Session issues with a PHP/MySQL search engine

Question

PHP/MySQL newbie here.

I've created a basic search engine that queries a MySQL table containing a number of values. However, the search form has MULTIPLE search buttons in order to limit a search based on a single value type (in this case, by region or magnitude, since we're working with earthquakes)

The initial search and displaying of results goes off without a hitch, but when I try to go to another page of the results using the pagination links, it can't pull up the next set of results, and when I click in my browser (I'm using Firefox, but I'll test this in other browsers), the session has expired, so I've managed to narrow the problem down to there (unless I'm barking up the completely wrong tree here)

What's going wrong here, and how do I fix it?

I feel like the answer is very simple, but I'm just not seeing it. Here's the code for the results page (it was hacked together from multiple tutorials I found on the Internet, but I do understand what's happening within the code), and I added the var_dump($_SESSION) commands to see what variables were being passed in the session and will be removed once this problem is fixed.

= $mag1) {
$query = "SELECT * FROM quake WHERE magnitude BETWEEN '$mag1' and '$mag2' LIMIT $start, $limit";
}
else{
$query = "SELECT * FROM quake WHERE magnitude BETWEEN '$mag2' and '$mag1' LIMIT $start, $limit";
}
}
else{
echo "No records found.";
exit;
}

$result = mysql_query($query) or die(mysql_error());;

/* Setup page vars for display. */
if ($page == 0) $page = 1;          //if no page var is given, default to 1.
$prev = $page - 1;                 //previous page is page - 1
$next = $page + 1;              //next page is page + 1
$lastpage = ceil($total_pages/$limit);      //lastpage is = total pages / items per page, rounded up.
$lpm1 = $lastpage - 1;                      //last page minus 1

/* 
    Now we apply our rules and draw the pagination object. 
    We're actually saving the code to a variable in case we want to draw it more than once.
*/
$pagination = "";
if($lastpage > 1)
{   
    $pagination .= "";
    //previous button
    if ($page > 1) 
        $pagination.= " previous";
    else
        $pagination.= " previous";  

    //pages 
    if ($lastpage < 7 + ($adjacents * 2))   //not enough pages to bother breaking it up
    {   
        for ($counter = 1; $counter <= $lastpage; $counter++)
        {
            if ($counter == $page)
                $pagination.= "$counter";
            else
                $pagination.= "$counter";                 
        }
    }
    elseif($lastpage > 5 + ($adjacents * 2))    //enough pages to hide some
    {
        //close to beginning; only hide later pages
        if($page < 1 + ($adjacents * 2))        
        {
            for ($counter = 1; $counter < 4 + ($adjacents * 2); $counter++)
            {
                if ($counter == $page)
                    $pagination.= "$counter";
                else
                    $pagination.= "$counter";                 
            }
            $pagination.= "...";
            $pagination.= "$lpm1";
            $pagination.= "$lastpage";       
        }
        //in middle; hide some front and some back
        elseif($lastpage - ($adjacents * 2) > $page && $page > ($adjacents * 2))
        {
            $pagination.= "1";
            $pagination.= "2";
            $pagination.= "...";
            for ($counter = $page - $adjacents; $counter <= $page + $adjacents; $counter++)
            {
                if ($counter == $page)
                    $pagination.= "$counter";
                else
                    $pagination.= "$counter";                 
            }
            $pagination.= "...";
            $pagination.= "$lpm1";
            $pagination.= "$lastpage";       
        }
        //close to end; only hide early pages
        else
        {
            $pagination.= "1";
            $pagination.= "2";
            $pagination.= "...";
            for ($counter = $lastpage - (2 + ($adjacents * 2)); $counter <= $lastpage; $counter++)
            {
                if ($counter == $page)
                    $pagination.= "$counter";
                else
                    $pagination.= "$counter";                 
            }
        }
    }

    //next button
    if ($page < $counter - 1) 
        $pagination.= "next ";
    else
        $pagination.= "next ";
    $pagination.= "
";       
}

?>



Recent Earthquakes


QUAKE SEARCH 


Search the latest quakes

";
// $row is array... foreach( .. ) puts every element
// of $row to $cell variable
foreach($row as $cell)
echo "";
echo "
";
?>
And here is the search form:




Recent Earthquakes


QUAKE SEARCH 


Search the latest quakes



Search by Magnitude:




  

Search by Region:



  



Source
EqID
Version
Date/Time
Latitude
Longitude
Magnitude
Depth
NST
Region

$cell

Marc B · Accepted Answer

What is the point of

if (mysql_real_escape_string($_POST['regbutton']) == submit||

You don't need to escape form data if you're not using it in an SQL operation - you're just going a comparison in PHP here, with no database in site for this particular line of code. As well, you're comparing your escaped form value against an undefined constant - note the lack of quotes around submit. PHP will politely treat that as an unquoted string, but will issue a warning - given you've not mentioned getting any warnings, you're probably working with display_errors off - meaning you're not going to see ANY problem reports from your code.

These errors are repeated throughout your code, so don't just fix this one line - fix the entire script.

Change that line to

if ($_POST['regbuttn'] == 'submit') || etc...

and then go read the PHP manual on how to enable display_errors. With this off, you're working in the dark and shooting yourself in the foot, repeatedly.

Plus... don't use the presence of a form field to determine if a POST has occured. It's unreliable. Use

if ($_SERVER['REQUEST_METHOD'] == 'POST') { ... }

instead, which is 100% reliable.

Session issues with a PHP/MySQL search engine

Answers (1)

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