如果我这样做:
var url="https://en.wikipedia.org/w/api.php?format=json&action=query&prop=categories&titles=Victory_Tests&callback=?";
$.getJSON(url,function(data){
$.each(data, function(i, item) {
console.log(item);
});
});
控制台告诉我:
{pages: {…}}
pages: 2347232:
categories: Array(8)0:
{ns: 14, title: "Category:Aftermath of World War II in the United Kingdom"}
1: {ns: 14, title: "Category:All articles with unsourced statements"}
2: {ns: 14, title: "Category:Articles with unsourced statements from August 2015"}
3: {ns: 14, title: "Category:Articles with unsourced statements from March 2009"}
4: {ns: 14, title: "Category:English cricket seasons from 1919 to 1945"}
5: {ns: 14, title: "Category:History of Test cricket"}
6: {ns: 14, title: "Category:Use British English from September 2011"}
7: {ns: 14, title: "Category:Use dmy dates from September 2011"}
length: 8
__proto__: Array(0)
ns: 0
pageid: 2347232
title: "Victory Tests"
__proto__:
Object__proto__:
Object__proto__: Object
我需要获得所有categories
我尝试了item.categories
和
data.parse.pages.map(function(val){
return val.categories[0];
}));
但它不对
答案 0 :(得分:2)
你应该得到多个页面,如果你想聚合一切,那就这样做:
var url="https://en.wikipedia.org/w/api.php?format=json&action=query&prop=categories&titles=Victory_Tests&callback=?";
$.getJSON(url,function(data){
var categories = Object.values(data.query.pages).map( (i) => i.categories.map( (c) => c.title) );
console.log(categories);
});
答案 1 :(得分:1)
假设已经返回了一个类似
的页面Object.values(data.query.pages)[0].categories.map(cat=>cat.title.substr(9))
应该做的工作:
var url="https://en.wikipedia.org/w/api.php?format=json&action=query&prop=categories&titles=Victory_Tests&origin=*";
$.getJSON(url,function(data){
var pages = Object.values(data.query.pages);
if(pages.length === 1)
console.log(pages[0].categories.map(cat => cat.title.substr(9)));
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
根据您所需的浏览器支持,您可能需要Object.values
的填充,例如https://github.com/tc39/proposal-object-values-entries/blob/master/polyfill.js。
(注意更改后的网址,在我看来使用&origin=*
代替&callback=?
是更好的方法,因为它不会返回JSONP但是有效的JSON,也可以通过{{1 }和XMLHttpRequest/fetch
。)
答案 2 :(得分:1)
使用API参数formatversion=2
:
var url="https://en.wikipedia.org/w/api.php?format=json&formatversion=2&action=query&prop=categories&titles=Victory_Tests&callback=?";
$.getJSON(url,function(data){
console.log($.map(data.query.pages, function(val){
return val.categories;
}));
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
(还利用jQuery.map
展平数组的能力。)