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! :)
CodePudding user response:
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}'))