Question

我在一张桌子上有一个房间清单以及它们的租金。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”中获得租金。我编写了以下代码，但不知何故它无法正常工作。有人可以帮我吗？

    <?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>

Answer 1

尽管您已经在同一文件中编写了PHP和JS，但仍然需要从PHP端返回数据并在JS中进行处理。

if ($ajax) return json_encode($dataTable)

从PHP端

dat = JSON.parse(data)

在JS中

Answer 2

创建一个采用POST / GET Request参数的JQuery AJAX函数，并在JQuery Event上调用该Ajax函数。 Ajax函数应该像

function LoadComponentPage( param ){
    $.ajax({
        type: "POST",
        url: "./controller/ajax/component_paginate.php",
        data: "page="+param,
        dataType: "text",
        success: function(resultData){
            let section = $('#ComponentsListing');
            section.empty();
            section.html(resultData);
        },
        error : function(e){
            console.log(e);
        }
    });
}

，并在事件发生时以onclick="LoadComponentPage(param)"的形式调用该函数。您可以后处理调用结果以显示结果或出现错误，如示例函数所示。

在变更事件上使用jQuery从数据库中获取价值

2 个答案: