Question

我正在尝试对“ http://www.omdbapi.com/?i=tt2975590&apikey=”进行API调用，我想知道是否有人可以向我展示如何通过javascript代码传递i参数。目前为i = tt2975590。

let detailMovie= "http://www.omdbapi.com/?i=tt2975590&apikey=<mykey>    
let request = new XMLHttpRequest()

request.open('Get',movieURL)
request.send()
request.onload = function() {
    if(request.status !=200){
        console.log("There is a problem")
    } else {
        let moviesResponse = JSON.parse(request.responseText)
        I want to be able to call http://www.omdbapi.com/?i=tt2975590&apikey=<mykey> but pass in the paramater(i) value.
    }
}

在此处输入代码

Answer 1

使用 ${}

let key = "1234";
let apikey = "878234";
let detailMovie = `http://www.omdbapi.com/?i=${key}&apikey=${apikey}`;
console.log(detailMovie);

或使用返回api的函数

function genApi(key, apikey) {
   return `http://www.omdbapi.com/?i=${key}&apikey=${apikey}`;
}

函数示例可以像您这样花费很多时间

const myApi = genApi("123", "23423"); 
// will be "http://www.omdbapi.com/?i=123&apikey=23423"

Answer 2

您可以使用axios简化api的调用。这里是如何将参数传递给get请求的代码段。

// Make a request for a user with a given ID

axios.get('/user?ID=12345')
  .then(function (response) {
    // handle success
    console.log(response);
  })
  .catch(function (error) {
    // handle error
    console.log(error);
  })
  .then(function () {
    // always executed
  });

// Optionally the request above could also be done as
axios.get('/user', {
    params: {
      ID: 12345
    }
  })
  .then(function (response) {
    console.log(response);
  })
  .catch(function (error) {
    console.log(error);
  })
  .then(function () {
    // always executed
  });

Answer 3

将i的值存储在某些变量中，例如：

var x ='tt2975590';

然后在您的网址中使用它，

let detailMovie =“” http://www.omdbapi.com/?i=“ + z +”＆apikey =“

使用上面的detailMovie变量打开您的请求。

Answer 4

很少有浏览器弃用XMLHttpRequest()，因此最好并强烈建议使用fetch API。 https://developer.mozilla.org/en-US/docs/Web/API/Fetch_API 您可以使用URLSearchParams(params)通过url传递值。例如

var url = new URL('https://sl.se')

var params = {lat:35.696233, long:139.570431} // or:
var params = [['lat', '35.696233'], ['long', '139.570431']]

url.search = new URLSearchParams(params)


fetch(url)
  .then(data => data.text())
  .then((text) => {
    console.log('request succeeded with JSON response', text)
  }).catch(function (error) {
    console.log('request failed', error)
  })

Answer 5

HttpUrl url1 = HttpUrl.parse("https://www.google.com").newBuilder() .addQueryParameter("search","parameter") .build();

将值传递给API

5 个答案: