Ajax 发布到 PHP 并获取返回数据 - Ajax Post to PHP and Get Return Data

Ajax Post to PHP and Get Return Data

本文关键字：获取返回数据 PHP Ajax | 更新日期: 2023-09-27

我正在开发一个滑块来设置我的预算，并希望实现我在滑块1.php中设置的值。但是，当我尝试使用以下代码时，我遇到了一个错误"注意：未定义的索引：幻灯片第 4 行 C：''xampp''htdocs''1204763e''slider1''slider1.php中的状态谢谢，PHP文件说"

在幻灯片.php中，我插入了这组代码：

     <html>
    <head>
    <script language="JavaScript" type="text/javascript">
    function ajax_post(val){
        var hr = new XMLHttpRequest();
        var url = "slider1.php";
        var ss = document.getElementById('sliderStatus').innerHTML = val;
        var vars = "sliderStatus="+ss;
        hr.open("POST", url, true);
        hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        hr.onreadystatechange = function() {
            if(hr.readyState == 4 && hr.status == 200) {
                var return_data = hr.responseText;
                document.getElementById("status").innerHTML = return_data;
            }
        }
        // Send the data to PHP now... and wait for response to update the status div
        hr.send(vars); // Actually execute the request
        document.getElementById("status").innerHTML = "processing...";
    }
    </script>
    </head>
    <body>
    <h2>Ajax Post to PHP and Get Return Data</h2>
    <input type="range" name="slide" min="0" max="100" value="50" step="2" onChange="ajax_post(this.value)" />
    <br /><br />
    <span id="sliderStatus">50</span>
    <br/><br/>
    <div id="status"></div>
    </body>
    </html>

In slider1.php, I inserted this set of code:
    <?php 
    echo 'Thank you '. $_GET['slideStatus'] . ', says the PHP file';
    ?>

您似乎尝试访问"slideStatus"，但您正在发布"sliderStatus"。