import os

letters = input("Enter the first 3 letters : ")
letters = letters.upper()
num_start = 1
zero = "00"
num_start_len = len(str(num_start))
print("num start length is ", num_start_len)

dir_look = (r"C:\Users\user\Documents\Client Drawings\\" + letters + zero + str(num_start))
check_for_folder = os.path.isdir(dir_look)
print(check_for_folder)

# working out how many zero digits required

if  num_start_len == 3 :
    zero = ""
    print("If" + zero)

elif num_start_len == 2 :
     zero = "0"
     print("elif" + zero)

else:
    print("else" + zero)

# make directory

if not check_for_folder :
        os.makedirs(dir_look)
        print("created folder : ", dir_look)

else:
    if check_for_folder:
        num_start += 1
        # os.makedirs(dir_look)
        print("Num start value", num_start)
        print(dir_look)

    else:
        print("end")

print(dir_look)

Output:
Enter the first 3 letters : aaa   
num start length is  1   
True   
else00   
Num start value 2   
C:\Users\user\Documents\Client Drawings\\AAA001   
C:\Users\user\Documents\Client Drawings\\AAA001   

Process finished with exit code 0   

Quote:I assumed if the nested variable updates the other would automatically.

name = "Sally"
mystring = "Hello, " + name
name = "Miguel" # Does not affect the mystring value
print(mystring) # prints "Hello Sally"

import os
from pathlib import Path

drawing_dir = Path("C:", "Users", "user", "Documents", "Client Drawings")
letters = input("Enter the first 3 letters: ")
letters = letters.upper()
for digits in range(1000):
    dir_name = drawing_dir / f"{letters}{digits:03}"
    try:
        os.mkdir(dir_name)
        print (f"Created {dir_name} as next free directory")
        break
    except FileExistsError:
        continue
else:
    print(f"All three-digit numbers taken for {letters}")

(Aug-27-2021, 07:42 PM)Teknohead23 Wrote: If I use a while loop, which I think I should, I get an infinite loop as it will always be true

import os

letters = input("Enter the first 3 letters : ").upper()
num_start = 1
for index,item in enumerate(letters, num_start):
    dir_look = fr"C:\Users\user\Documents\Client Drawings\{letters}{index:03}"
    print(dir_look)
    check_for_folder = os.path.isdir(dir_look)
    print(check_for_folder

Output:
Enter the first 3 letters : aaa
C:\Users\user\Documents\Client Drawings\AAA001
False
C:\Users\user\Documents\Client Drawings\AAA002
False
C:\Users\user\Documents\Client Drawings\AAA003
False

(Aug-27-2021, 08:34 PM)snippsat Wrote:

for index,item in enumerate(letters, num_start):

(Aug-27-2021, 08:34 PM)snippsat Wrote:

(Aug-27-2021, 07:42 PM)Teknohead23 Wrote: If I use a while loop, which I think I should, I get an infinite loop as it will always be true

A for loop is always the first choice in Python,here a example with enumerate and f-string which you should use.

import os

letters = input("Enter the first 3 letters : ").upper()
num_start = 1
for index,item in enumerate(letters, num_start):
    dir_look = fr"C:\Users\user\Documents\Client Drawings\{letters}{index:03}"
    print(dir_look)
    check_for_folder = os.path.isdir(dir_look)
    print(check_for_folder

Output:
Enter the first 3 letters : aaa
C:\Users\user\Documents\Client Drawings\AAA001
False
C:\Users\user\Documents\Client Drawings\AAA002
False
C:\Users\user\Documents\Client Drawings\AAA003
False

(Aug-27-2021, 08:01 PM)bowlofred Wrote:

Quote:I assumed if the nested variable updates the other would automatically.

No, that's not how python str()s work. They are just a sequence of characters. It's not a dynamic formula that updates. I wouldn't even call this a "nested" variable. This is just a static string that is created from the values of other strings. If the other strings change in the future, it doesn't affect the one you built.

name = "Sally"
mystring = "Hello, " + name
name = "Miguel" # Does not affect the mystring value
print(mystring) # prints "Hello Sally"

---

Something like this might work. I'm not sure what logic you were using for determining the number of digits that could be used. This one just hardcodes it to exactly 3. But making that dynamic on some input or other signal would not be difficult.

You can't create a directory with a conflicting name. As long as you're talking a small number of directories (1000 is still small), then it makes sense to just try to create the next one and handle the failure if it exists.

import os
from pathlib import Path

drawing_dir = Path("C:", "Users", "user", "Documents", "Client Drawings")
letters = input("Enter the first 3 letters: ")
letters = letters.upper()
for digits in range(1000):
    dir_name = drawing_dir / f"{letters}{digits:03}"
    try:
        os.mkdir(dir_name)
        print (f"Created {dir_name} as next free directory")
        break
    except FileExistsError:
        continue
else:
    print(f"All three-digit numbers taken for {letters}")