Ajax,PHP,MYSQL interaction


Ajax,PHP,MYSQL interaction

我有一个从page1链接的page2。page1向page2发送一个唯一的作业id。page2有如下所示的ajax代码。在第2页中,我有一个选择,它根据每个申请人获得的分数从第3页中获取面试申请人数。我将如何根据通过的工作id为申请者获取数据?我认为我的问题是在ajax中传递参数。请帮忙吗?这是我的代码,但不起作用。我在第2页遇到错误,一个未知的索引作业ID。

<?php $getid =$_GET['jobid'];?>
<html>
<head>
    <link href="css/bootstrap.min.css" rel="stylesheet">
<script>
function showSuccess(str,$getid) {
    if (str == "") {
        document.getElementById("txtHint").innerHTML = "";
        return;
    } else {
        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("POST","interview.php?q="+str);

        xmlhttp.send();
    }
}
</script>
</head>
<body>
  <?php
  echo $getid;
  ?>
<div class="col-md-4 col-md-offset-4">
<form>
    <?php 
     include('includes/conn.php'); 
     $row="SELECT * FROM jobs";  
     $query=mysqli_query($conn,$row) or die(mysqli_error($conn)); 
     while($row=mysqli_fetch_array($query)){
      $ref=$row['id'];
     }          
echo '<select name="names" onchange="showSuccess(this.value'.$getid.')" class="form-control">
  <option value="">Select a person:</option>
  <option value="5">5</option>
  <option value="0">0</option>
  <option value="10">10</option>
  <option value="15">15</option>
  <option value="20">20</option>
  <option value="25">25</option>
  <option value="30">30</option>
  <option value="35">35</option>
  <option value="40">40</option>
  <option value="45">45</option>
  <option value="50">50</option>
  <option value="55">55</option>
  <option value="60">60</option>
  <option value="65">65</option>
  <option value="70">70</option>
  <option value="75">75</option>
  <option value="80">80</option>
  <option value="85">85</option>
  <option value="90">90</option>
  <option value="95">95</option>
  <option value="100">100</option>
  </select>';
  ?>
  <!--<input type="text" class="form-control" name="number" placeholder='Enter the no of candidates' required>-->
</form>
</div>
<br>
<div id="txtHint" class="col-md-4 col-md-offset-4"><b>The candidates</b></div>
</body>
</html>
<?php $q = intval($_GET['q']);
?>
<?php $getid = intval($_POST['$getid']);?>
<?php
include('includes/conn.php');
$row="SELECT  idNo,id,name,jobTitle,SUM(points) AS total FROM shortlist WHERE job='$getid' GROUP BY id ORDER BY total DESC LIMIT $q";
$query=mysqli_query($conn,$row) or die(mysqli_error($conn));
while($row=mysqli_fetch_array($query))
{
  echo $row['name'];
}
mysqli_close($conn);
?>

您正在发送POST数据中的值

xmlhttp.open("POST","interview.php?q="+str);

但在你使用的第三个文件中,

<?php $getid = intval($_POST['$getid']);
?>

您没有在POST/GET参数中发送$getid