How can i make this code work? TY!
$site = '1'
$mysites = array('1', '2', '3', '4', '5', '6');
foreach($mysites as $mysite)
{
echo $mysites; **but not the site with value 1**
}
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4 Answers
A simple if will suffice:
$site = '1';
$mysites = array('1', '2', '3', '4', '5', '6');
foreach($mysites as $mysite)
{
if ( $mysite !== '1' )
{
echo $mysite;
}
}
or if you wan't to check against the $site variable:
$site = '1';
$mysites = array('1', '2', '3', '4', '5', '6');
foreach($mysites as $mysite)
{
if ( $mysite !== $site )
{
echo $mysite;
}
}
2 Comments
Shikiryu
Anyway,
$site = '1' isn't valid, $site = '1'; is. Plus, echo $mysites; won't display anything since $mysites is an array. Please, don't copy/paste bad code from the OP question just to be faster to answer.
Jan Hančič
@ClemDesm I've fixed those errors. It wasn't my intention to answer as quickly as possible, I just completely missed those errors. Thanks for pointing them out!
$site = '1'
$mysites = array('1', '2', '3', '4', '5', '6');
foreach($mysites as $mysite) {
if ($mysite == $site) { continue; }
// ...your code here...
}
2 Comments
Vladyslav at Prime Board
well this is a very small and simple piece of code, which directly answers the question asked, and such a simple code is pretty much self explanatory... If not, would you advice me for the future what would be a suitable explanation for this one.
Alex Tape
a small comment would be very helpful :) but who cares.. you are right. i apology.
Just use an if statement:
foreach($mysites as $mysite) {
if ($mysite !== $site) {
echo $mysite;
}
}
Comments
you can do this:
$mysites = array('1', '2', '3', '4', '5', '6');
foreach($mysites as $mysite)
{
if($mysite != 1){
echo $mysite;
}
}
2 Comments
Mohammed Abdul Nasser
it was !== this is was the problem and it fixed now
Mohammed Abdul Nasser
not the same result you can watch this link ibb.co/PxYBBsT