tinyurl

How to get livestream tiny URL with YouTube Data API v3

由 本秂侑毒 提交于 2021-01-29 07:36:48
问题 function authenticate() { return gapi.auth2.getAuthInstance() .signIn({scope: "https://www.googleapis.com/auth/youtube.readonly"}) .then(function() { console.log("Sign-in successful"); }, function(err) { console.error("Error signing in", err); }); } function loadClient() { gapi.client.setApiKey("YOUR_API_KEY"); return gapi.client.load("https://www.googleapis.com/discovery/v1/apis/youtube/v3/rest") .then(function() { console.log("GAPI client loaded for API"); }, function(err) { console.error(

How to get livestream tiny URL with YouTube Data API v3

由 点点圈 提交于 2021-01-29 07:31:09
问题 function authenticate() { return gapi.auth2.getAuthInstance() .signIn({scope: "https://www.googleapis.com/auth/youtube.readonly"}) .then(function() { console.log("Sign-in successful"); }, function(err) { console.error("Error signing in", err); }); } function loadClient() { gapi.client.setApiKey("YOUR_API_KEY"); return gapi.client.load("https://www.googleapis.com/discovery/v1/apis/youtube/v3/rest") .then(function() { console.log("GAPI client loaded for API"); }, function(err) { console.error(

Convert URL into TinyURL in iPhone

由 末鹿安然 提交于 2020-01-17 12:35:15
问题 I want to programmatically convert a URl to TinyURL in iPhone. How to do this? 回答1: Tiny URL has a simple API that you can use, it's very simple Just send this request with your URL http://tinyurl.com/api-create.php?url=http://yourURL.com/ It will return a tiny URL with your link Edit: here's a working example, this is a synchronous request though so it can make your app unresponsive if it takes too long. NSString *origUrl = @"http://stackoverflow.com"; NSURL *url = [NSURL URLWithString:

How does tiny url work

由 ￣綄美尐妖づ 提交于 2019-12-23 12:22:54
问题 I have been wondering how tiny url works. I would like to develop something similar for my site, but as most people, I use GUIDs for ids. When an object is created, should I then generate a 10 character random string to use as public id, or is there a smarter approach? Example of old url: www.mysite.com/default.aspx?userId={id} Example of new url: www.mysite.com/pwzd4r9niy 回答1: You can use any kind of random string generator or GUID for this. I don't think there is a much smarter approach.

Struggling with a series of variables

由 情到浓时终转凉″ 提交于 2019-12-12 04:46:59
问题 To avoid tweets becoming caught in the twitter spam filter I have some code that goes to tinyurl and creates a new short url each time the code is run for each of the original urls. What i want is each time 'u' is printed, it's value should be passed to a variable 'linkvar1', 'linkvar2', 'linkvar3' etc. This is the passed to the tweet submission later in the code: import simplejson import httplib2 import twitter import tinyurl print("Python will now attempt to submit tweets to twitter...")

How to get the complete URL address most efficiently?

由 穿精又带淫゛_ 提交于 2019-12-07 20:23:04
问题 I'm using a Java program to get expanded URLs from short URLs. Given a Java URLConnection , among the two approaches, which one is better to get the desired result? Connection.getHeaderField("Location"); vs Connection.getURL(); I guess both of them give the same output. The first approach did not give me the best results, only 1 out of 7 were resolved. Can the efficiency be increased by the second approach? Can we use any other better approach? 回答1: I'd use the following: @Test public void

How to use jQuery to produce TinyURL

由 依然范特西╮ 提交于 2019-12-06 03:10:21
问题 I'm trying to build a jQuery function that will allow me to produce a TinyURL from some other link for micro blogging reasons (yes, twitter)... I found this tutorial from James Padolsey, but am not getting a response back from the call. http://james.padolsey.com/javascript/create-a-tinyurl-with-jsonp/ function requestShortURL(longURL, success) { var API = 'http://reque.st/create.api.php?json&url=', URL = API + encodeURIComponent(longURL) + '&callback=?'; console.log('tweet apit url: ' + URL);

How to use jQuery to produce TinyURL

由 。_饼干妹妹 提交于 2019-12-04 09:13:13
I'm trying to build a jQuery function that will allow me to produce a TinyURL from some other link for micro blogging reasons (yes, twitter)... I found this tutorial from James Padolsey, but am not getting a response back from the call. http://james.padolsey.com/javascript/create-a-tinyurl-with-jsonp/ function requestShortURL(longURL, success) { var API = 'http://reque.st/create.api.php?json&url=', URL = API + encodeURIComponent(longURL) + '&callback=?'; console.log('tweet apit url: ' + URL); $.getJSON(URL, function(data){ success && success(data.url); }); } requestShortURL('http://www

Create TinyURL via Jquery Ajax call

由 眉间皱痕 提交于 2019-12-01 10:45:27
I've looked through simiilar questions on SO, but can't seem to find one addressing what seems like a simple call.. function TweetThis(url) { $.ajax({ url: "http://tinyurl.com/api-create.php?url=" + url, cache: false, success: function(data){ alert(data); } }); } Basically I want to call TinyURL with an Ajax call and a long URL and return the shortened URL.. The success never fires, but when I check the URL it builds it returns fine in a browser. Looking in Firebug it doesn't show response coming back.. what am I missing? Attempting to make a regular AJAX request is impossible because of same

Create TinyURL via Jquery Ajax call

由 牧云@^-^@ 提交于 2019-12-01 08:20:47
问题 I've looked through simiilar questions on SO, but can't seem to find one addressing what seems like a simple call.. function TweetThis(url) { $.ajax({ url: "http://tinyurl.com/api-create.php?url=" + url, cache: false, success: function(data){ alert(data); } }); } Basically I want to call TinyURL with an Ajax call and a long URL and return the shortened URL.. The success never fires, but when I check the URL it builds it returns fine in a browser. Looking in Firebug it doesn't show response
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