Question

任何人都可以帮我把我的三个片段放到一个可用的脚本中。

1) Get active window URL
2) Strip URL for ID only
3) Concatenate API request to include ID from URL

以下内容返回我当前的窗口网址。

chrome.tabs.query({active: true, currentWindow: true}, 
        function(tabs) {
        var tabURL = tabs[0].url;
        console.log(tabURL);
    });

实施例.... https://myapi.com/users/PLLFFR6

 function test() {
        var urlID = tabURL.split("/");
        urlID = urlID[urlID.length-1];
    }

这会拆分https://myapi.com/users/PLLFFR6 并仅返回＆＃34; PLLFFR6＆＃34;

var authorizationToken = "xxxxxxxxxxxxx";
function myapiRequest(endpoint, options) {
    $.ajax($.extend({}, {
    type: 'GET',
    dataType: "json",
    success: function(data) {
    $('.Name').html(data.user.name);
    $('.Email').html(data.user.email);
    $('.Address').html(data.user.teams[0].name);
},

    url: "https://api.myapi.com/" + endpoint,
    headers: {
      "Authorization": "Token token=" + authorizationToken,
      "Accept": "application/vnd.myapi+json;version=2"
    }
  },
  options));
}
myapiRequest('/users/' + urlID + '?include%5B%5D=contact_methods&include%5B%5D=teams');

基于此示例，myapiRequest应该=

**/users/PLLFFR6?include%5B%5D=contact_methods&include%5B%5D=teams**

Answer 1

这样的事情应该有效：

function currentUrl() {
  return new Promise(function (resolve) {
    chrome.tabs.query({
      active: true,
      currentWindow: true
    }, function(tabs) {
      resolve(tabs[0].url)
    })
  })
}

function userIdfromUrl(url) {
  var parts = url.split('/')
  return parts[parts.length - 1]
}

function apiRequest(endpoint, options) {
  return new Promise(function (resolve) {
    $.ajax(
      $.extend({}, {
        type: 'GET',
        dataType: 'json',
        success: function(data) {
          resolve(data)      
        },
        url: 'https://api.website.com/' + endpoint,
        headers: {
          Authorization: 'Token token='+authToken,
          Accept: 'application/vnd.myapi+json;version=2'
        }
      }, options)
    )
  })
}

/* example usage */
currentUrl()
  .then(function (url) {
    return userIdfromUrl(url)
  })
  .then(function (userId) {
    return apiRequest('users/' + userId + '?include%5B%5D=contact_methods&include%5B%5D=teams')
  })
  .then(function (data) {
    console.log(data.user.name)
    console.log(data.user.email)
    console.log(data.user.teams[0].name)
  })

请注意，由于您使用的是chrome.tabs，因此这只能用作浏览器中的Chrome扩展程序而不是。您还需要加载jQuery（使用$.ajax和$.extend）。

Jquery / JavaScript - 获取活动URL，拆分

1 个答案: