Question

我有以下代码：

function fetchDemo(url) {
  return fetch(url).then(function(response) {
    return response.json();
  }).then(function(json) {
    return json;
  });
}

fetchDemo(countriesUrl).then(function(result) {
  console.log(result);
});
fetchDemo(carMakesUrl).then(function(result2) {
  var obj = {};
  result2.forEach(function(ele) {
    obj[ele.make_country] = (obj[ele.make_country] || 0) + 1
  });
  for (var country in obj) {
    console.log(country + ", " + "produces the following number of cars: " + obj[country]);
  }
});

countriesUrl = https://rawgit.com/csabapalfi/20d74eb83d0be023205225f79d3e7964/raw/7f08af5c0f8f411f0eda1a62a27b9a4ddda33397/countries.json

[
{
"name": "Afghanistan",
"code": "AF"
},
{
"name": "Åland Islands",
"code": "AX"
},
{
"name": "Albania",
"code": "AL"
},
{
"name": "Algeria",
"code": "DZ"
},
{
"name": "American Samoa",
"code": "AS"
},
{
"name": "AndorrA",
"code": "AD"
},
{
"name": "Angola",
"code": "AO"
},
{
"name": "Anguilla",
"code": "AI"
},
{
"name": "Antarctica",
"code": "AQ"
} ....

carMakesUrl = https://rawgit.com/csabapalfi/20d74eb83d0be023205225f79d3e7964/raw/7f08af5c0f8f411f0eda1a62a27b9a4ddda33397/carmakes.json

[
{
"make_id": "abarth",
"make_display": "Abarth",
"make_is_common": "0",
"make_country": "Italy"
},
{
"make_id": "ac",
"make_display": "AC",
"make_is_common": "0",
"make_country": "UK"
},
{
"make_id": "acura",
"make_display": "Acura",
"make_is_common": "1",
"make_country": "USA"
},
{
"make_id": "alfa-romeo",
"make_display": "Alfa Romeo",
"make_is_common": "1",
"make_country": "Italy"
},
{
"make_id": "allard",
"make_display": "Allard",
"make_is_common": "0",
"make_country": "UK"
},
{
"make_id": "alpina",
"make_display": "Alpina",
"make_is_common": "0",
"make_country": "UK"
},.....

每个数组都有一个对象列表，我如何匹配来自不同fetch的数组。

我想比较结果和结果2

Answer 1

fetch工作异步，因此你必须堆叠fetch-requests。

fetchDemo(countriesUrl).then(function(result) {
  var countries = result;

  fetchDemo(carMakesUrl).then(function(result2) {
    var obj = {};
    result2.forEach(function(ele) {
      obj[ele.make_country] = (obj[ele.make_country] || 0) + 1
    });
    for (var country in obj) {
      console.log(country + ", " + "produces the following number of cars: " + obj[country]);
    }
  });
});

您有两个vars：countries和result2，您可以像往常一样进行比较（JavaScript - Compare two arrays）。

js比较来自fetch的2个对象数组

1 个答案: