Passing info from one file to another.

[abel122]

Hi all.

I have a php file that lists records from a table. Each record has a link next to it to /update.php?id=#. I have the update.php file but I can't get it to display the correct record. My question is:

How do I pass the id=# to a query in update.php? This is the beginning of the update.php file
<?
$dbcnx = @mysql_connect("**", "*", "");
if (!$dbcnx)
{
echo( "<p>Unable to connect to the " .

"database server at this time.</p>" );
exit();
}
if (! @mysql_select_db("*") )
{
echo( "<p>Unable to locate the " .

"database at this time.</p>" );

exit();
}
$result = @("SELECT FROM tbl** WHERE This is where I want to pass the id ");

if (!$result)

{

echo("<p>Error performing query: " . mysql_error() . "</p>");
exit();
}

Any response will be greatly appreciated.

[run2web]

$result = @("SELECT FROM tbl*** WHERE This is where I want to pass the id ");

Before your query try $id = 'your_id_#';

then in your query try

$result = @("SELECT FROM tbl*** WHERE $id ");

[gtibok]

$result = ("select ... where id = '$id' ");
...
<a href=update.php?id=$id>update</a>
*NOTE the $id variable must have a value in oreder for your script to work. For the firts select statement u could use some other field from the mysql table (like name or product ...)
than to set the $id:
$result = ("SELECT id, name, email, pet FROM friends where name ='monica' ");
$result_array = mysql_fetch_array("$result");
$id = $result_array['id'];