PHP if not equal(!=) and or (||) issue. Why doesnt this work?

Question

I know this is simple PHP logic but it just won\'t work...

 $str = \"dan\";
 if(($str != \"joe\") 
   || ($str != \"danielle\")
   || ($str != \"heather\")
          


        

Answer 1

Welcome to boolean logic:

$str = 'dan'

$str != "joe" -> TRUE, dan is not joe
$str != "danielle" -> TRUE, danielle is not dan
$str != "heather") -> TRUE, heather is not dan
$str != "laurie" -> TRUE, laurie is not dan
$str != "dan" -> FALSE, dan is dan

Boolean logic truth tables look like this:

and:

TRUE && TRUE -> TRUE
TRUE && FALSE -> FALSE
FALSE && FALSE -> FALSE
FALSE && TRUE -> FALSE

or:

TRUE || TRUE -> TRUE
TRUE || FALSE -> TRUE
FALSE || TRUE -> TRUE
FALSE || FALSE -> FALSE

Your statement boiled down to:

TRUE || TRUE || TRUE || TRUE || FALSE -> TRUE

Answer 2

try this

$str = "dan";

if($str == "joe" || $str == "daniella" || $str == "heather" || $str == "laurine" || $str == "dan"){ ... }

Answer 3

When you compare two strings you have to use strcmp().

Answer 4

I am not exactly sure what you want, but that logic will always evaluate to true. You might want to use AND (&&), instead of OR (||)

The furthest statement that is ever tested is ($str != "danielle") and there are only two possible outcomes as PHP enters the block as soon as a statement yields true.

This is the first:

$str = "dan";

$str != "joe" # true - enter block
$str != "danielle" #ignored
$str != "heather" #ignored
$str != "laurie" #ignored
$str != "dan" #ignored

This is the second:

$str = "joe";

$str != "joe" # false - continue evaluating
$str != "danielle" # true - enter block
$str != "heather" #ignored
$str != "laurie" #ignored
$str != "dan" #ignored

If the OR was changed to AND then it keeps evaluating until a false is returned:

$str = "dan";

$str != "joe" # true - keep evaluating
$str != "danielle" # true - keep evaluating
$str != "heather"  # true - keep evaluating
$str != "laurie" # true - keep evaluating
$str != "dan"  # false - do not enter block

The solution doesn't scale well though, you should keep an array of the exclude list and check against that do:

$str = "dan";
$exclude_list = array("joe","danielle","heather","laurie","dan")
if(!in_array($str, $exclude_list)){          
    echo " <a href='/about/".$str.".php'>Get to know ".get_the_author_meta('first_name')." &rarr;</a>";
}

Answer 5

Another approach is

$name = 'dan';
$names = array('joe', 'danielle', 'heather', 'laurie', 'dan');

if(in_array($name,$names)){  
    //the magic
}

Answer 6

Based on your comment in Glazer's answer, it looks like you want to enter the if block when $str is not one of the listed names.

In that case it would be more readable if you write it as

if( !( ($str == "joe") || ($str == "danielle") || ($str == "heather") || ($str == "laurie") || ($str == "dan") ) )

This actually reads as "if it's not one of these people..." to someone looking at your code. Which is equivalent to the slightly less obvious

if( ($str != "joe") && ($str != "danielle") && ($str != "heather") && ($str != "laurie") && ($str != "dan") )

The fact that they're equivalent is called DeMorgan's law in logic.