[英]how to pass the button id from one page to another page in php
我已经在我的php代码中创建了动态按钮,我需要将该创建的动态按钮ID转到php中的另一个页面,以从数据库中获取值。
我需要将该按钮的点击ID传递给SQL QUERY。
我如何在下面的代码创建动态按钮代码时传递值?
<?php
function dash()
{
include 'config.php';
$sql = "SELECT RoomNumber FROM roommaster ";
if ($result = mysqli_query($db, $sql)) {
$str = '';
while ($row = mysqli_fetch_array($result)) {
// generate array from comma delimited list
$rooms = explode(',', $row['RoomNumber']);
//create the dynamic button and set the value
foreach ($rooms as $v) {
$str .= "<input type='button' name='b1' id='btn' value='" . $v . "' />";
}
}
return $str;
} else {
echo "ERROR: Could not able to execute $sql. " . mysqli_error($db);
}
mysqli_close($db);
}
?>
<!Doctype html>
<html>
<head>
<meta charset="UTF-8">
<title>room boking</title>
<link href="css/bootstrap1.min.css" rel="stylesheet">
<link rel="stylesheet" href="css/front.css">
<script src='http://cdnjs.cloudflare.com/ajax/libs/jquery/2.1.3/jquery.min.js'></script>
<link rel="stylesheet" href="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/themes/smoothness/jquery-ui.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/jquery-ui.min.js"></script>
</head>
<body>
<form mathod="post" action="booking.php">
<div class=" row box col-md-4">
<div style="color:black">
<?php echo dash();?>
</div>
</div>
</form>
</body>
</html>
您可以在隐藏的输入类型内发送id值,如下所示:
<input type="hidden" value="btn" name="buttonID">
在booking.php中,使用$_POST['b1']
(其中b1
是输入的名称)来获取与任何其他表单输入相同的值。
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