CODING HELP :(.. pls'

[jushiro]

Im having trouble making a code for a feedback system.

Can anyone help me to make a code like this.

 

I have a table w/c contains

--------------------------------------------

| Itemname | Itemprice | Seller |

 

Laptop | 15000 | Name

Phone | 5000 | Name

-----------------------------------

 

Then.. i want a code that will query those values.. And have a button :

[bUY] and after the values.

 

Then after clicking POST FEEDBACK it will show the values of the SELLER name in another page.

 

HELP ME PLS

[Muddy_Funster]

your database is totaly insuficcient for the task you want to perform. You will need multiple tables and relational fields.

[jushiro]

What do you mean?

But what i want is to query just those 3 values from my database and have button for that.

 

 

[jushiro]

I dont know how to say it but. To simplify this :

 

I want to know how can i put values on the button?

 

what i did is make a array to output those values and have buttons. here.

 

if($result){
while($row = mysql_fetch_assoc($result)) 
    { 
    $itemname = $row['itemname']; 
$itemprice= $row['itemprice']; 
$owner =$row['owner']; 

$items[] = array($itemname,$itemprice,$owner);
    } 
}



echo("<form action=\"buy.php\" method=\"post\">\n");
for ($i=0;$i<count($items);$i++) {
	echo("".$items[$i][0]."<br>\n");
	echo("".$items[$i][1]."<br>\n");
	echo("".$items[$i][2]."<br>\n");

	echo "<input type=\"submit\" value=\"BUY\" name =\"buy\">";
	echo "<input type=\"submit\" value=\"POST FEEDBACK\" name=\"name\"></br><br></form>";

}

 

 

I succeeded in making those queries and button. but my problem is'

The button "POST FEEDBACK" i want it to have the value of $items[$i][2]..

So that when i turn to the next page i can echo the value of that POST FEEDBACK.

[scootstah]

He means that there is no way to reference that data from other tables. You'll need to use ID's and such so that you will be able to tell your feedback posts which item they belong to.