获取从另一个用户运行的应用程序的包标识符

Posted 2023-03-04

【中文标题】获取从另一个用户运行的应用程序的包标识符

【英文标题】：get the bundle identifier of an application running from another user

【发布时间】：2013-11-20 06:46:22

【问题描述】：

场景是这样的：“我从一个用户运行一个应用程序（比如 myproc），然后快速用户切换到第二个用户” 现在，当我尝试确定使用特定捆绑标识符运行的所有进程时（比如 com.ak.myproc）；对于从第一个用户运行的进程，我无法确定这一点。

我尝试了以下方法但徒劳无功：

[NSRunningApplication runningApplicationsWithBundleIdentifier:] [[NSWorkspace sharedWorkspace] runningApplications]，然后比较每个应用程序的包标识符 - 为第一个用户运行的应用程序甚至没有出现在此列表中。 使用sysctl()，然后遍历进程列表 - 这里，来自第一个用户的应用程序 pid 确实来了。在那之后： 当我尝试[NSRunningApplication runningApplicationWithProcessIdentifier:] 时，我得到了 nil。 当我尝试 GetProcessForPID() 后跟 ProcessInformationCopyDictionary() 时，我得到一个 nil 字典。 当我尝试 GetProcessForPID() 后跟 GetProcessInformation() 时，我在 ProcessInfoRec 中没有得到任何有用的信息。

有人可以帮忙吗？谢谢。

操作系统：Mac OS X 10.8.4 Xcode：4.6.2

【问题讨论】：

另外here 是来自类似问题的参考。

【参考方案1】：

您可以使用NSWorkspace 将进程名称映射到捆绑ID。

#include <sys/sysctl.h>
#include <pwd.h>
typedef struct kinfo_proc kinfo_proc;
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

    kinfo_proc *mylist =NULL;
    size_t mycount = 0;
    GetBSDProcessList(&mylist, &mycount);

    NSMutableArray *processes = [NSMutableArray arrayWithCapacity:(int)mycount];

    for (int i = 0; i < mycount; i++) 
        struct kinfo_proc *currentProcess = &mylist[i];
        struct passwd *user = getpwuid(currentProcess->kp_eproc.e_ucred.cr_uid);
        NSMutableDictionary *entry = [NSMutableDictionary dictionaryWithCapacity:4];

        NSNumber *processID = [NSNumber numberWithInt:currentProcess->kp_proc.p_pid];
        NSString *processName = [NSString stringWithFormat: @"%s",currentProcess->kp_proc.p_comm];
        if (processID)[entry setObject:processID forKey:@"processID"];
        if (processName)[entry setObject:processName forKey:@"processName"];
        if (processName)
        
            NSString *bunldeID = [self bundleIdentifierForApplicationName:processName];
            if (bunldeID)
                [entry setObject:bunldeID forKey:@"bundleId"];
        
        if (user)
            NSNumber *userID = [NSNumber numberWithUnsignedInt:currentProcess->kp_eproc.e_ucred.cr_uid];
            NSString *userName = [NSString stringWithFormat: @"%s",user->pw_name];

            if (userID)[entry setObject:userID forKey:@"userID"];
            if (userName)[entry setObject:userName forKey:@"userName"];
        
        [processes addObject:[NSDictionary dictionaryWithDictionary:entry]];
    
    free(mylist);

    return [NSArray arrayWithArray:processes];

+ (NSString *) bundleIdentifierForApplicationName:(NSString *)appName

    NSWorkspace * workspace = [NSWorkspace sharedWorkspace];
    NSString * appPath = [workspace fullPathForApplication:appName];
    if (appPath) 
        NSBundle * appBundle = [NSBundle bundleWithPath:appPath];
        return [appBundle bundleIdentifier];
    
    return nil;

【讨论】：

感谢@Parag，它适用于许多应用程序，如 Google Chrome、Wireshark、Perforce 和 Apple 应用程序，但不适用于 Adob​​e Reader 等应用程序。为什么会发生这种情况？

查看Adobe Reader的info.plist文件

我比较了 Safari 和 Adob​​e Reader 的 info.plist。他们看起来非常相似。它们是读者 plist 中的两个额外键，即“应用程序需要 Carbon 环境”和“应用程序需要本机环境”。我删除了它们，但结果仍然相同。无论如何，我已经为此打开了一个支持案例 - 一旦我得到答案就会更新。谢谢。