Question

我有两个文件test.html和test.php。我想要做的是使用java脚本将数据从PHP文件显示到HTML。 test.php的：

<?php
  $con = mysqli_connect('localhost','root','');
  if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
  }
  mysqli_select_db($con,"autofetch");
  $sql="SELECT * FROM sports ";
  $result = mysqli_query($con,$sql);
  $d =$_POST['iD'];
  if($d==1)
  {
    echo '<div id="container"></div>';
    echo '<div id="blocker"  style="display: none;"></div>';
    $response='';
    while($row=mysqli_fetch_array($result)) {
      $response = $response . " <li class='unread'>" .

        "<h4>". $row["id"] . "</h4>" . 
        "<p>" . $row["sportname"]  . 
        "</li>";
    }
    echo $response;
  }
?>

的test.html：

 <html>
 <head>
 <script>
   $( window ).load(function() {
     var url ='http://localhost/auto/test.php';
     $.get(url,function(data){
     $('#summary').html(data).show(1000);});
   });
</script>
</head>
<body>
  <div  id="summary">
  </div>
</body>
</html>

此外，我想在窗口加载时加载数据。我不知道自己哪里出错了。一切似乎都很好。

Answer 1

删除$d =$_POST['iD'];并且条件不是必需的。如果你想要的话。然后你必须在jquery中使用$ .post并将iD作为数据

<?php
$con = mysqli_connect('localhost','root','');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"autofetch");
$sql="SELECT * FROM sports ";
$result = mysqli_query($con,$sql);

echo '<div id="container"></div>';
echo '<div id="blocker"  style="display: none;"></div>';
$response='';
  while($row=mysqli_fetch_array($result)) {
$response = $response . " <li class='unread'>" .

"<h4>". $row["id"] . "</h4>" . 
"<p>" . $row["sportname"]  . 
"</li>";
 }
echo $response;

?>

使用javascript从其他php文件中获取数据

1 个答案: