解决双重网址问题..？ [英] Solving Dual URL Problem..?

问题描述

我正在使用cakephp，我有2个链接：

I am using cakephp I have 2 links:

<a href="#" tabindex="1" onclick="base_load_demo1('http://www.boxyourtvtrial.com/widget/beer/main/');" >beer</a>
<a href="#" tabindex="2" onclick="base_load_demo('http://www.boxyourtvtrial.com/widget/cocktail/main/');">cocktail</a>

使用以下JavaScript：

With the following JavaScript:

var Url1 = "http://www.boxyourtvtrial.com/widget/cocktail/main/";
var Url2 = "http://www.boxyourtvtrial.com/widget/beer/main/";
var Url3 = "http://www.boxyourtvtrial.com/widget/beer/mini/";

function base_load_demo(Url) {
  remoteCall(Url1,"","mainLeftContent");
  //remoteCall("SCRIPT_PATH","QUERY_STRING","TARGET_FUNCTION");
}

function base_load_demo1(Url2) {
  remoteCall(Url2,"","mainLeftContent");
  //remoteCall("SCRIPT_PATH","QUERY_STRING","TARGET_FUNCTION");
}

当我单击第一个链接时，它通过ajax调用显示其内容，但是当我单击第二个链接其给出的错误如下：

When I click on the first link it's showing its content through ajax call but when I click on the second link its giving error as follows:

缺少控制器

Missing Controller

错误：找不到Http：Controller。

Error: Http:Controller could not be found.

错误：在文件中创建以下类Http：Controller：app / controllers / http：controller.php

Error: Create the class Http:Controller below in file: app/controllers/http:controller.php

<?php
class Http:Controller extends AppController {

    var $name = 'Http:';
}
?>

注意：如果要自定义此错误消息，请创建应用程序/视图/错误/ missing_controller.ctp

Notice: If you want to customize this error message, create app/views/errors/missing_controller.ctp

并在FireFox控制台选项卡中

and in FireFox console tab

POST http://www.boxyourtvtrial.com /widget/beer/main/http://www.boxyourtvtrial.com/widget/cocktail/main/

我们如何解决

var xmlHttp;
var uri = "";
var callingFunc = "";
var sResponse = new Array();



function remoteCall(sUrl, sQueryStr, sCalledBy)
{
    alert(sUrl);
    var resStr = "";
var str = " { ";
if(sQueryStr != "") {
    var arr1 = new Array();
    arr1 = sQueryStr.split("&");
    if(arr1){
        for(i=0;i<=arr1.length;i++)
        {
            if(arr1[i] && arr1[i] != "") 
                              {
                var arr2 = new Array();
                arr2 = arr1[i].split("=");
                str += arr2[0]+":'"+arr2[1]+"' ,";
                
              }
        }
    }
}                   

    str += " tp: 'tp' } ";
      
      
    $.ajax({
        type: "GET",     
        url: sUrl,
        data: sQueryStr,
        dataType: "html",
        success: function(data) {
            $("#"+sCalledBy).html(data);
            //jih(sCalledBy,data);
        }
    });
      
  /*    $.get(sUrl,sQueryStr,function(data) {   
    jih(sCalledBy,data);
    });*/
}

function jih(divid,data)
{
         if(document.getElementById(divid))
                     document.getElementById(divid).innerHTML=data;
    }

推荐答案

加载的页面数：

<script type="text/javascript" src="http://www.boxyourtvtrial.com/widget/cocktail/main/js/common.js"></script>

。 common.js内部有一个名为 remoteCall 的函数，该函数将覆盖您的本地remoteCall函数。

in the header. Inside common.js is a function called remoteCall, which is overwriting your local remoteCall function.

common.js内部的remoteCall函数会添加

The remoteCall function inside common.js adds

var url= WIDGET_WEG_PATH+scr_url;

其中 WIDGET_WEG_PATH = http://www.boxyourtvtrial.com/widget/beer/main/

where WIDGET_WEG_PATH = "http://www.boxyourtvtrial.com/widget/beer/main/"

和 scr_url = http://www.boxyourtvtrial.com / widget / beer / main / （新remoteCall函数的第一个参数）

and scr_url = "http://www.boxyourtvtrial.com/widget/beer/main/" (the first parameter of the new remoteCall function)

这就是为什么获取url'加倍。

This is why you are getting the url 'doubled' in the post.

解决方案：

将本地remoteCall函数重命名为不同的名称。

Rename local remoteCall function to something that is distinct.

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