Ste*_*eve 6 email json ios mandrill
我有一个IOS应用程序,我想通过Mandrill发送电子邮件.我试图实现这一点,但它不起作用,我混淆了自己.
当按下按钮从IOS应用程序发送电子邮件时,我记录此错误消息:
{"status":"error","code":-1,"name":"ValidationError","message":"You must specify a key value"}
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我的代码是:
NSString *post = [NSString stringWithFormat:@"{\"key\": \"abcdefg123456\", \"raw_message\": \"From: me@mydomain.com\nTo: me@myotherdomain.com\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);
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请让我知道我哪里出错了.
它看起来你忘记了"后"的内容."
尝试编写"post"变量,如下所示:
NSString *post = [NSString stringWithFormat:@"{\"key\": \"abcdefg123456\", \"raw_message\": \"From: me@mydomain.com\nTo: me@myotherdomain.com\nSubject: Some Subject\n\nSome content.\"}"];
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我希望它有所帮助.
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