RE: mysql_insert_id() and JOIN

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> Hi again.
> 
> I dont understand how to use this two.
> 
> I read php.net and other sites, but dont understand...

http://dev.mysql.com/doc/
Available in the major languages, pick your filter and start reading.
Having documentation in your own language may help a lot towards
understanding.

> I have this four tables: table1, 2, 3 and 4
> 
> So the SELECT is like this?:
> 
> <?
> 
> $query_select = "SELECT * FROM table1 FULL JOIN table2 FULL JOIN table3
FULL
> JOIN table4 ON table1.id=table2.id=table3.id=table4.id ORDER BY id ASC";
> 
> mysql_query($query_select);
> 
> ?>
> 
> This is with FULL JOIN... How identificate the left table if I want to use
> LEFT JOIN?

The left table is the table to the left of the join. So:

SELECT id FROM table1 LEFT JOIN table2...

table1 is the left table and table2 is the right table (if you were to
use a RIGHT JOIN).

>> With my 4 tables the sql is like this?:
>> SELECT * FROM table1 LEFT JOIN table2 LEFT JOIN table3 LEFT JOIN table4
WHERE table1.id =  table2.id = table3.id = table4.id


> 
> How is the UPDATE?
> 
> And the INSERT I dont understand yet how is it?

UPDATEs and INSERTs follow the same procedure.

When you INSERT an entry into the main table, you then use the function
mysql_insert_id() which will give you the value of the last 'id' column
generated, which you can then use to insert data into the secondary
tables.

>> How is the the slq query with "mysql_insert_id()"?

> 
> I'm not to do a link between tables in PHPMyAdmin?
> 
>  
> 
> Thank you very... VERY... much!
> 

If you are using MySQL 4.1 or later you should use the mysqli extension
and not the older mysql one. UNLESS you are relying on the older style
passwords, which is a BAD idea.
>> I dont know what version I use =S

>> THANKS A LOT NEIL

--
Niel Archer



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