Question

我一直在寻找一种在iPhone上启动守护进程的方法，我通过学习ants application的源代码创建了一个使用Xcode的小测试应用程序，该代码告诉我应该使用launchctl但不幸的是它不起作用。

我已在/Applications/的iPod Touch上使用SSH安装了我的应用程序，然后我通过帐户mobile使用SSH启动它，我的日志说明了这一点：

Script started on Thu Feb 24 19:33:28 2011
bash-3.2$ ssh mobile@192.168.1.8
mobile@192.168.1.8's password: 
iPod-van-Henri:~ mobile$ cd /Applications
iPod-van-Henri:/Applications mobile$ cd DaemonUtility.app/
iPod-van-Henri:/Applications/DaemonUtility.app mobile$ ./DaemonUtility 
2011-02-24 19:35:08.022 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:09.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:10.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:11.021 DaemonUtility[1369:107] Read 0 bytes
Bug: launchctl.c:2367 (24307):13: (dbfd = open(g_job_overrides_db_path, O_RDONLY | O_EXLOCK | O_CREAT, S_IRUSR | S_IWUSR)) != -1
launchctl: CFURLWriteDataAndPropertiesToResource(/private/var/stash/Applications.pwn/DaemonUtility.app/com.developerief2.daemontest.plist) failed: -10
launch_msg(): Socket is not connected
2011-02-24 19:35:12.039 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:13.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:14.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:15.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:16.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:17.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:18.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:19.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:20.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:21.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:22.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:23.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:24.021 DaemonUtility[1369:107] Read 0 bytes
2011-02-24 19:35:25.021 DaemonUtility[1369:107] Read 0 bytes
^C
iPod-van-Henri:/Applications/DaemonUtility.app mobile$ exit
logout
Connection to 192.168.1.8 closed.

bash-3.2$ exit
exit

Script done on Thu Feb 24 19:34:49 2011

当我使用root（使用su执行）启动它时，我会让守护程序运行，但它不会执行任何操作。

我的守护进程应该在启动后每隔十秒显示UIViewAlert：

**main.m (Daemon)**
//
//  main.m
//  DaemonTest
//
//  Created by ief2 on 23/02/11.
//

#import <UIKit/UIKit.h>

@interface DAAppDelegate : NSObject <UIApplicationDelegate> {
    NSDate *_startupDate;
    NSTimer *_messageTimer;
}
@property (nonatomic, retain) NSDate *startupDate;
@end

@interface DAAppDelegate (PrivateMethods)
- (void)showMessage:(NSTimer *)timer;
@end

@implementation DAAppDelegate
@synthesize startupDate=_startupDate;

- (void)dealloc {
    [_startupDate dealloc];
    [_messageTimer dealloc];

    [super dealloc];
}

- (void)applicationDidFinishLaunching:(UIApplication *)theApplication {
    UIAlertView *myView;
    myView = [[UIAlertView alloc] initWithTitle:@"Daemon Launched" 
                                        message:@"The daemon was launched"
                                       delegate:nil
                              cancelButtonTitle:@"OK"
                              otherButtonTitles:nil];
    [myView show];
    [myView release];

    self.startupDate = [NSDate date];

    NSTimer *myTimer = [NSTimer scheduledTimerWithTimeInterval:10 
                                                        target:self
                                                      selector:@selector(showMessage:) 
                                                      userInfo:nil
                                                       repeats:YES];
    _messageTimer = [myTimer retain];
}

- (void)applicationWillTerminate:(UIApplication *)theApplication {
    [_messageTimer invalidate];

    UIAlertView *myView;
    myView = [[UIAlertView alloc] initWithTitle:@"Daemon Terminated" 
                                        message:@"The daemon was terminated"
                                       delegate:nil
                              cancelButtonTitle:@"OK"
                              otherButtonTitles:nil];
    [myView show];
    [myView release];
}

- (void)showMessage:(NSTimer *)timer {
    NSTimeInterval mySec;
    mySec = [self.startupDate timeIntervalSinceNow];
    NSString *format = [NSString stringWithFormat:
                        @"The daemon has been running for %llu seconds",
                        (unsigned long long)mySec];
    UIAlertView *myView;
    myView = [[UIAlertView alloc] initWithTitle:@"Daemon Message" 
                                        message:format 
                                       delegate:nil 
                              cancelButtonTitle:@"OK" 
                              otherButtonTitles:nil];
    [myView show];
    [myView release];
}
@end


int main(int argc, const char **argv) {
    NSAutoreleasePool *mainPool = [[NSAutoreleasePool alloc] init];

    UIApplicationMain(argc, (char **)argv, nil, @"DAAppDelegate");

    [mainPool drain];
    return 0;
}

可以在我的电脑上找到完整应用程序的源代码：
http://81.82.20.197/DaemonTest.zip

提前谢谢你，
ief2

Answer 1

你工作太辛苦了。您需要做的就是创建一个带有应用程序标识符和路径的.plist文件，并将其添加到/ System / Library / LaunchDaemon文件夹中。然后确保您的应用程序位于/ Applications文件夹中。重启，每次手机启动时都会有效。

Google“Chris Alvares daemon”并查看他的教程...

Answer 2

我认为launchD不能触发GUI级应用。任何“Aqua”级别都必须是“StartupItem”或“Login Item”。您仍然可以根据它们的起始位置以及它们所属的位置以root身份启动它们，但是launchd不会触及那些东西...我不认为如果要启动它，您甚至可以使用菜单栏图标处理它....

如果你谈论越狱iphone ...如果你想从“mobileterminal”打开一个GUI应用程序，你应该在Cydia中寻找应用程序那个“这样做“。它并不像启动可执行文件那么容易..有一些时髦的跳板交互..该实用程序负责。它已被调用......“ AppsThruTerm ”（bigboss repo）一旦安装..你用att blahblahblah

命令启动你的“app”

为越狱的iOS制作一个守护进程

2 个答案: