希望这可以解决问题:
- (void)openFile:(NSString *)filePath withTask:(NSTask *)task {
int pid = [task processIdentifier];
NSAppleEventDescriptor *target = [NSAppleEventDescriptor descriptorWithDescriptorType:typeKernelProcessID bytes:&pid length:sizeof(pid)];
const char *urlUTF8 = [[[NSURL fileURLWithPath:filePath] absoluteString] UTF8String];
NSAppleEventDescriptor *urlDescriptor = [NSAppleEventDescriptor descriptorWithDescriptorType:typeFileURL bytes:urlUTF8 length:strlen(urlUTF8)];
NSAppleEventDescriptor *event = [NSAppleEventDescriptor appleEventWithEventClass:kEventParamAppleEvent eventID:kAEOpen targetDescriptor:target returnID:kAutoGenerateReturnID transactionID:kAnyTransactionID];
[event setParamDescriptor:urlDescriptor forKeyword:keyDirectObject];
OSStatus err = AESendMessage([event aeDesc], NULL, kAENoReply | kAENeverInteract, kAEDefaultTimeout);
if (err != noErr) {
// Error handling goes here
}
// Activate the application
event = [NSAppleEventDescriptor appleEventWithEventClass:kAEMiscStandards eventID:kAEActivate targetDescriptor:target returnID:kAutoGenerateReturnID transactionID:kAnyTransactionID];
err = AESendMessage([event aeDesc], NULL, kAENoReply | kAENeverInteract, kAEDefaultTimeout);
}
您可能必须使用 an 启动应用程序NSTask
,然后向其发送适当的 open Apple Event。
实际上,您可以使用 an 启动
NSTask
,然后在
NSWorkspace
知道文件正在运行后通过它打开文件吗?或者这是否会在您的用户下启动应用程序的新实例?
原回复:
做
open -a SomeApplication SomeFile
工作?