我有一张table上的房间清单和租金。房间在下拉菜单中列出，我想在“输入”字段值、“页面加载”以及“下拉值更改”中获取租金。我写了下面的代码，但不知怎么的，它不是预期的工作。有人能帮我吗？
<?php
define("HOST", "localhost");
define("DB_USER", "root");
define("DB_PASS", "");
define("DB_NAME", "testdb");
$conn = mysqli_connect(HOST, DB_USER, DB_PASS, DB_NAME);
if (!$conn) {
die(mysqli_error());
}
$ajax = false;
$dbValue = 1; //or the default value of your choice - matched to the default selection value of the dropdown
if (isset($_GET['action']) && $_GET['action'] == 'ajax' && isset($_GET['dd'])) {
$dbValue = intval($_GET['dd']);
$ajax = true;
$res = mysqli_query($conn, "SELECT rent FROM `rooms` WHERE roomid = '$dbValue' limit 1");
$dataTable = '';
while ($data = mysqli_fetch_assoc($res)) {
$dataTable = $data['rent'];
}
}
// if ($ajax) return $dataTable;
?>
<html>
<head>
<title>jQuery Validation for select option</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
</head>
<body>
<select class="form-control" id= "roomid" name="roomid" required="">
<?php
$troom_sql = "SELECT roomid FROM rooms WHERE (isactive='y' AND isassigned='n' AND roomid NOT IN (SELECT roomid from roomalloc))";
$troom_rs = mysqli_query($conn, $troom_sql);
while ($troom_mem = mysqli_fetch_assoc($troom_rs)) {
?>
<option value="<?php echo $troom_mem['roomid']; ?>"><?php echo $troom_mem['roomid']; ?></option>
<?php
} ?>
</select>
<input type="text" placeholder="Monthly Rent" class="form-control" id="rent" name="rent" required>
<br>
</body>
<script>
$('#roomid').change(function()
{
var first = $('#roomid').val();
var req = $.get('getDB.php', {dd: first, action: 'ajax'});
req.done(function(data)
{
console.log("asdasd");
$('#rent').val("<?php echo $dataTable; ?>");
});
});
</script>
</html>
2条答案
按热度按时间23c0lvtd1#
创建一个jqueryajax函数，该函数接受post/get请求的参数，并在jquery事件上调用该ajax函数。ajax函数应该是这样的，
并在事件发生时调用该函数
onclick="LoadComponentPage(param)". 您可以对调用的结果进行后处理以显示结果或错误，如示例函数所示。vsaztqbk2#
虽然您已经在同一个文件中编写了php和js，但是仍然需要从php端返回数据并在js中处理。
从php端
在js中