what cause the error , when i'm trying to retreive data from database using PHP-CodePudding

I'm trying to get data from my database, I use XAMPP local server it doesn't display student table into html table it displays: '; } ?> and the empty table database name: test, User name: root, no password, table name: student any response will be appreciated here is my code:

<!DOCTYPE html>
<html>

<style>
    td,th {
        border: 1px solid black;
        padding: 10px;
        margin: 5px;
        text-align: center;
    }
</style>
 
 <?php
  $mysqli = new mysqli("localhost","root","","test")
           or die('Error connecting to MySQL server.');;
  $query = "SELECT * FROM student" ;
  $result = mysqli_query("test", $query);
  $row = mysqli_fetch_array($result);

  while ($row = mysqli_fetch_array($result)) 
    {
        echo $row['root'] . ' ' . $row[''] . '<br />';
    }
 
  ?> 
   
<body>
    <table>
        <thead>
            <tr>
                <th>Name</th>
                <th>Branch</th>
                <th>Roll Number</th>
            </tr>
        </thead>
        <tbody>
            <?php
               if(!empty($row))
               foreach($row as $rows)
              { 
            ?>
            <tr>
  
                <td><?php echo $rows['name']; ?></td>
                <td><?php echo $rows['branch']; ?></td>
                <td><?php echo $rows['roll_no']; ?></td>
  
            </tr>
           
        </tbody>
    </table>
</body>


</html>

CodePudding user response：

The foreach loop has no ending. You need to add a closure after the row ends:

    </tr>
    <?php
    }
    ?>
</tbody>

Also, you have two simicolons on line 1, but that shouldn't change anything.

Edit:

Take a look at this example file:

<div>
    <?php
    $arr = array(1, 2, 3, 4);
    
    foreach ($arr as $value) 
    {
    ?>
        <div><?php print($value * 2); ?></div>
    <?php
    }
    ?>
</div>

This correctly outputs:

<div>
    <div>2</div>
    <div>4</div>
    <div>6</div>
    <div>8</div>
</div>

But without the closing <?php } ?> for each number, there is a 500 error.

CodePudding user response：

This is how you should iterate the query result.

<!DOCTYPE html>
<html>

<style>
  td,
  th {
    border: 1px solid black;
    padding: 10px;
    margin: 5px;
    text-align: center;
  }
</style>

<body>
  <?php
  $mysqli = new mysqli("localhost","root","","test")
           or die('Error connecting to MySQL server.');;
  $query = "SELECT * FROM student" ;
  $result = mysqli_query("test", $query);

  ?>

    <table>
      <thead>
        <tr>
          <th>Name</th>
          <th>Branch</th>
          <th>Roll Number</th>
        </tr>
      </thead>
      <tbody>
          <?php
          while ($row = mysqli_fetch_array($result))  {
          ?>
          <tr>

            <td>
              <?php echo $row['name']; ?>
            </td>
            <td>
              <?php echo $row['branch']; ?>
            </td>
            <td>
              <?php echo $row['roll_no']; ?>
            </td>

          </tr>
          <?php
          }
          ?>
      </tbody>
    </table>

</body>

</html>

CodePudding user response：

Let's take a look at the first section


 <?php
  $mysqli = new mysqli("localhost","root","","test")
           or die('Error connecting to MySQL server.');;
  $query = "SELECT * FROM student" ;
  $result = mysqli_query("test", $query);
  $row = mysqli_fetch_array($result);

  while ($row = mysqli_fetch_array($result)) 
    {
        echo $row['root'] . ' ' . $row[''] . '<br />';
    }
 
  ?>

The function for mysqli_query takes in a MySQLI resource as the first parameter and the query as the second. It looks like you've got "test" instead of $mysqli

When you're calling $row = mysqli_fetch_array($result); right below, that's using the response and advancing the internal pointer to the next row, so the next time you call mysqli_fetch_array, it'll return back with the second row! After the while-loop, you've now gone through the entire result set, and $row will only be false (or null).

Finally, there is the case of a missing closing } for your foreach-loop.

Based on those, this is one possible way of showing the data, assuming 'name', 'branch' and 'roll_no' are columns on the student table.

<?php
$mysqli = new mysqli("localhost", "root", "", "test")
            or die('Error connecting to MySQL server.');

$query = "SELECT * FROM student";
$result = mysqli_query($mysqli, $query);
?>
<!DOCTYPE html>
<html>

<style>
    td,th {
        border: 1px solid black;
        padding: 10px;
        margin: 5px;
        text-align: center;
    }
</style>
<body>
    <table>
        <thead>
            <tr>
                <th>Name</th>
                <th>Branch</th>
                <th>Roll Number</th>
            </tr>
        </thead>
        <tbody>
            <?php while ($row = mysqli_fetch_array($result)): ?>
            <tr>
                <td><?= htmlspecialchars($row['name']) ?></td>
                <td><?= htmlspecialchars($row['branch']) ?></td>
                <td><?= htmlspecialchars($row['roll_no']) ?></td>
            </tr>
            <?php endwhile ?>
        </tbody>
    </table>
</body>
</html>