PHP if statement
Hi all,
I am trying to make do some simple database search functionality using php / MySQL. .. I am trying to search 3 different tables depending on what the user select in an oprtion field. So I basically have three querries one of which is called when a particular field is selected.
I am trying to use the an if statement to display the appropriate results on the result page but for some reason, it is showing all 3 section instead of one... what I have is something like:
*************************************
<?php
if ($option = 1) {
?>
Html with PHP Querry display for Option1
<?php
}
?>
<?php
if ($option = 2) {
?>
Html with PHP Querry display for Option2
<?php
}
?>
<?php
if ($option = 3) {
?>
Html with PHP Querry display for Option3
<?php
}
?>
***********************************
I would expect that if $option = 1 the page would only display Html with Querry display for Option1... Is that correct? ...if so then what am I doing wrong?
Thanks
Mizzy
Mark Hensler posted this at 17:52 — 17th August 2003.
He has: 4,048 posts
Joined: Aug 2000
= is the assignment operator
== is the comparison operator
Suzanne posted this at 18:09 — 17th August 2003.
She has: 5,507 posts
Joined: Feb 2000
To illustrate:
<?php
if ($option <strong>==</strong> 1) {
?>
Html with PHP Query display for Option1
<?php
}
?>
mizzy posted this at 18:42 — 17th August 2003.
They have: 47 posts
Joined: Jun 2001
oh...thats it
thanks
Busy posted this at 22:23 — 17th August 2003.
He has: 6,151 posts
Joined: May 2001
There are also other ways to do what your doing: if elseif else, and switch case, these two would give you a default, so if no option was choosen or the wrong option the main section would show up.
if($option == 1) { do stuff }
elseif ($option == 2) { do other stuff }
elseif ($option == 3) {do more stuff }
else { do the default }
or the other
switch ($option) {
case 0:
print "do something";
break;
case 1:
print "do something else";
break;
case 2:
print "do more somethings";
break;
default:
print "the default something";
}
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