RE: [PHP] Bizarre SQl, if issue....
| From: | Jay Blanchard | Date: | Tue, 29 Jul 2003 17:41:54 +0000 |
| Subject: | RE: [PHP] Bizarre SQl, if issue.... | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-157348@lists.php.net to get a copy of this message | ||
[snip]
Here's the SQL - I tested it via command line and it comes back empty
set.
SELECT ID FROM tblItems WHERE number = $place
>
> [snip]
> $person = mysql_query("Some Statement Here");
>
> if (!$person || $person == 0 || $person == '' || $person == '0') {
> $result =
> 1; }
> else { $result = 4; }
>
> If the SQL statement returns an empty set, shouldn't $result be equal
to
> 1?
> Instead, I'm getting $result = 4.
> [/snip]
>
> Actually I think we need to see the query == "Some Statement Here".
> $person in this case is either TRUE or FALSE
[snip]
So ...
$person = mysql_query("SELECT ID FROM tblItems WHERE number = $place");
$person IS TRUE because query executed (the OR statements do not count
as $person can only be TRUE or FALSE, so you can drop the OR's);
if(!$person){ // tests to see if person IS FALSE
$result = 1;
} else {
$result = 4;
}
$result will be 4
If you do this;
if($person){ // tests to see if person IS TRUE
$result = 1;
} else {
$result = 4;
}
$result would be 1
Now, if you used mysql_num_rows($person)
$thecount = mysql_num_rows($person);
if($thecount == 0){ // tests for zero
$result = 1;
} else {
$result = 4;
}
$result will be 1
if(!$thecount == 0){ // tests for other than zero
$result = 1;
} else {
$result = 4;
}
$result will be 4
HTH