HELP...!!! Been trying to get this working --- am new to php

[kiss848]

Short background... relatively new to PHP and was given a project to integrate it with a MySQL db basically just to produce reports on inventory. I had to run a query and return a value from a column. This value was then in turn passed through a hyperlink to another page as 'id'.

When I am on the new page --- I can echo $id and the value that was passed is correct. However when I try to run a mysql query using the $id variable I get nothing back. The minute I code in any text I know is in the table I am querying it works.!!!.. I have been working on this for hours and have tried every variation I could think of??? Could someone please direct me to something I am missing...

Here is the code I am using:

<?php

echo $id;


printf ("<div align=center>
<table width=600 border=1>
<tr align=center bgcolor=#999999><td>Part #</td><td>Quantity</td><td>Jobber</td><td>Extended</td><td>Part #</td><td>Quantity</td><td>Jobber</td><td>Extended</td><td>Part #</td><td>Quantity</td><td>Jobber</td><td>Extended</td>");

$result = mysql_query("SELECT * FROM `STORE_TEST` WHERE `ASC`='$id' ORDER BY `PART_NO`");

?>

as I said the previous page that links to this passes the $id through a hyperlink that is generated from a return on a query. I have tried "", [], removing quotes.. etc around the variable in the query but it will not work. I have also tried mysql_real_escape_string() etc just in hopes it might work. Have I done something wrong... I really need help because once I figure this out the project can be completed.

[ak1dnar]

if you are passing the id like this

http://www.yourdomain.com/your_second_page.php?id=10

from your_second_page.php you can get it and assign to a variable.
[PHP]
$id = $_GET['id'] ;[/PHP]

pass this $id to query string.

[ak1dnar]

when you post next time read the posing guidelines carefully. as a example your title is not suited for this thread. as a result of this bad title a search engine could not locate this thread when some search around.

[kiss848]

when you post next time read the posing guidelines carefully. as a example your title is not suited for this thread. as a result of this bad title a search engine could not locate this thread when some search around.

Sorry about that --- been working at figuring this out for a long time and read through the guidelines quickly. I will post better next time.

Thanks for your response but I did try to do this several times today:



$asc =$_GET['id];

then pass it to a query like this:

$result = mysql_query("SELECT * FROM `STORE_TEST` WHERE `ASC`='$asc' ORDER BY `PART_NO`");

but again when I attempt to output I get nothing in return. The id is being passed to the page by this code:

$result = mysql_query("SELECT `ASC`,`MANU`,SUM(`SUMJOBBER`) as totals FROM `STORE_TEST` WHERE `ASC` | 1 GROUP BY `ASC`");
$numofrows = mysql_num_rows($result);


for($i = 0; $i < $numofrows; $i++)
{
$row = mysql_fetch_array($result);
if($i % 2)
{
printf ("<tr class=trwh>\n");
} else {
printf ("<tr class=trgr>\n");
}
printf ("<td width=50><a href=detailrep.php?id=".$row['ASC']."</a>$row[ASC]</td><td padding=10> ".$row['MANU']."</td><td width=200>$ ".$row['totals']."</td>");}
?>

Here is a link to the site, if anyone as any ideas please let me know:

http://testing.thehappycomputer.com/testing4.php

Basically I am trying to pass the text of the link on this page to another php page. At the top of the second page the text seems to be passed as a value because that is an echo statement right below the title. But when I use the $var in a query it returns nothing. If I hardcode text in it works fine?

[kiss848]

Thanks... I got it...

I assigned a text value to a variable to test and that variable worked in the query so I thought I better check how the variable was being passed. I changed some code on the first page and now the variable works that is passed to a query. I think some characters were being passed that weren't showing up on an echo statement.

I appreciate everyone's help.