2016-12-29 37 views
0

到PHP我要提交包含其爲具有從另一個表的字段數據的動態複選框領域的引導形式,我想它來發送通過JQuery的AJAX到PHP。請告訴我如何讓AJAX功能得到引導modal.here數據是我的代碼。 HTML代碼發送動態複選框值通過AJAX的形式引導形式

<div class="form-group" id="myResponse"> 
       <label for="event1">Event</label> 
       <?php 
        $sql = "SELECT event_name FROM event1"; 
        $stmt = sqlsrv_query($conn, $sql); 
        if($stmt === false) 
        { 
         die(print_r(sqlsrv_errors(), true)); 
        } 

        $numFields = sqlsrv_num_fields($stmt); 

        while(sqlsrv_fetch($stmt)) 
        { 
         // Iterate through the fields of each row. 
         for($i = 0; $i < $numFields; $i++) 
         { 
          echo '<input type="checkbox" name="event[]"/>'." "; 
          echo sqlsrv_get_field($stmt, $i)." "; 

         } 
         echo "<br />"; 
        } 
       ?> 
      </div> 

PHP代碼

<?php 
//Database inclusion 
include_once 'db_connection.php'; 
//get values 
if(isset($_POST['addGuest'])) 
{ 
    $first_name  = $_POST['first_name']; 
    $last_name  = $_POST['last_name']; 
    $email   = $_POST['email']; 
    $event1   = $_POST['event1']; 

    //name can contain only alpha characters and space 
    /*if (isset($_POST['addRecord'])) 
    {*/ 
     $tsql = "INSERT INTO dbo.demo (first_name, last_name, email, event1) values (?, ?, ?, ?)"; 
     $var = array($first_name, $last_name, $email ,$event1); 
     $stmt = sqlsrv_query($conn, $tsql, $var); 
     if($stmt === false) 
     { 
      die(print_r(sqlsrv_errors(), true)); 
     } 
     echo $successmsg = "Successfully Registered!"; 
    /*}*/ 
} 

?>

+0

[用PHP jQuery的Ajax的POST示例](的可能的複製http://stackoverflow.com/questions/5004233/jQuery的Ajax的後例如與 - PHP) – MadisonTrash

回答

0
Hi first of all change this filed to and onsubmit call addform button 

    $event1   = $_POST['event']; 

    enter code here 

function addform() 
{ 
$.ajax({ 
      url : 'process.php', 
      type : 'POST', 
      data : $('#form').serialize(), 
      success: function(data){ 





      } 
}) 

}