error in SELECT statement using vars from $_GET array

[ashraf02]

can some one please help me i am getting this error but i cant seem to figure out wat is wrong on line 3 everything looks fine

[PHP]<?php
$conn = mysql_connect(" localhost","roo t","")
or die (mysql_error()) ;
mysql_select_db ("Shop_Test",$c onn) or die (mysql_error()) ;

$display_block = "<h1> My Store </h1>";

$get_item = "SELECT c.id as Cat_ID, c.cat_title, SI.Item_title,
SI.Item_Price, SI.item_desc, SI.item_Image FROM Store_Items AS SI LEFT JOIN
Store_Categorie s AS c on c.id =SI.Cat_ID WHERE SI.Store_ID = $_GET[Item_ID]";

$get_item_resul ts = mysql_query ($get_item) or die (mysql_error()) ;

if (mysql_num_rows ($get_item_resu lts)<1) {
$display_block .= "<p><em>Inv alid item selection.</em></p>";
} else {
$cat_id = mysql_result ($get_item_resu lts,0,'Cat_ID') ;
$cat_title = strtoupper (stripslashes (
mysql_result ($get_item_resu lts,0, 'cat_title')));
$item_title = stripslashes (mysql_result ($get_item_resu lts,0,'Item_tit le'));
$item_price = mysql_result ($get_item_resu lts,0,'Item_Pri ce');
$item_desc = stripslashes(my sql_result ($get_item_resu lts,0, 'item_Desc'));
$item_image = mysql_result ($get_item_resu lts,0,'item_Ima ge');

$display_block .= "<p><strong><em > Youre Viewing:</em><br><a href=\"catalogu e.php?Cat_ID=$C at_ID\">$cat_ti tle</a>
&gt; $item_title </strong></p>

<table cellpadding=3 cellspacing=3>
<tr>
<td valign=middle align=center><i mg src=\"$item_ima ge\"><td>
<td valign=middle>< p><strong>Descr iption:</strong><br>$ite m_desc</p>
<p><strong>Pric e:</strong> \$$item_price</p>";

$get_colours = "SELECT Item_Colour FROM Store_Item_Colo ur WHERE
Store_ID = $item_id ORDER BY Item_Colour";
$get_colour_res ults = mysql_result ($get_colours) or die (mysql_error()) ;

if (mysql_num_rows ($get_colour_re sults)>0) {
$display_block .= "<p><strong>Ava ilable Colours:</strong><br>";

while ($colours = mysql_fetch_arr ay($get_colour_ results)) {
$item_colour = $colours['Item_Colour'];
$display_block .= "item_colour<br >";
}
}

$get_sizes = "SELECT Item_size FROM store_item_size WHERE
Store_ID = $item_id ORDER BY Item_size";
$get_size_resul ts = mysql_query($ge t_sizes) or die (mysql_error()) ;

if (mysql_num_rows ($get_size_resu lts)>0) {
$display_block .= "<p><strong>Ava ilable Sizes:</strong></br>";

while ($sizes = mysql_fetch_arr ay ($get_size_resu lts)) {
$item_size = $sizes ['Item_size'];
$diplay_block .= "$item_size<br> ";
}
}

$display_block .= "
</td>
</tr>
</table>";
}
?>[/PHP]

Please help i have to do a project thanks in advance

[nathj]

Hi,

I've had a look over the code and I assume the first SQL is the one with the problem. Is that the case? If so then the following may be of help, if not then it's probably no good to you.

You have:
[php]
$get_item = "SELECT c.id as Cat_ID, c.cat_title, SI.Item_title,
SI.Item_Price, SI.item_desc, SI.item_Image FROM Store_Items AS SI LEFT JOIN
Store_Categorie s AS c on c.id =SI.Cat_ID WHERE SI.Store_ID = $_GET[Item_ID]";
[/php]

I have never managed to make the use of the $_GET array work directly in a SQL statement like that. So I would try:
[php]
$lnItemID = $_GET['Item_ID'] ;
$get_item = "SELECT c.id as Cat_ID, c.cat_title, SI.Item_title,
SI.Item_Price, SI.item_desc, SI.item_Image FROM Store_Items SI LEFT OUTER JOIN
Store_Categorie s c on c.id = SI.Cat_ID WHERE SI.Store_ID = $lnItemID";
[/php]
The other changes are just personal preferences but whenever I use that syntax it works first time.

The other potential issue I spotted was that you are not passing a link identifier (defined at the top of your code as $conn) to the mysql_query - this needs to be the second parameter.

I hope this helps you out.
nathj

[ronverdonk]

The other potential issue I spotted was that you are not passing a link identifier (defined at the top of your code as $conn) to the mysql_query - this needs to be the second parameter.

No entirely true. If the link identifier is not specified, the last link opened by mysql_connect() is assumed. See php documentation.

Ronald

[ashraf02]

Thanks Nathj for ur reply but i had no luck with the new code still coming up with the same error. any other suggestions everything seems fine ive been sifting through the code up and down but cant seem to find the error.

please help i need to complete this for my project.

[ronverdonk]

The following gets rid of the MySQL error in line 3:

$get_item = "SELECT c.id as Cat_ID, c.cat_title, SI.Item_title, SI.Item_Price, SI.item_desc, SI.item_Image 
FROM Store_Items AS SI 
LEFT JOIN Store_Categories AS c on c.id =SI.Cat_ID 
WHERE SI.Store_ID = '{$_GET['Item_ID']}'";

Ronald

[ashraf02]

Thanks once again ronald. uve been a great help

[ronverdonk]

You are welcome, until next time.

Ronald