如何彈出顯示模式，單擊一個按鈕嵌套在循環中

Question

我想要在循環中的目標按鈕。點擊每個按鈕應該會彈出一個模式，其中包含該特定發票的唯一細節。但是當任何按鈕被點擊時，我的代碼似乎只獲得第一張發票。

this is the loop with the modal 
 

 

 

 
<?php 
 
require 'connect.php'; 
 
require 'header.php'; 
 
    
 

 
?> 
 

 

 
    <!-- modal --> 
 

 
    <div class="modal fade print" tabindex="-1" role="dialog" aria-labelledby="myLargeModalLabel"> 
 

 
    <div class="modal-dialog" role="document"> 
 
     <div class="modal-content row" id="print"> 
 
     <address> 
 
      <img src="small.png"><br/> 
 
      #32 Adelabu Street,Uwani,Enugu<br/> 
 
      Phone: 08045454554 
 
      </address> 
 
     <address> 
 
      Date: 
 
      </address> 
 
     <table class="table table-bordered"> 
 
      <thead> 
 
      <tr> 
 
       <th>Customer Name</th> 
 
       <th>Product name</th> 
 
       <th>Quantity</th> 
 
       <th>price</th> 
 
       <th>total</th> 
 
       <th>amount paid</th> 
 
       <th>balance</th> 
 
      </tr> 
 
      </thead> 
 
      <tbody class="details"> 
 
      <tr> 
 
       <td></td> 
 
       <td></td> 
 
       <td></td> 
 
       <td></td> 
 
       <td></td> 
 
       <td></td> 
 

 
       <td></td> 
 

 

 

 
      </tr> 
 

 
      </tbody> 
 

 
     </table> 
 
     </div> 
 
    </div> 
 
    </div> 
 

 
    <div id="printarea"> 
 
    <table class="table table-bordered"> 
 

 
     <tbody class="details"> 
 
     <?php 
 
          $get_invoice = "SELECT * FROM invoice ORDER BY id DESC"; 
 
          $invoice_query = mysqli_query($connect,$get_invoice); 
 
          $invoice_rows = mysqli_num_rows($invoice_query); 
 
          while($full_rows=mysqli_fetch_array($invoice_query)) {?> 
 
     <tr> 
 
      <!-- the php code inside the id bracket is supposed to generate a unique id for each loop--> 
 
      <td><input type="hidden" value="<?php echo $full_rows['identify']; ?>" id="but<?php echo $full_rows['name']; ?>"></td> 
 
      <td> 
 
      <?php echo $full_rows['date']; ?> 
 
      </td> 
 
      <td> 
 
      <?php echo $full_rows['name']; ?> 
 
      </td> 
 
      <td> 
 
      <?php echo $full_rows['goods_description']; ?> 
 
      </td> 
 
      <td> 
 
      <?php echo $full_rows['quantity']; ?> 
 
      </td> 
 
      <td> 
 
      <?php echo $full_rows['price']; ?> 
 
      </td> 
 
      <td> 
 
      <?php echo $full_rows['total']; ?> 
 
      </td> 
 
      <td> 
 
      <?php echo $full_rows['amount_paid']; ?> 
 
      </td> 
 

 
      <td> 
 
      <?php echo $full_rows['balance'] ?> 
 
      </td> 
 
      <td><input type="button" class="btn btn-success" value="print"></td> 
 
      <td><input type="button" value="delete" id="add" class="btn btn-danger"></td> 
 

 
      <!-- the unique id generated is passed inside the onclick function to target specific invoice in the modal--> 
 
      <td><button type="button" class="btn btn-warning" data-toggle="modaSS" data-target=".print" onclick="modal('but<?php echo $full_rows['name'];?>')" id="view">View</button></td> 
 

 

 
      <!--DISPLAYING MODAL WITH INVOICE DETAILS--> 
 
      <script type="text/javascript"> 
 
      function modal(invoice) { 
 
       var identify = $("[id^='but']").val(); 
 
       var dataString = 'identify=' + identify; 
 
       $.ajax({ 
 
       type: "POST", 
 
       url: "modal.php", 
 
       data: dataString, 
 
       //cache: false, 
 
       success: function(html) { 
 
        alert(html); 
 
       } 
 
       }); 
 
       return false; 
 
      } 
 
      </script> 
 
     </tr> 
 
     <?php } ?> 
 
     </tbody> 
 

 
    </table> 
 
    </div> 
 

 
    </div>

this is the this is where the ajax request is been proccessed 
 

 

 
<?php 
 

 
$identify = $_POST['identify']; 
 
//connecting to server and selecting database 
 
$connect = mysqli_connect('localhost','root','','sidney'); 
 

 
if (isset($_POST['identify'])) { 
 
    $select = "SELECT * FROM `invoice` WHERE identify='$identify'"; 
 
    $query = mysqli_query($connect,$select); 
 
    $get_number = mysqli_fetch_array($query,MYSQLI_ASSOC); 
 
    if ($get_number) { 
 
    echo $get_number['identify']; 
 
    return $get_number['identify']; 
 
    }else{ 
 
    echo "There was trouble locating the number"; 
 
    } 
 
} 
 
?>

Answer 1

試試這個我剛纔給你如何應該可以打開模型並將數據發送到打開的模型的想法：

//Your Button HTML 
<button type="button" class="your_next_button_class btn btn-warning" 
data-name="<?php echo $full_rows['name'];?>" 
data-toggle="modaSS" data-target=".print" id="view">View</button> 

$('body').on('click', '.your_next_button_class', function(e) { 

    var name = $(this).data('name'); 
    //As abouve example you can get any data using data arrribute 
    // Now You can open a model on the same page you don't need to send data 

    $.ajax({ 
     type: "POST", 
     url: "modal.php", 
     data: dataString, 
     //cache: false, 
     success: function(response) { 
     // Here in response you get the $get_number['identify']; data. 
      alert(response); 

     //Now To Send this response to the model 
     // If you are adding this value to hidden input then 
     $('.hidden_input_class').val(response); 

     // Or you for display this in as text 
     $('.your_class_where_want_to_display').text(response); 


      // THis is how you open the model 
      $('.your_model_class').modal('show'); 

     } 
     }); 
}); 

I a假設你的查詢的正確和有效的數據是成功的。