Thanks for responding Steve,
My error reporting is turned on, because I always get errors. I found the
problem, I had two ;; at the end of one of the statements.
Thanks- Andrea
-Original Message-
From: Steve Cayford [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, March 20, 2002 10:27 AM
To: Andrea Caldwell
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] elseif statement syntax
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($num_results==0){
echo Sorry, nothing matched your search request. Please go back and
try
again.;
}
else {
echo pspan class=stdtext_boldNumber of Entries Found:
.$num_results./p/span;
}
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