php mysql form variable works one time

[baccaril]

Hello,

I have a form the post to a php script, the variable is passed to the
php script successfully. I then query MySQL and output 8 rows at a
time. So the problem I'm having is the first 8 rows are displayed on
the web page correctly, but now when I attempt to display rows 9-16
the variable that was posted from the form is now blank. How do I
ensure the variable does not disappear?

LouB.

=================================
Here are the three main part of my php script.
# post from form....
$sortby = $_REQUEST["sortby"];

$query = "select personal_identifier, first_name, last_name,
company_name, photo_path, custom_field_1 from $dbname order by $sortby
limit $offset, $rowsPerPage";

result = mysql_query ($query)
//or die ("Cannot execute query");
or die (mysql_error());

[BKDotCom]

ba******@gmail.com wrote:

Hello,

I have a form the post to a php script, the variable is passed to the
php script successfully. I then query MySQL and output 8 rows at a
time. So the problem I'm having is the first 8 rows are displayed on
the web page correctly, but now when I attempt to display rows 9-16
the variable that was posted from the form is now blank. How do I
ensure the variable does not disappear?

LouB.

something tells me you have a "click here to see rows 9-16" type link?
are you passing you var in the link?

[baccaril]

Yes I have links printed at the end of the first page for viewing
additional rows/page.
I did not believe I need to pass the vaiable. If needed how do I do
that?

Lou

Here is the code for links at end og page.
================================================== ==========
#
# print number of pages available at bottom of the web page

// how many rows we have in database
$query = "SELECT COUNT(*) AS numrows FROM $dbname";
$result = mysql_query($query) or die('Error, query failed');
$row = mysql_fetch_array($result, MYSQL_ASSOC);

$numrows = $row['numrows'];

// how many pages we have when using paging?
$maxPage = ceil($numrows/$rowsPerPage);

// print the link to access each page
$self = $_SERVER['PHP_SELF'];
$nav = '';

for($page = 1; $page <= $maxPage; $page++)
{
if ($page == $pageNum)
{
$nav .= " $page "; // no need to create a link to current page
}
else
{
$nav .= " <a href=\"$self?page=$page\">$page</a";
}
}

print "$nav";

================================================== ===

ba******@gmail.com wrote:

Hello,

I have a form the post to a php script, the variable is passed to the
php script successfully. I then query MySQL and output 8 rows at a
time. So the problem I'm having is the first 8 rows are displayed on
the web page correctly, but now when I attempt to display rows 9-16
the variable that was posted from the form is now blank. How do I
ensure the variable does not disappear?

LouB.

=================================
Here are the three main part of my php script.
# post from form....
$sortby = $_REQUEST["sortby"];

$query = "select personal_identifier, first_name, last_name,
company_name, photo_path, custom_field_1 from $dbname order by $sortby
limit $offset, $rowsPerPage";

result = mysql_query ($query)
//or die ("Cannot execute query");
or die (mysql_error());

[baccaril]

Thanks for the help. It was a issue where I was not passing the
secound var in the link. I corrected the problem with the following:

$nav .= " <a href=\"$self?page=$page&sortby1=$sortby\">$page</a";

BKDotCom wrote:

ba******@gmail.com wrote:

Hello,

I have a form the post to a php script, the variable is passed to the
php script successfully. I then query MySQL and output 8 rows at a
time. So the problem I'm having is the first 8 rows are displayed on
the web page correctly, but now when I attempt to display rows 9-16
the variable that was posted from the form is now blank. How do I
ensure the variable does not disappear?

LouB.

something tells me you have a "click here to see rows 9-16" type link?
are you passing you var in the link?

[SquishedOrange]

Thats very insecure script.

Have a look at: mysql_real_escape(), is_number(), isset()