Javascript AJAX从PHP后端加载的页面

我想做的是从不同的页面加载信息，而不必刷新整个主页

您能告诉我如何调整此代码以加载不同名称的文件，如about.html和project.html吗

注意：此代码仅用于加载“page..html”文件

以下是php文件：

if(!$_POST['page']) die("0");

$page = (int)$_POST['page'];

if(file_exists('pages/page_'.$page.'.html'))
echo file_get_contents('pages/page_'.$page.'.html');

else echo 'There is no such page!';

更换这条线

if(file_exists('pages/page_'.$page.'.html'))

用这个

if(file_exists('pages/'.$page.'.html'))

---

您可以发出多个ajax请求这是jquery加载方法：

$( "#result" ).load( "ajax/test.html", function() {
  alert( "Load was performed." );
});

这样做2倍，你的财富！ 请记住，要使其发挥作用，您将需要jquery库

我在我的网站上做了类似的事情，下面是我的代码：

    window.setInterval("check()",60000); 
//request information notice is inside a function called check() (it's not required to put inside function I only do this if I will be making the same request multiple time throughout the program)                      
                        function check() {
                            var request = $.ajax({
                                url: "file.php",
                                type: "POST",            
                                dataType: "html"
                            });

                            request.done(function(msg) {
//when request is done:
                                $(".wheretoputnewdata").html(msg);          
                            });

                            request.fail(function(jqXHR, textStatus) {
                                //if request failed do this:
    alert( "Request failed: " + textStatus );
                            });
                       }     

---

只需在页面名称前加上page_u，并确保它们位于页面文件夹中，我怀疑这与您的页面名称重复。不过，我认为，现在已经修复，它将重新打开。现在我将暂缓结束这个问题，但请不要在将来创建重复的问题-修复一个问题比再次提问更好。

    window.setInterval("check()",60000); 
//request information notice is inside a function called check() (it's not required to put inside function I only do this if I will be making the same request multiple time throughout the program)                      
                        function check() {
                            var request = $.ajax({
                                url: "file.php",
                                type: "POST",            
                                dataType: "html"
                            });

                            request.done(function(msg) {
//when request is done:
                                $(".wheretoputnewdata").html(msg);          
                            });

                            request.fail(function(jqXHR, textStatus) {
                                //if request failed do this:
    alert( "Request failed: " + textStatus );
                            });
                       }