[RESOLVED] Update on Duplicate Key Problem

dr1234

I have a problem with this code which I would like to insert a record into a table but if it already exists then update the existing record. This is my code so far:

if ((isset($POST["MM_insert"])) && ($POST["MM_insert"] == "form4")) {
$insertSQL = sprintf("INSERT INTO ProductsUpdated (Code, Web_Price, Overriden) VALUES (%s, %s, %s) ON DUPLICATE KEY UPDATE SET Web_Price=%s",
GetSQLValueString($POST['Code'], "text"),
GetSQLValueString($POST['Web_Price'], "text"),
GetSQLValueString($_POST['Overriden'], "int")); <Line 53

mysql_select_db($database_WebPriceUpdates, $WebPriceUpdates);
$Result1 = mysql_query($insertSQL, $WebPriceUpdates) or die(mysql_error());

I get the error message:

Warning: sprintf() [function.sprintf]: Too few arguments in C:\wamp\www\Test.php on line 53

Not sure what I am missing in this any help would be much appreciated.

Weedpacket

What it says: you have four "%s" in your string, but only three arguments to fill them.

dr1234

Just figured it out after I posted the thread, thanks for the reply

manicsurfer

I have been looking for a resolved to this myself. I am getting the same issue as you. May I ask please how you resolved this?

bradgrafelman

manicsurfer;11012601 wrote:
I have been looking for a resolved to this myself.

Unless "this" happens to mean that you're using the exact same code as the OP above, you should create a new thread (rather than hijacking someone else's) and show us, at a minimum, a) the relevant part(s) of your code, and b) any and all error messages you're receiving (or a description of what's happening versus what you expected to happen).

manicsurfer

Yes I am in fact using 'Exactly' the same code and strangely with a little more experimentation I too resolve it.

However, thank you for you informative answer. It reminds me why I like 'Help' forums so much.

bradgrafelman

Glad I could help!