AJAX不会使用Jquery从PHP文件中获取数据

I'm trying post data to PHP file but i can't receive any data from PHP file. Let me add codes. This is my jQuery function:

我正在尝试将数据发布到PHP文件,但我无法从PHP文件接收任何数据。我来添加代码。这是我的jQuery函数:

$(document).ready(function () {

$(function () {
$('a[class="some-class"]').click(function(){

   var somedata = $(this).attr("id");

   $.ajax({
      url: "foo.php", 
      type: "POST",
      data: "id=" + somedata,

      success: function(){
          $("#someid").html();
      },
      error:function(){
          alert("AJAX request was a failure");
      }
    });
    });
    });
});

This is my PHP file:

这是我的PHP文件:

<?php
$data = $_POST['id'];

$con = mysqli_connect('localhost','root','','somedatabase');
if (!$con) {
  die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"database");
$sql="SELECT * FROM sometable WHERE id = '".$data."'";
$result = mysqli_query($con,$sql);


while($row = mysqli_fetch_array($result)) {

 echo $row['info'];

 }

mysqli_close($con);

?>

This what i have in HTML file:

这就是我在HTML文件中的内容:

<p id="someid"></p>

<a href="#page2" class="some-class" id="1">Data1</a>
<a href="#page2" class="some-class" id="2">Data2</a>

Note: This website is horizontal scrolling and shouldn't be refreshed. When i'm clicking links (like Data1) it's going to another page without getting data from PHP file

注意:此网站是水平滚动,不应刷新。当我点击链接(如Data1)时,它将转到另一个页面而不从PHP文件中获取数据

5 个解决方案

#1

You have a few problems:

你有一些问题:

You are not using the data as mentioned in the other answers:
success: function(data){ $("#someid").html(data); },

您没有使用其他答案中提到的数据:success:function(data){$(“#someid”)。html(data); },

You are not cancelling the default click action so your link will be followed:
$('a[class="some-class"]').click(function(e){ e.preventDefault(); ...;

您没有取消默认点击操作,因此您的链接将被遵循:$('a [class =“some-class”]')。click(function(e){e.preventDefault(); ...;

As the id's are integers, you can use data: "id=" + somedata, although sending an object is safer in case somedata contains characters that need to be escaped:
data: {"id": somedata},;

由于id是整数,你可以使用数据:“id =”+ somedata,尽管如果somedata包含需要转义的字符,发送对象会更安全:data:{“id”:somedata} ,;

You have an sql injection problem. You should cast the variable to an integer or use a prepared statement:
$data = (int) $_POST['id'];;

你有一个SQL注入问题。您应该将变量强制转换为整数或使用预准备语句:$ data =(int)$ _POST ['id'] ;;

As also mentioned in another answer, you have two $(document).ready() functions, one wrapping the other. You only need one.

另外在另一个答案中提到,你有两个$(document).ready()函数,一个包装另一个。你只需要一个。

#2

success: function(){
          $("#someid").html();
      },

should be:

success: function(data){
          $("#someid").html(data);
      },

#3

You should add parameter in success

您应该成功添加参数

success: function(data){ //Added data parameter
      console.log(data);
      $("#someid").html(data);
  },

The data get the values what you echo in PHP end.

数据获取您在PHP结束时回显的值。

#4

This:

  success: function(data){
      $("#someid").html(data);
  },

and you have two document ready, so get rid of:

你准备好了两个文件,所以摆脱:

$(document).ready(function () { ...

});

#5

data: "id=" + somedata,

Change it to:

将其更改为:

data: { id : somedata }

#1