Question

我试图发布到我的rails api，但是当我发布到rails时，username参数是username =＆lt; UITextField。 Rails正在使用json。

当我使用卷曲时，它完美地运作

curl -H "Content-Type: application/json" -d '{"username":"test","password":"test"}' http://0.0.0.0:3000/api/v1/users/signin

目标C

-(IBAction)Login:(id)sender{

    NSMutableData *data = [[NSMutableData alloc] init];
    self.receivedData = data;

    NSURL *url = [NSURL URLWithString:@"http://0.0.0.0:3000/api/v1/users/signin"];

    NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[url standardizedURL]];

    //set http method
    [request setHTTPMethod:@"POST"];

    //initialize a post data
    NSString *postData = [NSString stringWithFormat:@"username=%@&password=%@", username, password];

    //set request content type we MUST set this value.
    [request setValue:@"application/json; charset=utf-8" forHTTPHeaderField:@"Content-Type"];

    [request setHTTPBody:[postData dataUsingEncoding:NSUTF8StringEncoding]];

    //initialize a connection from request
    NSURLConnection *connection = [[NSURLConnection alloc] initWithRequest:request delegate:self];
    [connection start];


}

Rails API

def authenticate

        user = User.find_by_username(params[:username])

        respond_to do |format|
          if user && user.authenticate(params[:password])
            format.json { render json: "Signed In"}
          else
            format.json { render json: "Wrong username or password"}
          end
        end
      end

Answer 1

用户名和密码似乎是文本字段，因此您无法提交对象。为此，您必须在对象

中提交文本

NSString *postData = [NSString stringWithFormat:@"username=%@&password=%@", username.text, password.text];

将目标c形式发布到rails api json响应

1 个答案: