function help

[kpdvx]

function userphoto($uid) {
$query = "SELECT photo FROM profiles WHERE userid = '$uid'";
$result = mysql_query($query);
@extract(mysql_fetch_array($result));

if ($photo != "&quot {
$photo_filename = $photo;

} elseif ($photo == "&quot {
$photo_filename = 'default.gif';

}

return $photo_filename;
}

$photo_filename has no value. Any idea why?

 

[KempCGDR]

Is the photo field in the database blank?

 

[sleipnir214]

Exactly where are you assigning a value to $photo before you test it? Want the best answers? Ask the best questions:

http://www.tuxedo.org/~esr/faqs/smart-questions.html

TANSTAAFL!

 

[kpdvx]

$photo is a field in the profiles table. It is assigned to $photo when I extract it. Yes, it is blank. If the field is blank, then I want $photo_filename = 'default.gif', and if it has a value, then I want $photo_filename to be equal to that value, like say, 'picture.jpg'.

 

[sleipnir214]

Is it blank, or is it NULL?
Want the best answers? Ask the best questions:

http://www.tuxedo.org/~esr/faqs/smart-questions.html

TANSTAAFL!

 

[kpdvx]

NOT NULL, so blank.

 

[danielhozac]

Well, you could simply change the

    } elseif ($photo == "") {

line with

    } else {

//Daniel

 

[sleipnir214]

Good recommendation. The additional test is unnecessary. Want the best answers? Ask the best questions:

http://www.tuxedo.org/~esr/faqs/smart-questions.html

TANSTAAFL!