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        通过ajax调用加载mysqli php数据

        时间:2023-07-29

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                  本文介绍了通过ajax调用加载mysqli php数据的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

                  问题描述

                  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');
                  

                  这篇关于通过ajax调用加载mysqli php数据的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持html5模板网!

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