Question

表单ajax.php包含一个dropdown和一个textarea。一个用于从数据库获取的员工姓名。我需要在更改下拉项目时显示员工的详细信息。这里是我的代码

<html>
<head>
    <meta http-equiv="content-type" content="text/html; charset=utf-8" />
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
$(document).ready(function() {
  $("#demo").change(function() {
    $empname = $('#demo').val();
    $.ajax({
      type: "post",
      url: "display.php",
      data: "empname=" + $empname,
      success: function(data) {
        if (data == '') {
          alert('no data');
        } else {
          $("#desc").html(data);

        }
      }
    });
  });
});
</script>
</head>
<body>
<?php
  $con=  mysqli_connect("localhost", "root", "", "process");
  if(!$con){
    echo 'not connected';
  }
?>
<label for="selector1" class="col-sm-2 control-label">Assignee</label>
 <div class="col-sm-8"><select name="emp" id="demo" class="form-control1">
<?php
  $sql=mysqli_query($con,"select * from employee"); 
  while($name=mysqli_fetch_array($sql)) {
?>
  <option value="<?php echo $name['id'];?>"><?php echo $name['empname'];?></option>
<?php } ?>
</select></div>
description <textarea id="desc"></textarea>
<input type="submit" name="sub" value="click" id="bt">
</body>
</html>

我的下一页即。display.php包含以下代码，

<?php

  $con=  mysqli_connect("localhost", "root", "", "process");
  if(!$con){
    echo 'not connected';
  }
  if( $_REQUEST["data"] ){
   $name = $_REQUEST['data'];
   $sql=mysqli_query($con,"select * from employee where id='$empname'");
   $name=mysqli_fetch_array($sql);
   echo $name['empname'];
 }

Answer 1

问题是您要在empname中传递数据并使用data进行重新更改：

<强> Display.php的

<?php

$con=  mysqli_connect("localhost", "root", "", "process");
if(!$con){
    echo 'not connected';
}
if( $_REQUEST["empname"] ){ //change here

   $empname= $_REQUEST['empname'];//change here
   $sql=mysqli_query($con,"select * from employee where id='$empname'");
   $name=mysqli_fetch_array($sql);
   echo $name['empname'];
}

下拉值无法使用ajax到达下一页

1 个答案: