Creating multiple form elements in html table to update SQL table

Question

I'm using a SELECT query to generate a html table (screenshot) which displays data from a sql table.

Every <tr> is generating an additional <tr> for the administrator (if ($userUid == "admin") {) with the sql entry's ID + <textarea> to edit the content in the SQL column public.

The second administrator <tr> includes a <form> element to edit the content. I assume the repeating name elements of the multiple <input>s / <button>s in the form element cause the problem, but the script action="includes/update.inc.php" just returns a blank page with the following URL and no php error:

http://localhost:10006/includes/update.inc.php?editId=18&editContent=adsasd&submit-edit=

Generated table:

<?php
<table> ... / table head...

foreach($result as $row) {

echo "<tr class='data-row'>";
  echo "<td>".$date_form."</td>
  echo "<td>".$time_form."</td>
  echo "<td>".$subject_form."</td>
  echo "<td>".$private."</td>
  echo "<td>".$public."</td>
  echo "<td>".$date_form."</td>
echo "</tr>";

if ($userUid == "admin") { ?>

<tr class='admin-row'>
  <form action="includes/update.inc.php" type="post">
    <td><input name="editId" type="text" value="<?php echo $Id; ?>"></td>
    <td class='p-2' colspan="4">
      <textarea name="editContent" class="content-update-textarea" type="text"></textarea>
    </td>
    <td><button name="submit-edit" type="submit">EDIT</button></td>
  </form>
</tr>

 <?php }}?>
</table>

This is the /includes/update.inc.php script called by the form element which is supposed to update the content using <textarea name="editContent"> of the SQL row with the ID: <input name='editId'>.

<?php
  include 'dbh.inc.php';

if (isset($_POST['submit-edit'])) {

  $id = $_POST['editId'];
  $editContent = $_POST['editContent'];


  $sql = "UPDATE lesson SET public = ? WHERE id = ?";
  $stmt = mysqli_stmt_init($conn);
  if (!mysqli_stmt_prepare($stmt, $sql)) {
    echo "Failed to prepare statement.";
    exit();
  } else {
    mysqli_stmt_bind_param($stmt, "ss", $editContent, $id);
    mysqli_stmt_execute($stmt);
    header ("Location: ../lesson.php?update=success");
  }
}

Askme23 · Accepted Answer · 2020-06-07 05:07:39Z

1

I think you need to change form's attribute "type" to "method". Your code after: <form action="includes/update.inc.php" method="post">

answered Jun 7, 2020 at 5:07

Askme23

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dspillai · Accepted Answer · 2020-06-07 04:53:40Z

0

First, you make sure that your value is correctly transported into the backend section.

<?php
  include 'dbh.inc.php';

if (isset($_POST['submit-edit'])) {

  $id = $_POST['editId'];
  $editContent = $_POST['editContent']; echo $editContent; //Verification step

  /*$sql = "UPDATE lesson SET public = ? WHERE id = ?";
  $stmt = mysqli_stmt_init($conn);
  if (!mysqli_stmt_prepare($stmt, $sql)) {
    echo "Failed to prepare statement.";
    exit();
  } else {
    mysqli_stmt_bind_param($stmt, "ss", $editContent, $id);
    mysqli_stmt_execute($stmt);
    header ("Location: ../lesson.php?update=success");
  }*/
}

answered Jun 7, 2020 at 4:53

dspillai

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