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从命令行启动 GUI 应用程序

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从命令行启动 GUI 应用程序

在 iOS 中，我使用启动命令从终端（通过 SSH）执行 GUI 应用程序。但是，它失败并出现以下错误：

$ launch com.apple.MobileSafari
2011-07-28 18:20:11.433 launch[6943:707] *** Assertion failure in -[UIApplication init], /SourceCache/UIKit/UIKit-1448.89/UIApplication.m:528
2011-07-28 18:20:11.443 launch[6943:707] *** Terminating app due to uncaught exception 'NSInternalInconsistencyException', reason: 'There can only be one UIApplication instance.'
*** Call stack at first throw:
(
    0   CoreFoundation                      0x325be64f __exceptionPreprocess + 114
    1   libobjc.A.dylib                     0x34f96c5d objc_exception_throw + 24
    2   CoreFoundation                      0x325be491 +[NSException raise:format:arguments:] + 68
    3   Foundation                          0x34ded573 -[NSAssertionHandler handleFailureInMethod:object:file:lineNumber:description:] + 62
    4   UIKit                               0x32144181 -[UIApplication init] + 128
    5   launch                              0x00001d90 main + 252
    6   launch                              0x00001bec _start + 312
    7   launch                              0x00001a88 start + 24
)
terminate called after throwing an instance of 'NSException'
Abort trap: 6

如何使用此命令执行应用程序？

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