php - 尝试使用 jQuery 从选定标记的选项中获取值并发布到 php 页面

Question

我正在尝试从所选选项中获取值(value)并在 Php 上发布。但我也有一个表格来发布数据。我试图将两者结合起来，但现在当我尝试运行代码时，我得到一个空白的查询页面。

message.php

    <form  id="messageEmployeeForm" class="form table-striped" method="POST" action="core/query.php">

<select  id='toEmployee' class='form-control formBlock' name='toEmployee_id' value='' type='text' class='form-control formBlock'>
                     <option value="">To...</option>
                       <?php
        $employeesQuery = $db->query("SELECT * FROM employees");
        while ($employee = mysqli_fetch_assoc( $employeesQuery)) {
            ?>
      <option id='toEmployees_id' name='toEmployees_id' value="<?php echo $employee['employee_id'] ;?>">

     <?php echo $employee['first_name'] ;?> <?php echo $employee['last_name'] ;?></option>
     <?php } ;?>
    </select> 


//other inputs here...

        <input name="sendMessage" id="submit" type="submit" class="formBlock btn btn-success"  value="Send Message"/>
    </form>

jquery

<script>    

    $(document).ready(function () {
            $("#toEmployee").bind("change", function (e) {
                $.ajax({
                    type: "POST",
                    data: { data: $("#toEmployee").val() }
                });
            });
        });
</script>  

query.php

if ($_SERVER['REQUEST_METHOD'] === 'POST' && isset($_POST['sendMessage']) && isset($_POST['data'])){
$var = $_POST['data'];
//other data to be posted and bind here using prepared statements...

]

Answer 1

您的 query.php 文件期望在请求期间出现两个 post key ，没有这些它不会运行内部代码。

isset($_POST['sendMessage']) && isset($_POST['data'])

因此在您的 jQuery 中，您需要额外发送 sendMessage 键，如下所示:

$.ajax({
    type: "POST",
    data: { 
        data: $("#toEmployee").val(),
        sendMessage: 'your value'
    },
    success: function(data) {
        // do something with output (data)
    }
});