Warning: 1 is not a MySQL link index in...

[Anon]

Anyone have this error before?

It occurrs only when I'm uploading a file the user has opted to upload. If no file, there is no error.

This is the line of code the error is referring to:

$fileid = mysql_insert_id($newresult);

I'm grabbing the id to do an update with the name of the uploaded file for that record. I can do an update with just straight text with no errors, seems to be related to the image upload.

Any ideas?

Thanks in advance for any help, will sum...

Chris

[Anon]

A link index error often reffers to a missing or incorrect 2nd argument in a mysql_query or mysql_inserted_id statement. If no 2nd argument is given (like in your case), PHP automatically uses your default open data connection. Have a look at this:

$db = mysql_connect("yourserver", "username", "pwd");
mysql_select_db("mydatabase", $db);

You now could say this (assuming that $newresults contains a valid result):

$fileid = mysql_insert_id($newresult);

This is the same as:

$fileid = mysql_insert_id($newresult, $db);

Your problem is (I think) that $db (or whatever you used) is not currently in your scope. This error occurs when you use a function like this:

// $db contains a db-link
function MyFunc() {
// $db is not set here!!
// use GLOBAL
global $db;
}

That global line do it. It could be you are using no function, I still think the problem is in your scope. Hope this will help you.

Greetz,

Vincent Driessen
Venlo, The Netherlands

[Anon]

Thanks,

I found an eventual workaround, but your help gave me some insight. I ended up not using the variable and set it earlier in the code from another source!

Thanks again

Chris

Vincent Driessen wrote:

A link index error often reffers to a missing or incorrect 2nd argument in a mysql_query or mysql_inserted_id statement. If no 2nd argument is given (like in your case), PHP automatically uses your default open data connection. Have a look at this:

$db = mysql_connect("yourserver", "username", "pwd");
mysql_select_db("mydatabase", $db);

You now could say this (assuming that $newresults contains a valid result):

$fileid = mysql_insert_id($newresult);

This is the same as:

$fileid = mysql_insert_id($newresult, $db);

Your problem is (I think) that $db (or whatever you used) is not currently in your scope. This error occurs when you use a function like this:

// $db contains a db-link
function MyFunc() {
// $db is not set here!!
// use GLOBAL
global $db;
}

That global line do it. It could be you are using no function, I still think the problem is in your scope. Hope this will help you.

Greetz,

Vincent Driessen
Venlo,....

[Anon]

Chris,

I have the exact same problem now - when the user is uploading a file it screws up the mySQL (giving the error message you quoted). However it appears to be still executing the SQL - but just giving an erroneous error message! How did you resolve this?

Rob.

Chris wrote:

Anyone have this error before?

It occurrs only when I'm uploading a file the user has opted to upload. If no file, there is no error.

This is the line of code the error is referring to:

$fileid = mysql_insert_id($newresult);

I'm grabbing the id to do an update with the name of the uploaded file for that record. I can do an update with just straight text with no errors, seems to be related to the image upload.

Any ideas?

Thanks in advance for any help, will sum...

Chris