使用AJAX发送单选按钮值以获取PHP文件，从而从MySQL中检索数据 - Sending radio button value using AJAX to get PHP file to retrieve data from MySQL

Sending radio button value using AJAX to get PHP file to retrieve data from MySQL

我希望能够用单选按钮的返回值编写一些mysql查询。这是我的代码

<html>
<head>
<script>
function showUser(str) {
if (str=="") {
document.getElementById("txtHint").innerHTML="";
return;
} 
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else { // code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
  document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
 }
}
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
                <?php include 'db_connector.php';
                $result = mysqli_query($con,"SELECT * FROM os_scope");
                while($row = mysqli_fetch_array($result)) {
                $row['name'];
                echo "<li><div class='toggle-btn-grp cssonly'>
                        <div><form><input type='radio' name='os' value=".$row['name']." id='myRadio' onchange='showUser(this.value)'>
                        <label class='toggle-btn'><div class='title'>".$row['name']."</form></div></label></div></div></li>";
            }
                ?>
<br>
<div id="txtHint"><b>Person info will be listed here.</b></div>
</body>
</html>

这是我的php文件getuser.php

  <?php
  $q = intval($_GET['q']);
  $con = mysqli_connect('localhost','root','abc123','mydb');
  if (!$con) {
  die('Could not connect: ' . mysqli_error($con));
  }
  mysqli_select_db($con,"ajax_demo");
  $sql="SELECT * FROM os ";
  $result = mysqli_query($con,$sql);
  echo "<table border='1'>
  <tr>
  <th>ID</th>
  <th>Name</th>
  <th>Scope</th>
  <th>Client</th>
  <th>Supplier</th>
  </tr>";
  while($row = mysqli_fetch_array($result)) {
  echo "<tr>";
  echo "<td>" . $row['id'] . "</td>";
  echo "<td>" . $row['name'] . "</td>";
  echo "<td>" . $row['scope'] . "</td>";
  echo "<td>" . $row['client'] . "</td>";
  echo "<td>" . $row['supplier'] . "</td>";
  echo "</tr>";
  }
  echo "</table>";
  mysqli_close($con);
  ?>

然而，当我运行代码时，我总是会收到以下错误：注意：未定义的索引：q这意味着q没有被发送到php文件，有什么想法可以解决这个问题吗？我真的很感激能得到的所有帮助。如果这是一个愚蠢的问题，我很抱歉，但我真的是新手，所以我真的很感激你的帮助。谢谢

更改

xmlhttp.open("GET","getuser.php?=q"+str,true);

至

xmlhttp.open("GET","getuser.php?q="+str,true);

q和=是错误的；）