iOS Square Register Popover OpenURL問題

Square最近刪除了OAuth商家授權的要求，如「在您的應用中啓動Square註銷事項」部分中的here所述。iOS Square Register Popover OpenURL問題

當我的應用程序彈出在通往廣場的註冊程序，一個警報彈出出現在廣場的應用說：

"API Error: No client ID specified"

儘管我清楚地列出我「的client_id」在我的應用程序時，我打電話開放廣場。我的代碼如下。 任何想法如何解決它？

-(void)squarePaymentWithName:(NSString *)name{ 
//Specify amount of money to charge 
float orderPriceFloat = [Order orderTotalPrice]; 
float orderPriceFloatCents = orderPriceFloat * 100; 
NSInteger orderPriceFloatCentsInteger = [[NSNumber numberWithFloat:orderPriceFloatCents] integerValue]; 
NSString *amountString = [NSString stringWithFormat:@"%ld", (long)orderPriceFloatCentsInteger]; 

NSDictionary *squareDictionary = @{@"callback_url": @"<CALLBACK_URL>", 
            @"client_id": @"<CLIENT_ID>", 
            @"version": @"1.2", 
            @"amount_money": 
             @{@"amount": amountString, 
             @"currency_mode":@"USD" 
             }, 
            @"options": 
             @{@"supported_tender_types": @[@"CREDIT_CARD", @"SQUARE_GIFT_CARD"], 
             @"auto_return": @"true" 
             } 
            }; 
NSString *jsonString = [NSString stringWithFormat:@"%@", squareDictionary]; 

NSString *encodedString = [jsonString stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]]; //encode the string 
NSString *scheme = [NSString stringWithFormat:@"square-commerce-v1://payment/create?data=%@", encodedString]; //input the string to the url 

UIApplication *application = [UIApplication sharedApplication]; 
NSURL *URL = [NSURL URLWithString:scheme]; 
BOOL canOpen = [application canOpenURL:URL]; //open the url 
[application openURL:URL]; 
}

jsonString的NSLog的是：

{ 
"amount_money" =  { 
    amount = 250; 
    "currency_mode" = USD; 
}; 
"callback_url" = CALLBACK_URL; 
"client_id" = CLIENT_ID; 
options =  { 
    "auto_return" = true; 
    "supported_tender_types" =   (
     "CREDIT_CARD", 
     "SQUARE_GIFT_CARD" 
    ); 
}; 
version = "1.2"; 
}

encodedString的NSLog的是：

而且，作爲一個說明， 'CALLBACK_URL' 和 'CLIENT_ID' 我的輸入是佔位符，因爲我不想把真正的價值。

謝謝！

來源

2016-12-15 mdimarca

您是否嘗試過使用我們的SDK？它使得你試圖做的事情變得相當簡單。 https://github.com/square/SquareRegisterSDK-iOS – tristansokol

我的原生iOS應用使用'SquareRegisterSDK'可可豆。請注意，我的應用程序是第三方應用程序，可將用戶放到Square進行付款，然後將其彈回到我的應用程序中。 – mdimarca

我想象的問題是將我創建的NSDictionary與數據轉換成字符串，但儘管修補了它仍然沒有得到它的工作。任何想法可能來自哪個問題？ – mdimarca

這是我使用的代碼工作：

NSDictionary *squareDictionary = @{@"callback_url": CALLBACK_URL, 
            @"client_id": CLIENTID, 
            @"version": @"1.2", 
            @"notes": notes, 
            @"state": name, 
            @"amount_money": 
             @{@"amount": amountString, 
             @"currency_code":@"USD" 
             }, 
            @"options": 
             @{@"supported_tender_types": @[@"CREDIT_CARD"], 
             @"auto_return": @"true" 
             } 
            }; 

NSError *error; 
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:squareDictionary 
                options:NSJSONWritingPrettyPrinted 
                error:&error]; // Pass 0 if you don't care about the readability of the generated string 

NSString *jsonString; 
if (! jsonData) { 
    NSLog(@"Got an error: %@", error); 
} else { 
    jsonString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding]; 

    NSString *encodedString = [jsonString stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]]; //encode the string 

    NSString *scheme = [NSString stringWithFormat:@"square-commerce-v1://payment/create?data=%@", encodedString]; //input the string to the url 

    UIApplication *application = [UIApplication sharedApplication]; 
    NSURL *URL = [NSURL URLWithString:scheme]; 
    BOOL canOpen = [application canOpenURL:URL]; //open the url 
    if (canOpen) { 
     [application openURL:URL]; 
    } else { 
     NSLog(@"couldn't open app"); 
    } 
}

來源

2016-12-17 20:59:40 mdimarca

iOS Square Register Popover OpenURL問題

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