Ajax 与 JQuery:200 正常,但不是“成功”

发布于 2024-10-15 02:45:37 字数 710 浏览 4 评论 0原文

我正在尝试使用 AJAX 向 Google 图书发送查询并在我的网站上显示结果。我使用 JQuery 发送请求并处理响应,如下所示:

var query = [formatted input from a form];
var URL = "http://books.google.com/books/feeds/volumes?q="+query+"&start-index=1&max-results=5";

$.ajax({
    type: "GET",
    url: URL,
    dataType: "xml",
    success: function(data, status){
        alert(status);
    }
});

目前,我只有脚本在收到响应时发出“成功”警报。如果我使用脚本将该查询发送到本地页面进行测试,则效果很好。但是,当我按照开发人员 API 页面上的说明将 URL 设置为上面列出的 Google URL 时,我从未看到该警报。根据 Firebug 的说法,我收到了响应和状态 200 ok,这是我应该做的,但它没有到达“成功”路径。有谁知道为什么?

编辑:我应该补充一点,如果我直接访问该网址,则为 http://books.google.com 等。使用一些随机 q,它可以毫无问题地显示提要 XML,因此查询不是问题。

I'm trying to use AJAX to send a query to Google Books and display the results on my website. I'm using JQuery to send the request and handling the response, like so:

var query = [formatted input from a form];
var URL = "http://books.google.com/books/feeds/volumes?q="+query+"&start-index=1&max-results=5";

$.ajax({
    type: "GET",
    url: URL,
    dataType: "xml",
    success: function(data, status){
        alert(status);
    }
});

Currently, I just have the script alerting "success" if a response is received. If I use my script to send that query to a local page for testing, this works just fine. But when I set the URL to the Google one listed above, as instructed on the Developer API page, I never see the alert. According to Firebug, I am receiving a response and a status of 200 ok as I should, but it's not getting to that "success" path. Does anyone know why?

Edit: I should add that if I follow the URL directly, to http://books.google.com etc. with some random q, it displays the feed XML with no problems, so the query is not the issue.

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评论(2

最丧也最甜 2024-10-22 02:45:37

在标准浏览器安全设置下,您无法使用 XMLHttpRequest 发出跨域请求。一种可能的解决方案是编写一个本地代理函数(假设您可以创建服务器端代码),将查询转发到外部站点,然后返回响应。

编辑:看起来 Google 也提供了 JavaScript API。我认为他们采用这种方式来避免跨域 XHR 问题。

http://code.google.com/apis/books/docs /js/devguide.html#execute

编辑:书籍的 JavaScript API 已弃用。虽然它不再实际有用,但您可以通过 Wayback Machine 存档查看原始引用的文档文本:http://web.archive.org/web/20120414070427/http://code.google.com/apis/books/docs/ js/devguide.html#execute

You can't make cross-domain requests using XMLHttpRequest under the standard browser security settings. One possible solution is to write a local proxy function (assuming you can create server-side code) that forwards the query to the external site, and then returns the response.

Edit: It looks like Google provides a JavaScript API as well. I would assume that they've crafted in such a way to avoid the cross-domain XHR issue.

http://code.google.com/apis/books/docs/js/devguide.html#execute

Edit: The JavaScript API for books was deprecated. While it's no longer practically useful, you can see the original referenced documentation text via the Wayback Machine archive: http://web.archive.org/web/20120414070427/http://code.google.com/apis/books/docs/js/devguide.html#execute

温柔戏命师 2024-10-22 02:45:37

这是 ajax 调用的跨域问题,因为浏览器具有基于域策略的安全模型。

如果您不想包含整个 Google Books API,您还可以使用 Google Ajax API 和 jsonp 进行跨域 ajax 调用。

此处的文档:

http://code.google.com/apis /books/docs/js/jsondevguide.html#basic_query

jQuery 示例

var query = 'jquery';
var URL = 'https://ajax.googleapis.com/ajax/services/search/books?v=1.0&q=' + query;

$.ajax({
    type: 'GET',
    url: URL,
    dataType: 'jsonp',
    success: function( data, status ){
        alert( data.responseData.results.length + ' results found!' );
    },
    error: function() {
        alert( 'Something goes wrong!' );
    }
});

Ciao!

It's a cross-domain problem with ajax calls because browsers have a security model based on a domain policy.

if you don't wan to include the whole Google Books API, you can also use Google Ajax API with jsonp for cross-domain ajax calls.

Docs here:

http://code.google.com/apis/books/docs/js/jsondevguide.html#basic_query

jQuery example

var query = 'jquery';
var URL = 'https://ajax.googleapis.com/ajax/services/search/books?v=1.0&q=' + query;

$.ajax({
    type: 'GET',
    url: URL,
    dataType: 'jsonp',
    success: function( data, status ){
        alert( data.responseData.results.length + ' results found!' );
    },
    error: function() {
        alert( 'Something goes wrong!' );
    }
});

Ciao!

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