Re: prepared statement syntax problem
all this code is SQL code...how it is going to be in PHP?
I tried this for example and I get other syntax problem
$connection->prepare('set @arg = ?%;
select u.name,u.lastname,t.user_type
from users u
join business_users b on u.user_id=b.crid
join buz_usertype t on b.bus_user_type=t.type_id
where u.lastname like=@arg')
Subject
Written By
Posted
September 08, 2015 05:58AM
September 08, 2015 10:45AM
September 08, 2015 11:13AM
September 08, 2015 01:30PM
Re: prepared statement syntax problem
September 08, 2015 09:58PM
September 08, 2015 10:13PM
September 08, 2015 10:23PM
September 09, 2015 12:58PM
September 10, 2015 10:26AM
Sorry, you can't reply to this topic. It has been closed.
Content reproduced on this site is the property of the respective copyright holders.
It is not reviewed in advance by Oracle and does not necessarily represent the opinion
of Oracle or any other party.