通过ajax调用加载mysqli php数据

2021-12-25 00:00:00 get jquery php mysqli ajax

我想做的是通过 ajax 和 php 调用一些数据库数据.但是ajax调用不起作用，网上也找不到解决办法.

What I'm trying to do is calling some database data via ajax and php. But the ajax call doesn't work, and I can't find out a solution on the web.

这是我的代码:

test.php

<?php

include_once 'db_class.php';

$cat = $_GET['cat'];  

$dbconn = new dbconn('localhost', 'root', 'somepsw', 'blog');

 $dbconn->set_query("select * from posts where category = '".$cat."'");

 echo '<br/>'.$dbconn->query.'<br/>';

 $result = $dbconn->result;

 $num = $dbconn->num_results;

 $array = mysqli_fetch_assoc($result);

 echo json_encode($array);
?>

如果我在浏览器上输入那个网址:http://127.0.0.1:82/blog/ws/test.php?cat=css

If i type that url on browser: http://127.0.0.1:82/blog/ws/test.php?cat=css

jsonEncode 返回的数据是正确的，但是当我用 jquery 在 html 页面上加载它时，他无法读取数据.

The data returned via jsonEncode is correct, but when i'm loading it on a html page with jquery he can't read the data.

test.html

<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<script>
function ajaxCall() {

var css;

$.ajax({                                      
      url: 'test.php',
      type: "GET",     
      data: {cat: css},              
      dataType: 'json',    
      success: function(rows)         
      {

     alert(rows);

      },
      error: function() { alert("An error occurred."); }

    });

    }

    ajaxCall();

</script>
</head>
<body></body>
</html>

提前致谢.

推荐答案

你的变量 css 没有价值.您想使用 string 'css'.也许您也希望能够加载其他类别.因此，将您的 ajaxCall 函数更改为

Your variable css has no value. You wanted to use the string 'css'. Maybe you want to be able to load other categories, too. So change your ajaxCall function to

function ajaxCall(category)
{
    $.ajax({
        url: 'test.php',
        type: "GET",
        data: {cat: category},
        dataType: 'json',    
        success: function(rows) {
           alert(rows);
        },
        error: function() {
           alert("An error occurred.");
        }
    });
}

并使用

ajaxCall('css');

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