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我需要一些解决方案来在我的 jquery ajax 调用中提供相对路径，因为现在我正在使用诸如 ../ 之类的 soem 东西。任何建议都会被高度赞赏。

代码

$.ajax({
        type: 'get',
        url: '../../MyPage/getDetails',
        success: function (data) {
            if (data > 0) {
4

3 回答 3

1

只要您网站的层次结构保持不变，您给出的路径绝对是相对路径。

您在 ajax 调用中提供的 URL 将始终向上移动两个文件夹并调用在 MyPage 中定义的 getDetails 方法

于 2012-05-22T18:24:21.197 回答
1

这有点骇人听闻，但是您是客户端，所以:)

这段代码做了一些假设——比如myscript.js存在的脚本标签

假设您的页面上有一些标记，如下所示：\

<script src='../js/myscript.js' type="text/javascript"></script>
<div>hiya</div>

好的，现在我们将使用脚本标签并从中获取基本 url :)

让我们创建一个我们可以使用的 Util 类，添加一个 get base url 方法并警告该值：

此处的工作副本：http: //jsfiddle.net/SmCLy/

function Util() { /*...Nothing...*/
}

Util.getBaseURL = function() {
    //<summary>
    //    Returns the application base URL (well, a URL that is
    //    equivalent to such - it may have some "backtracking", 
    //    i.e. "../" at
    //    the end of the URL to simulate the root...)
    //</summary>    
    // 1) Find the script include for this JavaScript file:
    var scriptElements = document.getElementsByTagName("script"),
        getScriptPathFragmentPosition = Util.getBaseURL._getScriptPathFragmentPosition,
        getScriptPathFragmentPositionResult = getScriptPathFragmentPosition(scriptElements, "js/myscript.js"),
        posScriptPathFragment = getScriptPathFragmentPositionResult.posScriptPathFragment;

    // 2) Create the "base" URL by taking the current URL,
    // stripping the page from the end, and appending
    // sufficient "../" sequences to
    // construct the equivalent of the application's root URL:
    var scriptElementSrcLower = getScriptPathFragmentPositionResult.scriptElementSrcLower;
    var backPathToRoot = ((scriptElementSrcLower !== "") ? scriptElementSrcLower.substring(0, posScriptPathFragment) : "");
    var currentPath = location.href;
    var currentPathWithoutPage;
    var PAGE_TOKEN = ".aspx";
    var posPageToken = currentPath.toLowerCase().indexOf(PAGE_TOKEN);
    if (posPageToken > -1) {
        var trimmedPath = currentPath.substring(0, posPageToken + PAGE_TOKEN.length);
        currentPathWithoutPage = trimmedPath.substring(0, trimmedPath.lastIndexOf("/") + 1);
    } else {
        currentPathWithoutPage = currentPath.substring(0, currentPath.lastIndexOf("/") + 1);
    }
    return (currentPathWithoutPage + backPathToRoot);
};

Util.getBaseURL._getScriptPathFragmentPosition = function(scriptElements, scriptPathFragment) {
    var scriptElementsIndex = (scriptElements.length - 1),
        scriptElementSrc = "",
        scriptElementSrcLower = "",
        posScriptPathFragment = -1;

    while (scriptElementsIndex >= 0) {
        scriptElementSrc = scriptElements[scriptElementsIndex].getAttribute("src");
        if (typeof(scriptElementSrc) != "undefined" && scriptElementSrc !== null && scriptElementSrc !== "") {
            scriptElementSrcLower = scriptElementSrc.toLowerCase();
            posScriptPathFragment = scriptElementSrcLower.indexOf(scriptPathFragment);

            if (posScriptPathFragment >= 0) {
                break;
            }
        }

        scriptElementsIndex--;
    }

    var result = {
        posScriptPathFragment: posScriptPathFragment,
        scriptElementSrcLower: scriptElementSrcLower
    };

    return result;
};

alert(Util.getBaseURL());

所以，在您的页面上，您可以执行以下操作：

var myajaxUrl = getBaseURL()+"MyPage/GetDetails";
于 2012-05-22T19:10:42.780 回答
0

试试这个定义<?php $site_root="WWW.example.com/" ?>

var URL="<?php echo $site_root;?>folder/your_action.php";

并触发 Ajax 调用

于 2017-01-28T12:03:13.107 回答