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Home/ Questions/Q 5984581
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T22:23:53+00:00 2026-05-22T22:23:53+00:00

Ok, so i am working on some software where users can submit tickets for

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Ok, so i am working on some software where users can submit tickets for bugs found on a website. I am using multiple checkboxes so users can check all browsers that are affected by the bug. The bug bugId, title, type, etc are stored in one table called bugs and the affected browsers are stored in another table called affectedbrowsers. The commonality between the two tables is the bugId. I have the form submitting everything correctly.

My problem is returning the data. I have an html table that has one row for each bug in the bugs database table. There is one column called “Affected Browsers” that i would like to populate with the data from the affectedbrowsers table. I tried using a while loop to loop through the bugs and echo out the rows in the html table and using a second while loop within that first while loop that would query the affectedbrowsers table and find all the records that have the same bugId. This isn’t returning any data in the second while loop. I would like to use JOINS if possible but i am not really familiar with them. What are your thoughts?

My Code:

<?php
    echo "<table>";
    $resultBug = mysql_query("SELECT * FROM bugs WHERE projectId = '$projectId' ORDER BY bugId ASC");
    echo "<tr> <th>Case Title</th> <th>Affected Browsers</th> </tr>";
    while($rowBug = mysql_fetch_array( $resultBug )){
        $bugId = $_POST['bugId'];
        echo "<tr><td>";
        $rowBugTitle = htmlspecialchars($rowBug['title']);
        echo $rowBugTitle;
        echo "</td><td>";
        $resultAffectedBrowsers = mysql_query("SELECT * FROM affectedbrowsers WHERE bugId = '$bugId' ORDER BY id ASC");
        while($rowAffectedBrowsers = mysql_fetch_array( $resultAffectedBrowsers )){
            $affectedBrowsers = $rowAffectedBrowsers['label'];
            echo $affectedBrowsers . " - ";
        }
        echo "</td></tr>"; 
    }
    echo "</table></div>";
?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T22:23:54+00:00Added an answer on May 22, 2026 at 10:23 pm

    You want to get the browsers for the current bug, so you should have

    $bugID = $rowBug['bugId'];
    

    instead of

    $bugID = $_POST['bugId'];
    
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