PHP How to show fetched results from the DB in console log

user_9771157 0
Viewed 24

I'm currently using the following code to fetch some data from my database table and show it into a table. It's using PDO for database connection.

<tbody>
  <?php $sql = "SELECT * from users 
        ORDER BY id DESC";
            $query = $dbh->prepare($sql);
            $query->execute();
            $results = $query->fetchAll(PDO::FETCH_OBJ);
            $cnt = 1;
            if ($query->rowCount() > 0) {
                foreach ($results as $result) {

            
                    ?>
  <form action="manage-users.php" method="post">
    <input type="hidden" name="uid" value="<?= $result->id ?>" />
    <?php
                        $status = 1;
                        if ($result->status == 1) {
                            $status = 0;
                        }
                        ?>
      <input type="hidden" name="status" value="<?= $status ?>" />
      <tr>

        <td>
          <?php echo htmlspecialchars($cnt); ?>
        </td>
        <td>
          <?php echo htmlspecialchars($result->full_name); ?>
        </td>
        

      </tr>
  </form>
  <?php $cnt = $cnt + 1;


                }
            } ?>
</tbody>

But I wanted to console log the fetched results, so I modified my code as below.

<tbody>
  <?php $sql = "SELECT * from users 
        ORDER BY id DESC";
            $query = $dbh->prepare($sql);
            $query->execute();
            $results = $query->fetchAll(PDO::FETCH_OBJ);
            $cnt = 1;
            if ($query->rowCount() > 0) {
                foreach ($results as $result) {

                    echo '<script type="text/javascript">' .
                        'console.log(' . $results . ');</script>';

                    ?>
  <form action="manage-users.php" method="post">
    <input type="hidden" name="uid" value="<?= $result->id ?>" />
    <?php
                        $status = 1;
                        if ($result->status == 1) {
                            $status = 0;
                        }
                        ?>
      <input type="hidden" name="status" value="<?= $status ?>" />
      <tr>

        <td>
          <?php echo htmlspecialchars($cnt); ?>
        </td>
        <td>
          <?php echo htmlspecialchars($result->full_name); ?>
        </td>


      </tr>
  </form>
  <?php $cnt = $cnt + 1;


                }
            } ?>
</tbody>

But the issue is it's not showing any data or error in the console. Any help would be much appreciated.

php
1 Answers

You can use json_encode() to convert the PHP array into a JavaScript literal.

  <?php 
    $sql = "SELECT * from users 
        ORDER BY id DESC";
    $query = $dbh->prepare($sql);
    $query->execute();
    $results = $query->fetchAll(PDO::FETCH_OBJ);
    $results_js = json_encode($results);
    ?>
    <script>
    console.log(<?php echo results_js; ?>);
    </script>
    <?php
user_14918950
Related