PHP (Database Help)

daryllsheridan · 12-03-2007 06:00PM #1

Need some help with this php code, i have setup a system where when people register and login for my site (its just for a college project so no security issues here) that their username is shown at the top with some account options. Kinda like dailymotion or you tube do

Anyway a further option people can do is register as a seller, and if they do that there user id is entered into a seller table (hopefully somebody can folloow this)

What the code below is trying to do is check if the persons username has a corresponding entry in the sellers table (basically check are they registered as a seller - the SQL is correct cause i have that part tested) (if seller_id != 0) and if they have it displays the link to sell a book and if their not registered as a seller they get the option to register as a seller)

Code keeps throwing up errors and im not sure if the logic is correct either,
any1 have any pointers?

Cheers

<?php
include("db.php");

$check1 = mysql_query("SELECT seller_id FROM seller s,user u WHERE u.id = s.seller_id and u.username ='$username'")or die("error");

while($info1 = mysql_fetch_array( $check1))
{
if ($seller_id != 0($info1["seller_id"))
{
?>
<a href="sell.php">Sell a Book</a>

<?php
}

else {
?>
<a href="register_seller.php">Become a seller</a>

<?
}
}
?>

Regards
Daryll

Ziycon · 12-03-2007 06:23PM

Try changing the query line to this:

$check1 = mysql_query("SELECT seller_id FROM seller s,user u WHERE u.id = s.seller_id and u.username ='".$username."'")or die("error");

daryllsheridan · 12-03-2007 07:17PM

Thanks for the response

In this case i dont think thats the answer

Ive used that query in another part (but for a different purpose) and it works fine

I think its somewhere in the logic below that is the problem

What im looking for is

If the query returns a result then

<a href="sell.php">Sell a Book</a>

is displayed

and if the query returns a null set then

<a href="register_seller.php">Become a seller</a>

is returned

louie · 12-03-2007 07:49PM

is the error from the query made to the database or after?

try this:
[php]
<?php
include("db.php");

$check1 = mysql_query("SELECT seller_id FROM seller s,user u WHERE u.id = s.seller_id and u.username ='$username'")or die("error");

if ($info1 = mysql_fetch_array( $check1)){
echo "<a href=\"sell.php\">Sell a Book</a>";
}else{
echo "<a href=\"register_seller.php\">Become a seller</a>";
}
?>
[/php]

kdouglas · 12-03-2007 07:57PM

if ($seller_id != 0($info1["seller_id"))

the above line looks suspect??

id imagine you meant
if($seller_id != 0 && $seller_id == $info1["seller_id"])

im gonna guess your writing code in a plain text editor?
i recently started using the eclipse ide with the phpeclipse plugin, it's pretty good and will pick up on errors similar to above while you type

daryllsheridan · 12-03-2007 10:38PM

louie wrote:

is the error from the query made to the database or after?

try this:
[php]
<?php
include("db.php");

$check1 = mysql_query("SELECT seller_id FROM seller s,user u WHERE u.id = s.seller_id and u.username ='$username'")or die("error");

if ($info1 = mysql_fetch_array( $check1)){
echo "<a href=\"sell.php\">Sell a Book</a>";
}else{
echo "<a href=\"register_seller.php\">Become a seller</a>";
}
?>
[/php]

Genius

That worked perfect

Cheers!

PHP (Database Help)

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