如何在API中循环浏览多个页面? [英] How do I loop through multiple pages in an API?
问题描述
我正在使用《星球大战》 API https://swapi.co/我需要输入星际飞船信息,星舰的搜索结果跨越4页,但是get调用每页仅返回10个结果.如何遍历多个页面并获取所需的信息?
I am using the Star Wars API https://swapi.co/ I need to pull in starships information, the results for starships span 4 pages, however a get call returns only 10 results per page. How can I iterate over multiple pages and get the info that I need?
我已经使用fetch api获取星际飞船的第一页,然后将这个10的数组添加到我的totalResults数组中,然后创建了While循环来检查'next!== null'(next是数据中的下一页属性,如果我们正在查看最后一页,即第4页,则next将为null"next" = null)因此,只要next不等于null,我的While循环代码应获取数据并将其添加到我的totalResults数组.我在最后更改了next的值,但它似乎永远循环并崩溃.
I have used the fetch api to GET the first page of starships and then added this array of 10 to my totalResults array, and then created a While loop to check to see if 'next !== null' (next is the next page property in the data, if we were viewing the last page i.e. page 4, then next would be null "next" = null) So as long as next does not equal null, my While loop code should fetch the data and add it to my totalResults array. I have changed the value of next at the end, but it seems to looping forever and crashing.
function getData() {
let totalResults = [];
fetch('https://swapi.co/api/starships/')
.then( res => res.json())
.then(function (json) {
let starships = json;
totalResults.push(starships.results);
let next = starships.next;
while ( next !== null ) {
fetch(next)
.then( res => res.json() )
.then( function (nextData) {
totalResults.push(nextData.results);
next = nextData.next;
})
}
});
}
代码不断循环,这意味着我的'next = nextData.next;'增量似乎不起作用.
Code keeps looping meaning my 'next = nextData.next;' increment does not seem to be working.
推荐答案
您必须在while循环中 await
响应,否则循环将同步运行,而结果将异步到达,换句话说while循环永远运行:
You have to await
the response in a while loop, otherwise the loop runs synchronously, while the results arrive asynchronously, in other words the while loop runs forever:
async getData() {
const results = [];
let url = 'https://swapi.co/api/starships/';
do {
const res = await fetch(url);
const data = await res.json();
url = data.next;
results.push(...data.results);
} while(url)
return results;
}
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