[SOLVED] MYSQL Replace error.

[JasonO]

Hi.

 

Is there anything wrong with this syntax? I get an error and it refuses to work :S

 

mysql_query("REPLACE INTO currentcust WHERE id=$cust_id
(id, name, email) VALUES ('$cust_id', '$cust_name', '$cust_email') ")
or die(mysql_error());

 

The error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE id=34 (id, name, email) VALUES (34,' at line 1

 

All the variables are right and contain the correct values.

 

Any ideas?

 

Thanks,

JasonO

[mmarif4u]

Try this one:

mysql_query("REPLACE INTO currentcust WHERE id='$cust_id'
(id, name, email) VALUES ('$cust_id', '$cust_name', '$cust_email') ")
or die(mysql_error());

 

[JasonO]

Nope, didn't work. Same error but with '34' instead of 34

[eRott]

I've never heard of any REPLACE function in mySQL. There is however and UPDATE function. Perhaps you are trying to do that?

 

Try this:

 

mysql_query("UPDATE currentcust SET id = '$cust_id', name = '$cust_name', email = '$cust_email' WHERE id=$cust_id") 
or die(mysql_error());  

 

EDIT:

 

Ok, try this:

 

mysql_query("REPLACE INTO currentcust (id, name, email) VALUES ('$cust_id', '$cust_name', '$cust_email') WHERE id='$cust_id'")
or die(mysql_error());

[mmarif4u]

@ eRott

Yeh there is replace into function in mysql , Read the manual here.

http://dev.mysql.com/doc/refman/4.1/en/replace.html

 

About this query i am not sure about the where clause where it will go in replace into function.

[JasonO]

I tried eRott's alternative function and that did the trick.

 

Thanks for all your help, and everyone that contributed to helping me

 

Topic solved

[eRott]

@mmarif4u

 

Yea, I just googled it. Thanks

 

@JasonO

 

I edited your code. Perhaps placing the WHERE at the end might work?

[mmarif4u]

Yes in update function where clause will go at very end of the query.

[eRott]

I was referring to his REPLACE INTO query .