Populating html table with database data through user selection in php - php

Question

So, I have a checkbox that allows the user to select the columns they want to view. So, if the user wants for example to select only 2 columns (say, TicketID and Category) it will only show the data of the ID and the category of the ticket. Here's what I have so far:
form:
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<ul>
<li><input type="checkbox" name="filter[]" value="TicketID" >Ticket ID</li>
<li><input type="checkbox" name="filter[]" value="Category" />Category</li>
<li><input type="checkbox" name="filter[]" value="Priority" />Priority</li>
<li><input type="checkbox" name="filter[]" value="Status" />Status</li>
<li><input type="checkbox" name="filter[]" value="InitialDescription" />Initial Description</li>
<li><input type="checkbox" name="filter[]" value="SubmittedDate" />Submitted Date</li>
<li><input type="checkbox" name="filter[]" value="Description" />Status Description</li>
<input type="submit" value="Refresh Filters">
</ul>
</form>
</td>
<td>
<?php
viewTicketTable($_SESSION['userID'],$_POST['filter']);
?>
viewticketTable function:
function viewTicketTable($userID,$columns) {
/* Accepts $userID which will identify the tickets related with the user and
$columns which will filter only the columns that are required and outputs the
table with the columns passed through. */
/*
$columns_array = explode(',', $columns);
foreach($array as $array){
echo $array;
} */
foreach($columns as $filter) {
$filter = $filter . ',' ;
}
$filter = substr($filter, 0, -1);
$query = mysql_query("
SELECT $filter
FROM Ticket
LEFT JOIN TicketHistory
ON Ticket.TicketID = TicketHistory.TicketID
WHERE CustomerID = $userID;
");
/* Creation of the table */
echo '
<table border="1">
<thead>
<tr>';
foreach($columns as $tableHeader) {
$tableHeader = '<th scope="col">' . $tableHeader . '</th>' ;
echo $tableHeader;
}
echo '
</tr>
</thead>
<tbody>
';
/*Looping through the script to print out all the information.*/
while ($row = mysql_fetch_array($query) or die(mysql_error())) {
echo '
<tr>';
foreach($columns as $field) {
echo '<td>' . $row[$field] . '</td>';
}
echo '</tr>
';
}
echo ' </tbody>
</table>';
}
The problem resides in the while loop. row[$field] is only populating the last column instead of populating all of them. Any help?

Answer 1

You have an error in your SQL syntax - you have an extra ; before the " closes
Also, you should use mysql_fetch_assoc in place of mysql_fetch_array as using mysql_fetch_array without specifying a method is going to return a double array.

Answer 2

I fixed it. I'll post the answer in case someone has the same issue.
The problem was fixed here:
$filter = "";
foreach($columns as $c) {
$filter = $filter . $c . ',' ;
}
$filter = substr($filter, 0, -1);
The loop was not assigning the values to the $filter variable so I had to create an external variable. There was also anther error in the foreach loop that I had to fix with an if statement (not the best way of doin it, but it works):
foreach($columns as $field) {
if ($field == 'Ticket.TicketID'){
$field = 'TicketID';
}
echo '<td>' . $row[$field] . '</td>';
}
Ticket.TicketID wasn't accepted in the $row array, so I had to change it to TicketID
Thanks everyone for the help anyway.