Update для таблицы

286

15 июня 2017, 05:21

Помогите сделать update для таблицы, delete реализован.

    <table class="table table-bordered table-hover table-striped" style="table-layout: fixed">
    <thead>
    <tr>
        <th>User Id</th>
        <th>User Name</th>
        <th>User E-mail</th>
        <th>User Pass</th>
        <th>Delete User</th>
    </tr>
    </thead>
    <?php
    include("database/db_conection.php");
    $view_users_query="select * from users";//select query for viewing users.
    $run=mysqli_query($dbcon,$view_users_query);//here run the sql query.
    while($row=mysqli_fetch_array($run))//while look to fetch the result and store in a array $row.
    {
        $user_id=$row[0];
        $user_name=$row[1];
        $user_pass=$row[2];
        $user_email=$row[3];
    ?>       
    <tr>
        <td><?php echo $user_id;  ?></td>
        <td><?php echo $user_name;  ?></td>
        <td><?php echo $user_email;  ?></td>
        <td><?php echo $user_pass;  ?></td>
        <td><a href="delete.php?del=<?php echo $user_id ?>"><button class="btn btn-danger">Delete</button></a></td> <!--btn btn-danger is a bootstrap button to show danger-->
    </tr>
    <?php } ?>
</table>
    </div>

Delete реализован так

include("database/db_conection.php");
$delete_id=$_GET['del'];
$delete_query="delete  from users WHERE id='$delete_id'";//delete query
$run=mysqli_query($dbcon,$delete_query);
if($run)
{
//javascript function to open in the same window
    echo "<script>window.open('view_users.php?deleted=user has been deleted','_self')</script>";
}

Answer 1

<form action="" method="post"> 
  <table class="table table-bordered table-hover table-striped" style="table-layout: fixed"> 
      <thead> 
 
      <tr> 
 
          <th>User Id</th> 
          <th>User Name</th> 
          <th>User E-mail</th> 
          <th>User Pass</th> 
          <th>Delete User</th> 
      </tr> 
      </thead> 
 
      <?php 
      include("database/db_conection.php"); 
      $view_users_query="select * from users";//select query for viewing users. 
      $run=mysqli_query($dbcon,$view_users_query);//here run the sql query. 
 
      while($row=mysqli_fetch_array($run))//while look to fetch the result and store in a array $row. 
      { 
          $user_id=$row[0]; 
          $user_name=$row[1]; 
          $user_pass=$row[2]; 
          $user_email=$row[3]; 
      ?>        
      <tr> 
 
          <td><input type="text" name="user_id" value="<?php echo $user_id; ?>"></td> 
          <td><input type="text" name="user_name" value="<?php echo $user_name; ?>"></td> 
          <td><input type="text" name="user_email" value="<?php echo $user_email; ?>"></td> 
          <td><input type="text" name="user_pass" value="<?php echo $user_pass; ?>"></td> 
          <td><a href="delete.php?del=<?php echo $user_id ?>"><button class="btn btn- danger">Delete</button></a></td> <!--btn btn-danger is a bootstrap button to show danger--> 
          <td><button type="submit" name="update" class="btn">Update</button></td> 
      </tr> 
 
      <?php } ?> 
 
  </table> 
</form> 
 
<?// следующий код можно оформить как функцию 
  // если нажали на update 
  if (isset($_POST['update']) { 
    $user_id = mysql_real_escape_string($_POST['user_id']); 
    $user_name = mysql_real_escape_string($_POST['user_name']); 
    $user_email = mysql_real_escape_string($_POST['user_email']); 
    $user_pass = mysql_real_escape_string($_POST['user_pass']); 
    $update_query = "UPDATE `users` SET     `user_id`='$user_id',`user_name`='$user_name',`user_email`='$user_email' 
            ,`user_pass`='$user_pass' WHERE `id`='$user_id'"; 
     $run=mysqli_query($dbcon,$update_query); 
  } 
?>

В кратце: form'a и input'ы для отправки через post(такие вещи как email, !пароль! - лучше отправлять скрыто); не уверен, что Вам нужно updatить id( дело Ваше)