Member Avatar for oamunoz

Hello!

I've got the code below which takes data from my db and displays it to the page listing reports submitted by the user. At the end of each row I have button that is supposed to launch a PDF generator so that the user can view their report. I have multiple rows, yet only the bottom row button will work correctly. If I click the submit button on any of the rows prior to the last row, the data for the last report will be displayed. I'd appreciate any help with this. Thanks!

<form name="vin_report_table" action="reportPDF.php" method="post">
        <table>
<?php
    $color = "#cccccc";
    echo "<tr bgcolor=\"" . $color . "\"><td><strong>VIN</strong></td>";
    echo "<td><strong>Year </strong></td>";
    echo "<td><strong>Model </strong></td>";
    echo "<td><strong>Date </strong></td><td></td></tr>";
    echo "<br>";
    
    if(mysql_num_rows($result)){
    while($row = mysql_fetch_assoc($result)){
            
            if ($color == "#cccccc") {
                $color = "#ffffff";
            } else {
                $color = "#cccccc";
            }
            
            echo "<tr bgcolor=\"" . $color . "\"><td>".$row['vin_full']. "</td>
          <td>" . $row['vin_model'] . "</a></td>
              <td>" . $row['vin_year'] . "</a></td>
                  <td>". $row['vreport_timestamp'] . "</a></td>
                      <td><input type=\"hidden\" name=\"u\" value=\"".$username."\"/>
                          <input type=\"hidden\" name=\"rid\" value=\"".$row['report_id']."\"/>
                          </><input type=\"submit\" value=\"View Report\" /></td></tr>\n";
            


        }
    } else {
        echo "<tr><td>No reports on record</td></tr>";
    }
?>
        </table>
    </form>

Thanks!

Edited by oamunoz because: clarified statement
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Dani AI

Generated

Short diagnosis and recap: the form in the original post wraps every row, so multiple controls share the same names and the POST data clashes — PHP will end up with a single value for each non-array name (which is why the last row appeared). and pointed to separate forms or distinct names, and confirmed turning each row into its own form fixed the immediate issue.

A tidy alternative that keeps one form and avoids duplicate field names is to give each row its own submit button whose name/value carry the report id. Only the clicked submit button’s name/value is sent, so the server can detect which report to show:

<!-- per table row -->
<td>
  <button type="submit" name="view_report" value="<?php echo (int)$row['report_id']; ?>">View</button>
</td>

<!-- in reportPDF.php -->
if (isset($_POST['view_report'])) {
    $rid = (int) $_POST['view_report'];
    // validate ownership, then generate/stream the PDF
}

Other workable approaches: keep a single submit and use JS to set a hidden rid before submitting, or keep one small <form> inside the table cell for each row (the approach that solved this thread). Note on markup: wrapping <tr> with <form> can be invalid under some doctypes — place the form inside a <td> or use HTML5’s form="id" attribute on buttons for valid markup.

Security and maintainability reminders: move away from deprecated mysql_* calls to PDO or mysqli with prepared statements; always validate that the logged-in user owns the requested report id before generating the PDF; cast inputs to integers and escape any echoed DB content with htmlspecialchars; protect sensitive actions with CSRF checks.

Recommended Answers

All 3 Replies

Member Avatar for aquilax

Either use new form for each row, or change the field names for each row.

Member Avatar for broj1

Your form has form elements with the same name in each row so $_POST array contains only the last set of values I guess. Try to enclose form elements that belong to one row in <form></form> tags (so each form has only one submit button).

Member Avatar for oamunoz

Great tips! I went and made each row into a Form and that worked perfectly. Thanks for the help!

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