使用 Python 从 Windows 开始菜单获取可执行路径

Question

I'm developing a user-friendly program from college, and we wanted our Python script to open a program for the user. Most users won't know where their executables are stored precisely, so we were wondering if there was a way for us to get the paths we need through the Windows Start Menu? (Every program that shows up when you search for it on Windows has a shortcut saved on the Start Menu). Thanks for your time! :)

Answer 1

I was able to come up with a crude solution based on some other answers:

def get_shortcut_path(path: str) -> str:
    target = ''

    with open(path, 'rb') as stream:
        content = stream.read()
        # skip first 20 bytes (HeaderSize and LinkCLSID)
        # read the LinkFlags structure (4 bytes)
        lflags = struct.unpack('I', content[0x14:0x18])[0]
        position = 0x18
        # if the HasLinkTargetIDList bit is set then skip the stored IDList 
        # structure and header
        if (lflags & 0x01) == 1:
            position = struct.unpack('H', content[0x4C:0x4E])[0] + 0x4E
        last_pos = position
        position += 0x04
        # get how long the file information is (LinkInfoSize)
        length = struct.unpack('I', content[last_pos:position])[0]
        # skip 12 bytes (LinkInfoHeaderSize, LinkInfoFlags, and VolumeIDOffset)
        position += 0x0C
        # go to the LocalBasePath position
        lbpos = struct.unpack('I', content[position:position+0x04])[0]
        position = last_pos + lbpos
        # read the string at the given position of the determined length
        size= (length + last_pos) - position - 0x02
        temp = struct.unpack('c' * size, content[position:position+size])
        target = ''.join([chr(ord(a)) for a in temp])
        
        return target

def get_exe_path(app: str) -> str:
    username = getpass.getuser()
    start_menu = f'C:\\Users\\{username}\\AppData\\Roaming\\Microsoft\\Windows\\Start Menu\\Programs'
    
    for subdir, dirs, files in os.walk(start_menu):
        if not(subdir.startswith('Windows')):
            for file in files:
                if file.endswith('.lnk'):
                    print(get_shortcut_path(f'{subdir}/{file}'))