]

Check your sql statements, and always use an IF statement to check if your mysql_query() was successful. Print the mysql_error() if it fails.

Anon

Alright - enough. This was driving me nuts, and no, it wasn't a problem I fixed by checking errors, etc.

This is a very common problem (just do a search for the error string in Google and see how many others have had thier own problems documented in the cache for all to see.) Stems from a 3 to 4 conversion, and here's a quick fix I found when you're looking to use results from one query after the other. For a more detailed solution, see the answer I found on another PHP board.

Step 1 - Get rid of your mysql_insert_id statement.

Step 2 - Where you would have placed your var in your INSERT statement, use LAST_INSERT_ID() instead (no quotes, etc.)

Step 3 - Read the rest of this post if you need more help and that for some reason doesn't work.

This was a very simple fix with a slew of different answers avail - it needs to be addressed at some point in greater detail. Hopefully this will help some.

If you want to use the ID that was generated for one table and insert it
into a second table, you can use SQL statements like this:

INSERT INTO foo (auto,text)
VALUES(NULL,'text'); # generate ID by inserting NULL
INSERT INTO foo2 (id,text)
VALUES(LAST_INSERT_ID(),'text'); # use ID in second table

...found here:
http://www.mysql.com/doc/G/e/Getting_unique_ID.html

It works even without inserting the NULL value for some reason 😉
The following is great for monitoring:
$new_id = mysql_insert_id();
print "New id: $new_id\n";