objective-c - 我需要从 Swift 调用 Objective C 方法

Question

我在 Objective C 中有这个接口(interface)，我试图从 Swift 调用它，但它似乎不属于完成代码。不知道我错过了什么!

@interface WPLogin : NSObject

/*!
 * @brief Logs into the application returning success or failure with an error     object.
 * If login is successful the credential is automatically stored
 *
 * @param serverURLString   Address to login to
 * @param username  Username to login with
 * @param password  Password for user
 * @param completion Completion block to be called after attempting login
 */
- (void)loginToURL:(NSString *)serverURLString
      withUsername:(NSString *)username
              password:(NSString *)password
        completion:(void (^)(WPLoginStatus success, NSError * error))completion;

@end

Swift 的调用函数看起来像这样......

    var uid = "test"
    var pwd = "test"
    var url = "http://www.google.com"
    var loginAuth = WPLogin();

        loginAuth.loginToURL(url, withUsername: uid, password: pwd, completion: { (status:WPLoginStatus, error:NSError!) -> Void in
            println("Inside Login")
    })

桥接文件

#import <Foundation/Foundation.h>
#import <UIKit/UIKit.h>
#import "WPLogin.h"

适用于 Objective C 的版本

if (!self.login) {
    self.login = [WPLogin new];
}

[self.login loginToURL:url
                       withUsername:uid
                           password:pwd
                         completion:^(WPLoginStatus status, NSError *error) {
                                NSLog(@"complete");
                          }];

Answer 1

问题在于，在您的 Swift 版本中，WPLogin 对象是一个本地变量 (var loginAuth)。因此，它在有机会做任何事情之前就死了。将其设置为属性，如您的 Objective-C 版本 (self.login):

class MyClass {
    var login : WPLogin = WPLogin()
    func myMethod () {
        var uid = "test"
        var pwd = "test"
        var url = "http://www.google.com"
        self.login.loginToURL(url, withUsername: uid, password: pwd, completion: { 
            (status:WPLoginStatus, error:NSError!) -> Void in
            println("Inside Login")
        })
    }
}

这个对象需要随着时间的推移采取行动 - 上网，登录，调用你的完成处理程序 - 如果它立即死亡，它就无法做到这一点，就像你的 Swift 代码中那样。它必须持续很长时间。