onChange Update php id auswählen und dann Abfrage mysql

Question

ich mit einem Informations modal/Schaltfläche, um eine ausgewählte, dass ich eine Verbindung zu dinamically wollen, wenn das Auswahl geändert wird

Mein Code wie dieser

<select name="name_select" id="my_select"> 
    <option value="1">Rossi</option> 
    <option value="2">Skunk</option> 
    <option value="3">Ceres</option> 
</select> 

<button type="button" class="btn btn-primary btn-lg" data-toggle="modal" data-target="#myModal"> Information </button> 

<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel"> 
    <div class="modal-dialog" role="document"> 
    <div class="modal-content"> 
     <div class="modal-header"> 
     <button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span></button> 
     <h4 class="modal-title" id="myModalLabel">Personal information</h4> 
     </div> 
     <div class="modal-body"> 

// query mysql 
<?php (...) SELECT name, address, phone FROM table WHERE id = #my_select ?> 

<input type="text" name="name" value="<?php echo $name ?>"> 
<input type="text" name="address" value="<?php echo $adress ?>"> 
<input type="text" name="phone" value="<?php echo $phone ?>"> 

     </div> 
     <div class="modal-footer"> 
     <button type="button" class="btn btn-default" data-dismiss="modal">Close</button> 
     <button type="button" class="btn btn-primary">Save changes</button> 
     </div> 
    </div> 
    </div> 
</div> 

Wie zu Implementieren ein jquery post/bekommen, um das zu lösen?

Answer 1

Versuchen mit Ajax

Das Script für diese Seite wäre:

$(function() { 
    $(document).on("change", "#my_select", function() { 
     $.ajax({ 
      url: 'path/to/ajaxfile.php', 
      type: 'GET', 
      dataType: 'json', 
      data: {id: $(this).val()}, 
     }) 
     .done(function(response) { 
      if (response.status) { 
       $("#myModal").find(".modal-body").html(response.html); 
      } else { 
       alert(status.message); 
      } 
     }) 
     .fail(function(data) { 
      alert("Something went wrong please try again later."); 
      console.log(data.responseText); 
     }); 

    }); 
}); 

And The path/to/ajaxfile.php Code:

/** 
* Connect to the database and do other initialization if required. 
* 
* Assuming Procedual MySQLi. 
*/ 

if (empty($_GET['id'])) { 
    echo json_encode(array("status" => false, "message" => "There is no User Selected.")); 
    exit; 
} 

$id = mysqli_real_escape_string($connection, $_GET['id']); 

$query = "SELECT name, address, phone FROM table WHERE id = '{$id}'"; 
$result = mysqli_query($connection, $query); 

if (mysqli_num_rows($result) > 0) { 
    $row = mysqli_fetch_assoc($result); 
    $inputs = '<input type="text" name="name" value="'. $row['name'] .'"> 
     <input type="text" name="address" value="'. $row['address'] .'"> 
     <input type="text" name="phone" value="'. $row['phone'] .'">'; 

    echo json_encode(array("status" => true, "html" => $inputs)); 
    exit; 
} else { 
    echo json_encode(array("status" => false, "message" => "There is no user with that id.")); 
    exit; 
} 

Ich hoffe, das hilft.