[SOLVED] drawing data from the sql database with mutliple rows in php

[trampolinejoe]

Hello all you very helpful people,

 

if you guys have been online for the last few days you may have noticed i have been trying to build a shopping cart.

 

I have a table in my database that stores all the products that are ordered. Also, it gives each product a order to belong to.

 

basically:

A customer HAS A Order and a order HAS A product

 

I am trying to withdraw the data from the database but the thing is a customer can have multiple products. My code only works if the customer has a single product. My query brings back multiple rows as the customer has multiple products ordered.

 

Ill show you what I have:

$order = mysql_query("SELECT * FROM OrderProducts WHERE OrderId =" . $connote );

while($row = mysql_fetch_array($order))
	  {	
			$product = $row['ProductId'];
			     
	  }

 

Iam sure there is away in php to add each product to some kind of product array. I tried afew ideas but it got errors since I don't know the correct syntax, but i am guessing i need some kind of loop that appends the product to the array for each product that there is. No idea how to actually do it. I am sure somebody here can teach me how to do it.

 

Cheers,

Joe.

 

[waynew]

$order = mysql_query("SELECT * FROM OrderProducts WHERE OrderId =" . $connote );

   while($row = mysql_fetch_array($order))
        {   
            $product = $row['ProductId']; //overwriting $product each time!!!!!!!!
                 
        }

 

I am guessing that this is because you are overwriting $product each time:

 

Try

 

$order = mysql_query("SELECT * FROM OrderProducts WHERE OrderId =" . $connote );
$products = array();
   while($row = mysql_fetch_array($order))
        {   
            array_push($products,$row['ProductId']); //place product ids in array
                 
        }
//output array
$i = 0;
while($i < sizeof($products)){
echo $products[$i];
$i++;
}

[trampolinejoe]

hello waynewex

 

dosnt seem to be working, getting the same error actually

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

 

for this line

   while($row = mysql_fetch_array($order))

 

I see myself overwriting product, so i see what you had done to add to the array using array_push.

But yeah, if i run the query directly in the DB it works perfectly fine.

 

Cheers,

Joe.

[waynew]

It's happening because your query is failing somehow. Could you use this and tell me what it outputs?

 

$order = mysql_query("SELECT * FROM OrderProducts WHERE OrderId =" . $connote );
echo "<h2>SELECT * FROM OrderProducts WHERE OrderId =" . $connote."</h2>";
$products = array();
   while($row = mysql_fetch_array($order))
        {   
            array_push($products,$row['ProductId']); //place product ids in array
                 
        }
//output array
$i = 0;
while($i < sizeof($products)){
echo $products[$i];
$i++;
}

[Alexhoward]

I'd use sessions to hold the info, then put it in to SQL when they continue to the checkout

 

try using: foreach

 

http://uk3.php.net/foreach

[trampolinejoe]

waynewex,

 

I found the problem,

I needed the ' around $connote.

 

It works now, thanks alot

$order = mysql_query("SELECT * FROM OrderProducts WHERE OrderId =" . "'$connote'" );

 

instead of

$order = mysql_query("SELECT * FROM OrderProducts WHERE OrderId =" . $connote );

 

Thank you so much for helping be solve my problem, next time i will know what to do.

 

Cheers,

Joe.