Question

我有一个HTML文件，显示“PRESENT”一词。

当我点击该单词时，我想将AJAX请求发送到PHP文件，该文件在屏幕上和数据库中将单词更改为“ABSENT”。

有两个问题：

PHP永远不会回复HTML文件。如果确实如此，那么changeState（）函数会发出一条警告“服务器已响应！”
即使调用了changeState（）函数，PHP文件也从不执行SQL语句。如果是这样，数据库将被更新，而不是。

的index.html

<html>

  <head>
    <title>Attendance System</title>
  </head>

  <body>

    <div id="1">
      <span onclick="changeState(this)">PRESENT</span>
    </div>

    <script>

    function changeState(elem) {

      var oldValue = elem.innerHTML;
      var newvalue;
      var itemID = elem.parentNode.getAttribute('id');

      if (oldValue == 'PRESENT') {
        newvalue = 'ABSENT';
      } else {
        newvalue = 'PRESENT';
      }

      alert("new value is " + newvalue + ", itemID is " + itemID);

      var xmlhttp;

      if(window.XMLHttprequest) {
        xmlhttp = new XMLHttpRequest();
      } else {
        xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
      }

      xmlhttp.onreadystatechange = function() {
        if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
          alert("Server has responded!");
          elem.innerHTML = xmlhttp.responseText;
        }
      }

      xmlhttp.open("GET", "updateState.php?newValue=" + newvalue + "&id=" + itemID, true);
      xmlhttp.send();
    }

    </script>

  </body>
</html>

updateState.php

<?php

// Login procedure code here...

// I tried to substitute this palce with fixed dummy values before and it successfully updated the database. But the code doesn't work when it is like this.
$value = $_GET['newValue'];
$id = $_GET['id'];

// Construct SQL query
$query = "UPDATE attendanceList SET attendOrNot = '$value' WHERE id = '$id'";

// Execute SQL query
mysqli_query($conn, $query) or die('Query execution failed! ' . mysqli_error($conn));

// Print result text
print $value

?>

Answer 1

index.html

中存在语法错误

 if(window.XMLHttprequest) {
    xmlhttp = new XMLHttpRequest();
 } else {
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
 }

“XMLHttprequest”应该用大写字母R拼写为“XMLHttpRequest”。

很抱歉让所有人感到困惑。

为什么PHP文件不响应我的AJAX请求？

1 个答案: