can get else statment to work

[contra10]

im trying to get my else statement to work if there is nothing in the query

 

<?php
 mysql_connect("localhost", "root", "") or die(mysql_error()); 
mysql_select_db("registration") or die(mysql_error());

$queryi = "SELECT * FROM `users` WHERE `username` = '$usernamec'";
$resulti = mysql_query($queryi);
if($comcount = mysql_num_rows($resulti)){
while($rowid = mysql_fetch_assoc($resulti))
{
$com= "{$rowid['comment']}";
$cname= "{$rowid['cusername']}";
echo"$cname thinks $com";
}
}else{
echo "no comments have been made";
}

?>

theres nothing in my query so the second line should show but it shouws the first echo without the variables present

 

[contra10]

or how can i place the else statement so that if there is nothing to query it shows that comment

[niranjnn01]

 

check n see what the mysql_num_rows is returning...

 

I guess in some cases, if there are no rows in the result set, mysql_num_rows will return 0 . this means that ur if will work.

 

rewrite it as if(mysql_num_rows ($query ) ! = 0)

 

[contra10]

my statment shows but i get error with this

 

<?php	
 mysql_connect("localhost", "root", "") or die(mysql_error()); 
mysql_select_db("registration") or die(mysql_error());

$queryi = "SELECT * FROM `users_should_comment` WHERE `username` = '$usernamec'";
$resulti = mysql_query($queryi);
if($comcount = mysql_num_rows($resulti)){
while($rowid = mysql_fetch_assoc($resulti))
{
$com= "{$rowid['comment']}";
$cname= "{$rowid['cusername']}";
echo"$cname thinks that $username should $com";
}
}else{
echo "no comments have been made to this user";
}

?>

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\wamp\www\profile\index.php on line 205

[kenrbnsn]

When you get that error it means your query has a problem. Rewrite your code to be something like this:

<?php
mysql_connect("localhost", "root", "") or die(mysql_error()); 
mysql_select_db("registration") or die(mysql_error());
$queryi = "SELECT * FROM `users_should_comment` WHERE `username` = '$usernamec'";
$resulti = mysql_query($queryi) or die("Problem with the query: $queryi<br>" . mysql_error());
$comcount = mysql_num_rows($resulti);
if ($comcount > 0)
   {
while($rowid = mysql_fetch_assoc($resulti))
{
                 $com= $rowid['comment'];
                 $cname= $rowid['cusername'];
                 echo "$cname thinks that $username should $com";
}
   }else{
echo "no comments have been made to this user";
}
?>

 

Ken

[contra10]

your certainly right ken...and after searching and searching...i finally decided to start from the beginning and i found the problem...i was 1 letter of in my query...now i have a heache...lol thanks