David Robley wrote:
Aviation Coding wrote:
Hi all,
I am having problems with a connection to a mysql database.
I am using
----
function con()
{
mysql_connect("localhost","user","pass") or die(mysql_error());
mysql_select_db("tava") or die(mysql_error());
}
----
Now, when I call the _function_ (!)
----
con() or die("no con");
----
I get the "no con" output.
When I call the mysql_connect and mysql_select directly before executing a
query, I get some DB output. But that won't work when I am using the
function...
Any ideas would be greatly appreciated.
Cheers!
Chris
I think you need to return something from the function, like true if the
connection/select worked, false if not.
Cheers
You are correct.
function foo() {
// does something
}
var_dump(foo()); // returns NULL
why? because you don't explicitly return anything. If you did, that'd be
the return value. So if you did:
function bar() {
// does something
return true;
}
var_dump(bar()); // return true
Now, your script assumes a return-value:
baz() or somethingElse();
is an expression. This basically says:
if(!baz()) {
somethingElse();
}
Now, return (implicitly) null will result in (trough lazy comparison) a
false value (*null == false*, null !== false), which then triggers your
die() condition. Returning a meaningful value will get rid of that.
Other than that, choose 1 of 2 techniques:
1. error out inside the function itself (like you do now)
OR
2. error out based on the return-value of the function (like you do now
ASWELL)
don't combine 1 and 2, stick with either, but not both.
function con()
{
mysql_connect("localhost","user","pass") or die(mysql_error());
mysql_select_db("tava") or die(mysql_error());
}
con();
works. The following would work too:
function con()
{
$r = mysql_connect("localhost","user","pass");
if(!$r) {
return false;
}
$r2 = mysql_select_db("tava");
if(!$r2) {
return false;
}
return true;
}
con() or die('no conn');
would also work correctly. So, stick with either, don't mix em.
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