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Ios 在Google+上共享文本、图像和Url;未登录_Ios_Objective C_Sfsafariviewcontroller_Qquickitem_Nsurlcomponents - Fatal编程技术网

Ios 在Google+上共享文本、图像和Url;未登录

Ios 在Google+上共享文本、图像和Url;未登录,ios,objective-c,sfsafariviewcontroller,qquickitem,nsurlcomponents,Ios,Objective C,Sfsafariviewcontroller,Qquickitem,Nsurlcomponents,我想在不登录iOS应用程序的情况下,在Google+上与Url共享图像 我使用了: NSString *temp = @"Hello World"; // Construct the Google+ share URL NSURLComponents* urlComponents = [[NSURLComponents alloc] initWithString:@"https://plu

我想在不登录iOS应用程序的情况下,在Google+上与Url共享图像

我使用了:

NSString *temp = @"Hello World";

// Construct the Google+ share URL
NSURLComponents* urlComponents = [[NSURLComponents alloc]                                                      initWithString:@"https://plus.google.com/share"];
                urlComponents.queryItems = @[[[NSURLQueryItem alloc]
                                              initWithName:@"text"
                                              value:temp]];
NSURL* url = [urlComponents URL];

if ([SFSafariViewController class]) {
    // Open the URL in SFSafariViewController (iOS 9+)
    SFSafariViewController* controller = [[SFSafariViewController alloc]                                                               initWithURL:url];
    controller.delegate = self;
    [self presentViewController:controller animated:YES completion:nil];

} else {
    // Open the URL in the device's browser
    [[UIApplication sharedApplication] openURL:url];
}

我不知道如何用同样的方法发送图像

回答这个问题可能太晚了,但有人会有用的

NSURL *shareURL = [NSURL URLWithString:@"https://example.com/image.jpg"];
urlComponents.queryItems = @[
        [[NSURLQueryItem alloc] initWithName:@"url" value:[shareURL absoluteString]],
        [[NSURLQueryItem alloc] initWithName:@"text" value:text]
    ];

回答这个问题可能太迟了,但有人会有用的

NSURL *shareURL = [NSURL URLWithString:@"https://example.com/image.jpg"];
urlComponents.queryItems = @[
        [[NSURLQueryItem alloc] initWithName:@"url" value:[shareURL absoluteString]],
        [[NSURLQueryItem alloc] initWithName:@"text" value:text]
    ];
Swift 4.1

 let strKeyValue: String =  "https://example.com/image.jpg"

guard var urlComponents = URLComponents(string: "https://plus.google.com/share") else {
                return nil
   }    
 urlComponents.queryItems = [URLQueryItem(name: "url", value:strKeyValue),URLQueryItem(name: "text", value:"This is testing text")]

  let browserViewController = SFSafariViewController(url: urlComponents.url)

  browserViewController.delegate = self 

  present(browserViewController, animated: true)
Swift 4.1

 let strKeyValue: String =  "https://example.com/image.jpg"

guard var urlComponents = URLComponents(string: "https://plus.google.com/share") else {
                return nil
   }    
 urlComponents.queryItems = [URLQueryItem(name: "url", value:strKeyValue),URLQueryItem(name: "text", value:"This is testing text")]

  let browserViewController = SFSafariViewController(url: urlComponents.url)

  browserViewController.delegate = self 

  present(browserViewController, animated: true)

我还面临着只共享url的问题,一旦它将我重定向到safari,它不会显示POST按钮,它会直接将我重定向到Google Plus页面。我还面临着只共享url的问题,一旦它将我重定向到safari,它不会显示POST按钮,它会直接将我重定向到Google Plus页面。谢谢你的回答!请格式化代码,使其显示为代码,确保每行至少缩进四个空格,这是最简单的方法?(我自己也看了一眼,不知道这些线到底是怎么包装的。)你能解释一下你为了制作这个Swift 4.1所做的修改吗?谢谢代码中没有太大的变化,但是语法肯定会有变化。谢谢你的回答!请格式化代码,使其显示为代码,确保每行至少缩进四个空格,这是最简单的方法?(我自己也看了一眼,不知道这些线到底是怎么包装的。)你能解释一下你为了制作这个Swift 4.1所做的修改吗?谢谢代码中没有太大的变化,但语法肯定有变化