1
我想打开一个应用程序,并从它返回一些返回值,如成功或失败。怎么做?如何使用x-callback-url?
从文档
[scheme]://[host]/[action]?[x-callback parameters]&[action parameters]
问题1:
我应该在[动作参数]什么地方呢?这是强制性的吗?
发送应用甲
- (IBAction)openReceivingAppBButtonPressed:(id)sender {
NSString *xcallBack = @"x-callback-url/payment?&amount=1.00";
NSString *URLEncodedText = [xcallBack stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *ourURL = [NSURL URLWithString:[@"receivingAppB://" stringByAppendingString:URLEncodedText]];
if([[UIApplication sharedApplication] canOpenURL:ourURL]){
[[UIApplication sharedApplication] openURL:ourURL];
}
}
接收应用B
- (BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)url{
[self returnToSendingAppAWithResponse];
return true;
}
-(void)returnToSendingAppAWithResponse{
NSString *xcallBackSuccess = @"success";
NSString *URLEncodedText = [xcallBackSuccess stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *ourURL = [NSURL URLWithString:[@"sendingAppA://" stringByAppendingString:URLEncodedText]];
if([[UIApplication sharedApplication] canOpenURL:ourURL]){
[[UIApplication sharedApplication] openURL:ourURL];
}
}
问题2:
在接收的应用程序,是它正确调用另一个的OpenURL调用发送应用程序A返回成功消息?
我能够实现我想要的。但只是怀疑这是否是使用x-callback-url的正确方式。 x-callback-url似乎对我无用。