Reputation: 133
The output of code is :
Connected and SQL Problem
Which means that it connects but the desired databse i.e malala is not being selected.Please help?
$con = mysqli_connect("localhost", "username", "password") or die("Connection Problem" . mysqli_connect_error($con));
echo "Connected and ";
$database = mysqli_select_db($con,malala) or die("SQL Problem".mysqli_connect_error($con));
echo "Selected";
Upvotes: 0
Views: 53
Reputation: 74232
Because, malala
in ($con,malala)
needs to be quoted. It's being treated as a constant.
I.e.:
($con, "malala")
Or place it inside
$con = mysqli_connect("localhost", "savehand_tallal", "&aBE+9NxZ6^c")
as
$con = mysqli_connect("localhost", "savehand_tallal", "&aBE+9NxZ6^c", "malala")
so you won't need to use
$database = mysqli_select_db($con,malala)...
Having used error reporting though, would have thrown you an undefined constant malala notice.
Reference for mysqli_connect()
:
Also make sure that the malala
database was indeed created and doesn't hold any whitespace or foreign characters.
Upvotes: 2