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问将id从MySQL传递给modal并通过Ajax传递
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Stack Overflow用户

提问于 2021-05-24 17:35:25

回答 1查看 35关注 0票数 1

我从数据库获取id并将其传递给modal，然后通过ajax在另一个页面上回显它，但问题是它总是返回相同的id。

        <?php if(mysqli_num_rows($select_invoice_query)>0) {
        while($invoice_result = mysqli_fetch_assoc($select_invoice_query)) { ?>
        <tr>
        <td><a href="#send-invoice-modal" id="send_invoice" data-toggle="modal" data-invoiceid="<?php echo $invoice_result['order_id']; ?>" data-placement="top" title="Send"><i class="fa fa-envelope"></i></a></td>
        </tr>
        <?php  }} ?>

引导模式

         <div id="send-invoice-modal" class="modal fade" role="dialog">
         <div class="modal-dialog">
         <div class="modal-content">
         <div class="modal-header">
         <h4 class="modal-title">Send Invoice</h4>
         </div>
         <div class="modal-body">
         <div class="form-group">
         <input type="email" id="invoice_to" class="form-control" placeholder="Email"> 
         </div>
         <div class="form-group">
         <input type="text" id="invoice_subject" class="form-control" placeholder="Subject">  
         </div>
         <div class="form-group">
         <textarea class="form-control" id="invoice_message" rows="3" placeholder="Message"></textarea>  
         </div>
         </div>
         <div class="modal-footer">
         <button type="button" class="btn btn-secondary" data-dismiss="modal">Close</button>
         <button type="button" class="btn btn-primary" data-dismiss="modal" id="send_invoice_btn">Send</button>
         <div class="error_message"></div>
         </div>
         </div>
         </div>
         </div>

   $('#send_invoice_btn').on('click',function(){
      var invoice_id = $('#send_invoice').data('invoiceid'); // this is collecting the id from the anchor tag 
      var invoice_to = $('#invoice_to').val(); 
      var invoice_subject = $('#invoice_subject').val(); 
      var invoice_message = $('#invoice_message').val(); 
   
      $.ajax({
         url: "includes/send-invoice.inc.php"+invoice_id,
         type: 'POST',
         data: {invoice_id:invoice_id,invoice_to:invoice_to,invoice_subject:invoice_subject,invoice_message:invoice_message},
         success: function(send_invoice_result){
             $('.error_message').html(send_invoice_result);
         }
      })    
   });

send-invoice.inc.php文件

<?php 


include 'db-connection.php';

echo $invoice_id = $_GET['invoice_id'];


?>

由于某种未知的原因，它一直返回相同的id。谁能告诉我我做错了什么？

php

mysql

回答 1

Stack Overflow用户

发布于 2021-05-24 19:02:41

这对我很管用

       $('.send_invoice').on('click',function(){
      var invoice_id = $(this).attr('data-invoiceid');
      
      $('#send_invoice_btn').on('click',function(){
         var invoice_to = $('#invoice_to').val();
         var invoice_subject = $('#invoice_subject').val();
         var invoice_message = $('#invoice_message').val();
          
        $.ajax({
           url: 'includes/send-invoice.inc.php?invoice_id='+invoice_id,
           type: 'POST',
           data: {invoice_to:invoice_to,invoice_subject:invoice_subject,invoice_message:invoice_message},
           success: function(send_invoice_result){
               
           }
        });
      });    
   });