mysql count rows

[uwannadonkey]

<?php
include('inc/header.php');

echo"--- Food---<br>";


$result=mysql_query('SELECT * FROM food WHERE owner = $user->ID and name = Apple'); 
if (mysql_num_rows($result) >= 1)
{
$apple = mysql_num_rows($result);
echo " Apples($apple)";
}









include('inc/footer.php');
?>

 

thats the code for counting apples, but this comes out:

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/trukki4/public_html/averyel/inventory.php on line 8

[uniflare]

replace

$result=mysql_query('SELECT * FROM food WHERE owner = $user->ID and name = Apple'); 

with

$result=mysql_query('SELECT * FROM food WHERE owner = $user->ID and name = Apple') or die(mysql_error()); 

 

and tell us what it says

[uwannadonkey]

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '>ID and name = Apple' at line 1

[uniflare]

silly me i should of noticed, you are using ' quotes instead of " for your query, you cannot include $variables without concatenating with single quotes, you have 2 options:

 

either:

 

$result=mysql_query("SELECT * FROM food WHERE owner = $user->ID and name = Apple"); 

 

or concatenate:

 

$result=mysql_query('SELECT * FROM food WHERE owner = '.$user->ID.' and name = Apple'); 

[uwannadonkey]

Unknown column 'Apple' in 'where clause'

 

but there is a column apple..

 

this is a copy of the mysql table:

ID  owner  name  effect  bonus

1 1             Apple 5 energy

2 1             Apple 2 energy

3 1               Apple 5 energy

[uwannadonkey]

fixed

 

$apple1= Apple;

$result=mysql_query("SELECT * FROM food WHERE owner = $user->ID AND name = 'Apple'")or die(mysql_error()); 

if (mysql_num_rows($result) >= 1)

{

$apple = mysql_num_rows($result);

echo " Apples($apple)";

}

 

apple needed a 'apple' to work