Problems With Else If Statments

Question

Ok so what I want to happen is when a user is logged out my script should check the url for the code 2 and display to the user they have been logged out. If 1 There password or username was wrong and 3 Unsure how they got there. It pulls the correct number but will only display the message for 1. I don't know whats wrong :/

<?php
$err = $_GET['err'];
echo $err;//Proof the code is getting number
if ($err = "1") {
echo '<center><span class="error-box">'; echo "Whoops! Wrong Username or Password. Please try again.</span></center>";
} 
elseif ($err = '2') {
echo '<center><span class="info-box">'; echo "You Have Been Succesfully Logged Out.</span></center>";
} 
else {
echo '<center><span class="info-box">'; echo "Whoops! Somthing funny happened and were not quite sure what your trying to do. Please try again.</span></center>";
}
?>

you need to change all the = because it's an assignment operator there is == comparison operator and it compares only the value without caring about the type like 1 as integer when you use == to "1" as a string it returns ture === it compares the value and the type both like $var1 = 1; and $var2 = "2"; if you used === it will return false, and if you used == it will return true — Mohamed Hassan
– Mohamed Hassan, Commented Apr 19, 2012 at 0:05

zerkms · Accepted Answer · 2012-04-19 00:02:10Z

6

Because comparison operator is ==, not =.

So you need to have

if ($err == "1") {

answered Apr 19, 2012 at 0:02

zerkms

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5 Comments

iMysak Over a year ago

also in php or c many people recommend use if ("1" == $err) instead if ("1" == $err)

Tyler Radlick Over a year ago

There we go, Thank you so much!

zerkms Over a year ago

@iMysak: I wouldn't say that many, but indeed, some people recommend doing that (though it is a protection from a very obvious mistake that usually detected very soon)

zerkms Over a year ago

@Tyler Radlick: a typo-like mistake of using = instead of ==

nickb Over a year ago

@TylerRadlick - It works because if( "1" == $var) is valid syntax, but if( "1" = $var) means you'd assign a variable to a constant, which is not valid. It's a way to represent if statements to guarantee fixing these typos. That being said, I still prefer the other way, as IMO it is more readable.

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