Resource id #6 error

[willwill100]

from this code:

[code]$namen = mysql_query("SELECT * FROM `$compl` WHERE `Name`='$name'");
echo ("<pre>" . $namen . "</pre>");[/code]

i get the <pre> displaying:

[code]Resource id #6[/code]

what is the reason for this?

[kenrbnsn]

The mysql_query() function returns a pointer to the rows selected not the rows. To get individual rows, you need to use one of the mysql_fetch functions, usually in a loop.

Ken

[willwill100]

ok, ive ammended my code but i still get a negative result:

[code]////if their name does not already exist in the REAL table, insert it////
$q = "SELECT COUNT(*) FROM `$compl` WHERE `Sail Number`='$snum'";
echo ("<pre>" . $q . "</pre>");
$rs = mysql_query($q) or die('Problem with query: ' . $q . '<br>' . mysql_error());

if (mysql_fetch_assoc($rs)==0){
    $b = "INSERT INTO `$compl` ( `Name` , `Sail Number` , `Class`) VALUES ( '$name', '$snum', '$boattype')";
  
}else{
echo ("<br>No user updated!");
}[/code]

I have inputted $q into phpmyadmin and a value of "0" is returned, why then is the "if statement" taken to be false and "No user updated!" outputted??

[redbullmarky]

on this line:

[code]
if (mysql_fetch_assoc($rs)==0){
[/code]

you're checking if an entire array equals 0. even if there are no rows, it still wont return the number 0.

what you need is
[code]
if (mysql_num_rows($rs) == 0) {
[/code]

[willwill100]

okay that makes sense but i still get the error.

any ideas, im really stuck on how to get my table to check if there is already a user in it and enter one if the user is not in the table.....

[keeB]

[!--quoteo(post=357078:date=Mar 21 2006, 08:31 PM:name=WillWill)--][div class=\'quotetop\']QUOTE(WillWill @ Mar 21 2006, 08:31 PM) [snapback]357078[/snapback][/div][div class=\'quotemain\'][!--quotec--]
okay that makes sense but i still get the error.

any ideas, im really stuck on how to get my table to check if there is already a user in it and enter one if the user is not in the table.....
[/quote]

[code]<?php

$q = "SELECT COUNT(*) FROM `$compl` WHERE `Sail Number`='$snum'";

$result = mysql_query($q);

    if ( mysql_num_rows($query) == 0 ) {
        $q2 = "INSERT INTO `$compl` ( `Name` , `Sail Number` , `Class`) VALUES ( '$name', '$snum', '$boattype')";
        mysql_query($q2);
    }

//more code here or whatever

?>[/code]

[craygo]

why not just do this

[code]////if their name does not already exist in the REAL table, insert it////
$q = "SELECT * FROM `$compl` WHERE `Sail Number` = '$snum'";
$rs = mysql_query($q) or die('Problem with query: ' . $q . '<br>' . mysql_error());
  $num_rows = mysql_num_rows($rs);
if ($num_rows == 0){
    $b = "INSERT INTO `$compl` ( `Name` , `Sail Number` , `Class`) VALUES ( '$name', '$snum', '$boattype')";
  
}else{
echo ("<br>No user updated!");
}[/code]

Ray

[willwill100]

thats great, thanx for the help guys!