SQL query; Please help me I have spent 2 days on this with no progress

[suprsnipes]

In basic terms I want to

 

UPDATE table 1 column 2

 

with the data from table 2 column 2

 

WHERE table 1 column 1 = table 2 column 1

 

After spending 2 days on this and after trying all various methods I'm stumped., it just won't work.

 

table 1 = t1

table 2 = t2

 

t1 column 1 = id

t1 column 2 = type

 

t2 column 1 = id

t2 column 2 = type

 

Can anyone help me?

[Michdd]

$result = mysql_query('Select type FROM `t2` WHERE id="' . $id . '" LIMIT 1');
$row = mysql_fetch_assoc($result);
mysql_query('UPDATE `t1` SET type="' . $row['type'] . '" WHERE id="' . $id . '"');

[suprsnipes]

Thanks for the reply Michdd.

 

Sorry my definitions of the columns on t2 were slightly incorrect could you please update your suggested code for me.

 

table 1 = t1

table 2 = t2

 

t1 column 1 = id

t1 column 2 = type

 

t2 column 1 = id1

t2 column 2 = type1

[roopurt18]

http://dev.mysql.com/doc/refman/5.0/en/update.html

update t1, t2 set t1.type=t2.type1 where t1.id=t2.id1

[suprsnipes]

I'm getting an notice: Undefined variable:id on line 9 and line 11.

 

9.   $result = mysql_query('Select `type` FROM `t2` WHERE `id`="'. $id . '" LIMIT 1');
10. $row = mysql_fetch_assoc($result);
11. mysql_query('UPDATE `t1` SET type="' . $row['type'] . '" WHERE id="' . $id . '"');

 

 

[Ken2k7]

Well you haven't given us any code to work with so Michdd put up an example. You have to define the variable $id if you actually need that WHERE clause case.