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到PHP我要提交包含其为具有从另一个表的字段数据的动态复选框领域的引导形式,我想它来发送通过JQuery的AJAX到PHP。请告诉我如何让AJAX功能得到引导modal.here数据是我的代码。 HTML代码发送动态复选框值通过AJAX的形式引导形式
<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, $i)." ";
}
echo "<br />";
}
?>
</div>
PHP代码
<?php
//Database inclusion
include_once 'db_connection.php';
//get values
if(isset($_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!";
/*}*/
}
?>
[用PHP jQuery的Ajax的POST示例](的可能的复制http://stackoverflow.com/questions/5004233/jQuery的Ajax的后例如与 - PHP) – MadisonTrash