如何在按钮上单击打开instagram应用程序 [英] how to open instagram app on button click

问题描述

1. 大家好，我都想在单击按钮时打开instagram应用，但不是 能够在plist中将Url方案设置为instagram

   NSString *instagramURL = @"instagram://app"; 
   NSURL *ourURL = [NSURLURLWithString:instagramURL]; 
   if ([[UIApplication sharedApplication]canOpenURL:ourURL]) {
       [[UIApplication sharedApplication]openURL:ourURL];
   
   } else {
       //The App is not installed. It must be installed from iTunes code.
       NSString *iTunesLink = @"//Some other Url goes here";
       [[UIApplication sharedApplication] openURL:[NSURL URLWithString:iTunesLink]];
   
       UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"URL error"
             message:[NSString stringWithFormat: @"No custom URL defined for %@", ourURL]
             delegate:self cancelButtonTitle:@"Ok" otherButtonTitles:nil];
       [alert show];
   

   我确实喜欢这样，但是应用未对iOS全新开放，可以提供任何帮助 赞赏

解决方案

您可以按用户名打开instagram应用，

 NSURL *instagramURL = [NSURL URLWithString:@"instagram://user?username=USERNAME"];
if ([[UIApplication sharedApplication] canOpenURL:instagramURL]) {
    [[UIApplication sharedApplication] openURL:instagramURL];
}

您可以参考 Instagram的iPhone挂钩，以了解更多方法和信息关于api的详细信息！！！

更新:

替换行下方

 NSString *instagramURL = @"instagram://app"; 

使用

   NSURL *instagramURL = [NSURL URLWithString:@"instagram://app"];

您直接将字符串分配为url !!

1. Hi all am trying to open instagram app on a button click but am not able to i set Url scheme as instagram in plist also

   NSString *instagramURL = @"instagram://app"; 
   NSURL *ourURL = [NSURLURLWithString:instagramURL]; 
   if ([[UIApplication sharedApplication]canOpenURL:ourURL]) {
       [[UIApplication sharedApplication]openURL:ourURL];
   
   } else {
       //The App is not installed. It must be installed from iTunes code.
       NSString *iTunesLink = @"//Some other Url goes here";
       [[UIApplication sharedApplication] openURL:[NSURL URLWithString:iTunesLink]];
   
       UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"URL error"
             message:[NSString stringWithFormat: @"No custom URL defined for %@", ourURL]
             delegate:self cancelButtonTitle:@"Ok" otherButtonTitles:nil];
       [alert show];
   

   i did like this but app not opening am new to iOS any help can be appreciated

解决方案

You can open instagram app by username like,

 NSURL *instagramURL = [NSURL URLWithString:@"instagram://user?username=USERNAME"];
if ([[UIApplication sharedApplication] canOpenURL:instagramURL]) {
    [[UIApplication sharedApplication] openURL:instagramURL];
}

You can refer iPhone Hooks of Instagram for more ways and details about api!!!

Update :

Replace below line,

 NSString *instagramURL = @"instagram://app"; 

with

   NSURL *instagramURL = [NSURL URLWithString:@"instagram://app"];

You are assigning directly string as url!!

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