我需要在iOS中的twilio documentation发送与此相当的POST请求以发送短信
curl -XPOST https://api.twilio.com/2010-04-01/Accounts/AC5ef8732a3c49700934481addd5ce1659/Messages.json \
-d "Body=Jenny%20please%3F%21%20I%20love%20you%20<3" \
-d "To=%2B15558675309" \
-d "From=%2B14158141829" \
-u 'AC5ef8732a3c49700934481addd5ce1659:{AuthToken}'
这就是我所拥有的:
NSString *kTwilioSID = @"AC5ef8732a3c49700934481addd5ce1659";
NSString *kTwilioSecret = @"SECRET";
NSString *kFromNumber = @"+14158141829";
NSString *kToNumber = @"+15558675309";
NSString *kMessage = @"Jenny please?! I love you <3";
NSString *twilioURL = [NSString stringWithFormat:@"https://%@:%@@api.twilio.com/2010-04-01/Accounts/%@/Messages.json", kTwilioSID, kTwilioSecret, kTwilioSID];
NSDictionary *params = @ {@"account" :kTwilioSID, @"from":kFromNumber, @"to":kToNumber, @"body":kMessage };
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
[manager POST:twilioURL parameters:params
success:^(AFHTTPRequestOperation *operation, id responseObject)
{
NSLog(@"Success: %@", responseObject);
}
failure:
^(AFHTTPRequestOperation *operation, NSError *error) {
NSLog(@"Error: %@", error);
}];
错误失败,&#34;请求失败:错误请求(400)&#34;,并且短信也没有被发送 我该如何解决这个问题?
答案 0 :(得分:1)
我建议你不要使用
[manager POST:twilioURL parameters:params ....
这种方法对我不起作用。相反,这里是一个有效的代码示例:
NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:@"%@", @"http://myurl/actionToperform"]];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
//1
[request setHTTPMethod:@"POST"];
[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
NSMutableData *postBody = [NSMutableData data];
[postBody appendData:[[NSString stringWithFormat:@"{ \"name\" : \"%@\", \"polygon\" : [%@]}",[alertView textFieldAtIndex:0].text, concatCoord] dataUsingEncoding:NSUTF8StringEncoding]];
//post
[request setHTTPBody:postBody];
AFHTTPRequestOperation *operation = [[AFHTTPRequestOperation alloc] initWithRequest:request];
operation.responseSerializer = [AFJSONResponseSerializer serializer];
[operation setCompletionBlockWithSuccess:^(AFHTTPRequestOperation *operation, id responseObject) {
//Do things if succesfull
NSLog(@"%@", responseObject);
} failure:^(AFHTTPRequestOperation *operation, NSError *error) {
//Do things when you've got an error
NSLog(@"error: %@", error);
}];
[operation start];
请注意,我在此代码中使用JSON来格式化我的参数(1)。
祝你好运。答案 1 :(得分:0)
此问题可以修复,但您必须在serializerRequest
对象上设置AFHTTPRequestOperationManager
。
如果我更正了此错误的原因是您没有设置标题字段。做这样的事情。
AFJSONRequestSerializer *requestSerializer = [AFJSONRequestSerializer serializer];
NSString pairString = [NSString stringWithFormat: @"%@", someRandomId];
NSData pairData = [pairString dataUsingEncoding NSUTF8StringEncoding];
NSString *serializerVal = [NSString stringWithFormat:@"%@", [pairData base64EncodedStringWithOptions:0];
[requestSerializer setValue: serializerVal forHTTPHeaderField:@"Auth"];
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
[manager setRequestSerializer:requestSerializer];