IOS JSON從Mandrill發送郵件

我有一個IOS應用程序，我想通過Mandrill發送一封電子郵件。我試圖實現這一點，但它不工作，我迷惑了自己。IOS JSON從Mandrill發送郵件

按下按鈕時，從IOS應用程序，我記錄此錯誤消息發送一封電子郵件：

{"status":"error","code":-1,"name":"ValidationError","message":"You must specify a key value"}

我的代碼是：

NSString *post = [NSString stringWithFormat:@"{\"key\": \"abcdefg123456\", \"raw_message\": \"From: [email protected]\nTo: [email protected]\nSubject: Some Subject\n\nSome content.}"]; 
NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES]; 

NSString *postLength = [NSString stringWithFormat:@"%d", [postData length]]; 

NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init]; 
[request setURL:[NSURL URLWithString:@"https://mandrillapp.com/api/1.0/messages/send-raw.json"]]; 
[request setHTTPMethod:@"POST"]; 
[request setValue:postLength forHTTPHeaderField:@"Content-Length"]; 
[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"]; 
[request setHTTPBody:postData]; 
    NSLog(@"Post: %@", post); 

NSURLResponse *response; 
NSData *POSTReply = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:nil]; 
NSString *theReply = [[NSString alloc] initWithBytes:[POSTReply bytes] length:[POSTReply length] encoding: NSASCIIStringEncoding]; 
NSLog(@"Reply: %@", theReply);

請讓我知道我錯了。。

來源

2014-07-13 Steve

看起來你忘了\內容」後「」。

嘗試寫你的「後」變量如下：

NSString *post = [NSString stringWithFormat:@"{\"key\": \"abcdefg123456\", \"raw_message\": \"From: [email protected]\nTo: [email protected]\nSubject: Some Subject\n\nSome content.\"}"];

我希望它能幫助

來源

2014-07-13 17:50:20 scollaco

感謝我做在內容後錯過了\我還需要在每個\ n之前添加一個\ – Steve

IOS JSON從Mandrill發送郵件

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