[Resolved] problems with "question-mark" - IF - statement

[northk]

can anyone tell me, why i got a PARSE ERROR here:

//generate search-string
      $author    ? $query[0] = "pub_author LIKE %'$author'%" : unset $query[0];
      $title     ? $query[1] = "pub_title LIKE %'$title'%"   : unset $query[1];
      $projectid ? $query[2] = "pub_projectid_ = $projectid" : unset $query[2];

thanks in advance for your help!

[BuzzLY]

The unset is giving the parse error. I'm not sure it's allowed inside such a construct. It is better practice to use the ternary operator like so:

$query[0] = ($author ? "pub_author LIKE %'$author'%" : NULL);
$query[1] = ($title ? "pub_title LIKE %'$title'%" : NULL);
$query[2] = ($projectid ? "pub_projectid_ = $projectid" : NULL);

If it's important to you to use unset instead of assigning NULL values to the array, then use full if-else statements:

if ($author) {
   $query[0] = "pub_author LIKE %'$author'%";
} else {
   unset($query[0]);
}

Of course, in that case, if $author doesn't contain anything, then $query[0] won't even exist. In my first example, it will exist, and contain nothing. I can see where you would want to use unset.

When building your query, however, you may just want to do this:

unset($query);
if ($author) $query[] = "pub_author LIKE %'$author'%";
if ($title) $query[] = "pub_title LIKE %'$title'%";
if ($projectid) $query[] = "pub_projectid_ = $projectid";

That way, $query only contains your search terms, which is what I think you were trying to do in the first place, right?

[northk]

i already solved my problem with full IF-ELSE statements..

anyhow, thank you very much..
..cause, i was quite interested why it didn't work the other way...