No record found

joeywoodbury

Ok I am having another problem. I am trying to check to see if a record has been found, if not I want it to display a message that says no record found. Here is the code I have so far. If not record is found then it displays a blank page. Instead of the message. Thanks in advance for any help.
Code is below:

//Connect
$db = mysql_connect($dbhost,$dbuser,$dbpass);
@mysql_select_db( "$dbname",$db) or die( "Unable to select database");

//error message (not found message)
$XX = "No Record Found";

//Query
$query = mysql_query("SELECT * FROM $usertable WHERE companyname LIKE '$companyname%' ");

while ($myrow = mysql_fetch_array($query))
{
print ("\nCompany Name: ");
if ($myrow[web]) {
print ("\n<a href=\"http://");
echo "$myrow[web]";
print ("\" target=_blank>");
}
echo "$myrow[companyname]";
print ("</a>");
print ("\n Name: ");
echo "$myrow[firstname] ";
echo "$myrow[lastname] ";
if (!$myrow[indkey]) {
// throw message if no record
print ("$XX");
exit;
}
}

end

Anon

Hello,
you can use mysql_num_rows($query). It returns the number of rows returned by SQL statement. Or maybe you should use is_empty($myrow[indkey])or is_set($myrow[indkey]) instead of !$myrow[indkey].

Zdenek

Anon

Okay, here is the problem: you are running a while loop to recurse through all the result rows and print them out. BUT: if the argument inside the while loop returns false, it BREAKS OUT of the loop. BUT: your if statement to check this false statement is INSIDE the while loop, so it never get's executed. Modify your code to this:

$firstrow = mysql_fetch_array($query);
if($firstrow) {
print $firstrow["web"]; // print the first row
while($nextrow = mysql_fetch_array()) {
// print the subsequent rows
}
}
else {
print "Sorry, no rows"; // print your $XX thing here
}

hope this helps!
-sridhar

joeywoodbury

Thanks to both of you for the help. I have it working now! Yeah!