You probably have error reporting turned off in your php.ini file or you
may be directing errors into your log file, try checking there.
Why are you tacking "or die(mysql_error())" on the end of each of your
query assignments? You're just assigning a string to a variable so
there's nothing to fail, or at least no mysql errors.
-Steve
On Wednesday, March 20, 2002, at 10:41 AM, Andrea Caldwell wrote:
> Hi All, I'm pretty new at this, so go easy on me please ;-)
>
> What is wrong w/ this syntax? If the search results are 0, it just
> displays
> a blank screen instead of echoing the error message if numresults ==0
> or the
> mysql_error message. If data is found, everything is fine. Thanks in
> advance for your help!
>
> if($searchterm){
> $query = "select directory.realname, directory.phone, directory.ext,
> directory.phone2, directory.email, directory.location from directory
> where
> realname like '%".$searchterm."%'" or die (mysql_error());
> }
> elseif($location){
> $query = "select directory.realname, directory.phone, directory.ext,
> directory.phone2, directory.email, directory.location from directory
> where
> location like '%".$location."%'" or die (mysql_error());
> }
> else{
> $query = "select directory.realname, directory.phone, directory.ext,
> directory.phone2, directory.email, directory.location from directory
> where
> location like '%".$searchloc."%'" or die (mysql_error());
> }
>
> $result = mysql_query($query) or die (mysql_error());
> $num_results = mysql_num_rows($result)or die (mysql_error());;
>
> if($num_results==0){
> echo "Sorry, nothing matched your search request. Please go back and
> try
> again.";
> }
>
> else {
> echo "<p><span class=stdtext_bold>Number of Entries Found:
> ".$num_results."</p></span>";
> }
>
>
>
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