Easy PHP fix?????

Discussion in 'PHP' started by SearchBliss, Jun 3, 2009.

0
  1. #1
    I have a PHP code snippet that gets ignored, there is no parsing error, and I can't fiqure out why? Please help me if you can. The result of (getIt($mySite) returns a number between 0 and 2, yet this code is ignored???

    if (getIt($mySite) == "0")
echo "<img src=0.jpg>";

elseif (getIt($mySite) == "1")
echo "<img src=1.jpg>";

else
echo "<img src=2.jpg>";
    Code (markup):
    Thanks!
     
    SearchBliss, Jun 3, 2009 IP
  2. jsamdirect

    jsamdirect Peon

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    #2
    what is getIt? Is that a custom function or something that is getting the $mySite data from $_GET or $_POST? Can you paste the code for that?
     
    jsamdirect, Jun 3, 2009 IP
  3. James Barcellano

    James Barcellano Active Member

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    #3 
    
if (getIt($mySite) == "0") {
    echo "<img src=0.jpg>";
} elseif (getIt($mySite) == "1") {
    echo "<img src=1.jpg>";
} else {
    echo "<img src=2.jpg>";
}
    PHP:
    Format your code correctly please, also what is getIt()
     
    James Barcellano, Jun 3, 2009 IP
  4. SearchBliss

    SearchBliss Well-Known Member

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    #4
    getIt is a custom function...and I'm an ASP coder and new to PHP so thanks for the format correction...I need it.
     
    SearchBliss, Jun 3, 2009 IP