我只想打开一个应用程序实例一次,所以我就像
NSArray *apps = [NSRunningApplication runningApplicationsWithBundleIdentifier:@"com.my.app"];
在 main.m 中一样。
当我从控制台打开应用程序时,应用程序数组的计数将为 0。但是,当我双击应用程序时,它将是1
那么谁能告诉我双击和控制台打开有什么区别?或者给我另一种方法来检查是否已经有一个实例在运行?
问问题
289 次
2 回答
1
该命令查询工作区并“延迟”更新
当应用程序从 finder 启动时,它会通过启动NSWorkspace
,因此工作区会立即更新
当应用程序通过控制台/xcode 启动时,它不会通过 启动 NSWorkspace
,因此该类在开始时返回 0。在NSApplication
您的进程启动后,将通知工作区及其 1。
=> 它总是正确的- (void)applicationDidFinishLaunching:(NSNotification *)aNotification
所以要么等待 NSApplication 启动然后杀死它(就像你现在做的那样,但稍后)
或者
请参阅防止 Linux 上的多个进程实例,以了解无需可可的方法
或者
你看看可以做到这一点的launchd :) http://developer.apple.com/library/mac/#documentation/MacOSX/Conceptual/BPSystemStartup/Chapters/CreatingLaunchdJobs.html
于 2012-12-28T01:53:13.040 回答
-1
您可以使用以下代码。 getBSDProcessList
函数将返回NSArray
正在运行的进程。[1]
static int GetBSDProcessList(kinfo_proc **procList, size_t *procCount)
// Returns a list of all BSD processes on the system. This routine
// allocates the list and puts it in *procList and a count of the
// number of entries in *procCount. You are responsible for freeing
// this list (use "free" from System framework).
// On success, the function returns 0.
// On error, the function returns a BSD errno value.
{
int err;
kinfo_proc * result;
bool done;
static const int name[] = { CTL_KERN, KERN_PROC, KERN_PROC_ALL, 0 };
// Declaring name as const requires us to cast it when passing it to
// sysctl because the prototype doesn't include the const modifier.
size_t length;
// assert( procList != NULL);
// assert(*procList == NULL);
// assert(procCount != NULL);
*procCount = 0;
// We start by calling sysctl with result == NULL and length == 0.
// That will succeed, and set length to the appropriate length.
// We then allocate a buffer of that size and call sysctl again
// with that buffer. If that succeeds, we're done. If that fails
// with ENOMEM, we have to throw away our buffer and loop. Note
// that the loop causes use to call sysctl with NULL again; this
// is necessary because the ENOMEM failure case sets length to
// the amount of data returned, not the amount of data that
// could have been returned.
result = NULL;
done = false;
do {
assert(result == NULL);
// Call sysctl with a NULL buffer.
length = 0;
err = sysctl( (int *) name, (sizeof(name) / sizeof(*name)) - 1,
NULL, &length,
NULL, 0);
if (err == -1) {
err = errno;
}
// Allocate an appropriately sized buffer based on the results
// from the previous call.
if (err == 0) {
result = malloc(length);
if (result == NULL) {
err = ENOMEM;
}
}
// Call sysctl again with the new buffer. If we get an ENOMEM
// error, toss away our buffer and start again.
if (err == 0) {
err = sysctl( (int *) name, (sizeof(name) / sizeof(*name)) - 1,
result, &length,
NULL, 0);
if (err == -1) {
err = errno;
}
if (err == 0) {
done = true;
} else if (err == ENOMEM) {
assert(result != NULL);
free(result);
result = NULL;
err = 0;
}
}
} while (err == 0 && ! done);
// Clean up and establish post conditions.
if (err != 0 && result != NULL) {
free(result);
result = NULL;
}
*procList = result;
if (err == 0) {
*procCount = length / sizeof(kinfo_proc);
}
assert( (err == 0) == (*procList != NULL) );
return err;
}
+ (NSArray*)getBSDProcessList
{
NSMutableArray *ret = [NSMutableArray arrayWithCapacity:1];
kinfo_proc *mylist;
size_t mycount = 0;
mylist = (kinfo_proc *)malloc(sizeof(kinfo_proc));
GetBSDProcessList(&mylist, &mycount);
int k;
for(k = 0; k < mycount; k++) {
kinfo_proc *proc = NULL;
proc = &mylist[k];
NSString *fullName = [[self infoForPID:proc->kp_proc.p_pid] objectForKey:(id)kCFBundleNameKey];
NSLog(@"fullName %@", fullName);
if (fullName != nil)
{
[ret addObject:fullName];
}
}
free(mylist);
return ret;
}
+ (NSDictionary *)infoForPID:(pid_t)pid
{
NSDictionary *ret = nil;
ProcessSerialNumber psn = { kNoProcess, kNoProcess };
if (GetProcessForPID(pid, &psn) == noErr) {
CFDictionaryRef cfDict = ProcessInformationCopyDictionary(&psn,kProcessDictionaryIncludeAllInformationMask);
ret = [NSDictionary dictionaryWithDictionary:(NSDictionary *)cfDict];
CFRelease(cfDict);
}
return ret;
}
查看技术问答 QA1123(获取 Mac OS X 上所有进程的列表)
于 2012-12-28T06:23:41.460 回答