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why for loop test check only first item in dictionary

this user check only first item in dectionary
def update_details():# here the user choose which feild he want to update
input_details = input("please select which field you want to change: ")
for item in store_group_details: #check if feild exist in decionary
if item == input_details: # removed ['type']
update_all_feature(input_details) # send input_details to update_all_feature() function
break
else:
print ("your input dose not exist? ")
update_details()
following dictionry i want to eterate over it
'app_group2': ['slack', ' Discord', 'zoom', 'vs code'], 'Date': '2222-02-12'}

It's happening because you have the wrong if/else statement. When the first item is not in the list you break from the loop and run the whole function again without checking other items.
You have to put else one indentation before (to make for/else statement):
for item in store_group_details:
if item == input_details:
update_all_feature(input_details)
break
else:
print("your input does not exist? ")
update_details()

sharing a data between two functions in python

so i have a code which loops until the user types cancel and allows the user to input things into a list and then see the list if they want to. but i have a problem wherein if i want to see the list the user has to input again and the data i typed when i added to the list is gone. here is the code pls help me im pretty new to python.
answer=""
def addlist():
list=input("what would you like to add? ")
print("added successfully")
return list
def showlist():
list=addlist()
print(list)
while answer !="cancel":
answer=input("what would you like to do?, 1 for add to list, 2 for show list, cancel to close")
if answer=="1":
addlist()
elif answer=="2":
showlist()
else:
print("wrong value")

So we've got a few problems here:
addlist doesn't actually create a list, it creates a string.
showlist depends on addlist, when really we should be looking for a common list that we toss everything into (assuming you actually want to work with a list)
Your indentation in your while loop is incorrect
def addlist():
item = input("what would you like to add?")
return item
def showlist(mylist):
print(mylist)
mylist = []
answer = ""
while answer != "cancel":
answer = input("what would you like to do?, 1 for add to list, 2 for show list, cancel to close")
if answer == "1":
add_item = addlist()
mylist.append(add_item)
print("added successfully")
elif answer == "2":
showlist(mylist)
else:
print("wrong value")
That above seems to do the trick.
You seem to have a grasp on return statements, so how about you pass each of those functions a common list as I did mylist and have them do stuff with that.
I changed addlist to actually just get the item we want to add to the list and then return that item and append it outside of the function. In showlist I pass mylist to it via: showlist(mylist) and then print the list I get in that function.

You should create a list element in order to append your items in it. You can see more here. One way to do what you want is like this:
answer = ""
temp_list = []
def addlist():
item = input("what would you like to add? ")
temp_list.append(item)
print("added successfully")
def showlist():
print(temp_list)
while answer != "cancel":
answer = input(
"what would you like to do?, 1 for add to list, 2 for show list, cancel to close")
if answer == "1":
addlist()
elif answer == "2":
showlist()
else:
print("wrong value")

Adding items to shopping List Python

I'm learning Python and I'm trying to make shopping List
where you can add items
first it will ask you to add the items if the shopping list is empty
it will be added automatically if not it will ask you where you would like
to put the item (index)
but also I'm trying to make the program exit in certain condition like DONE
or HELP or SHOW but even that i put a condition for that but it's not working can anyone help me with this
hope I explained enough
import os
shopping_list =[]
# Function for clearing the screen
def clear_screen():
os.system("cls" if os.name == "nt" else "clear")
def show_help():
print("Enter 'Done' if you finish adding item \n Enter 'Show' to show your items \n Enter 'Help' toshow this help ")
# Function to show the items that you've entered to the list
def show_item():
clear_screen()
index = 1
for item in shopping_list:
print("{} {}.".format(index,item))
index += 1
# Function to add items to the list
def add_to_list():
while True:
new_item = input("Please enter the item that you would like to add to your shopping list ")
if shopping_list and ((new_item.upper() != "DONE") or (new_item.upper() != "HELP") or (new_item.upper() != "SHOW")):
position = input("Where you would like to add {} to the list \n press 'Enter' if you want to add to the end of the list".format(new_item))
position = abs(int(position))
shopping_list.insert(position - 1 , new_item)
show_item()
else:
if new_item.upper() == "DONE":
break
elif new_item.upper() == "SHOW":
show_item()
continue
elif new_item.upper() == "HELP":
show_help()
continue
else:
shopping_list.append(new_item)
show_item()
show_help()
add_to_list()

Welcome to stackoverflow. I think your logic statement is wrong, you need and instead of or. Right now all you need for the statement in the parentheses to be true, is that new_item.upper() is at least not one of those three words. It actually can't equate to False since two of the three are always true.
((new_item.upper() != "DONE") or (new_item.upper() != "HELP") or (new_item.upper() != "SHOW"))
If you have for example done the first statement is False ,but the other two are True, adding up to True in or-Statements.
>>> new_item = 'done'
>>> print((new_item.upper() != "DONE") or (new_item.upper() != "HELP") or (new_item.upper() != "SHOW"))
True

I'm having issues looping through a tuple

I'm writing a program to simulate a bank scenario. It allows users (once they have logged in) to display their balance, withdraw and deposit money, and will eventually contain a transfer subroutine.
My problem is the login subroutine. I am having trouble getting my code to loop through both accounts in the tuple that saves them. I can log in to the first account and perform all the code's functions with that account but when I try to access the second account, I get the error message telling me that the account is not found. I am very certain that this is to do with the way I have written my loop but I cannot figure out how to fix it.
If someone could provide me with a solution that I can use to fix this error then I would be very grateful. If it is possible, could the solution also stick to the general format of my code and not be one where I have to create new functions and change the entire body of my LogOn function?
Here is the code:
accounts=[
["Luke",'00001','1234',1337],
["Louis",'00002','4321',420],
]
name = ""
x = []
def LogOn():
global name
global x
print ("Welcome to the Bank of Bailey")
accnum = raw_input("-please enter your account number-")
if len(accnum) != 5:
print("That number is not valid, please enter a valid number")
LogOn()
else:
for x in range(0,len(accounts)):
if accnum in accounts[x][1]:
name = accounts[x][0]
print ("Account found")
pin = raw_input("-please enter your pin-")
if pin in accounts[x]:
print ("Pin found")
MainMenu()
else:
print("Pin not found")
print("Please try again")
LogOn()
break
else:
print("Account not found")
print("Please try again")
LogOn()

The problem is:
if accnum in accounts[x][1]:
do something
else:
print("Account not found")
print("Please try again")
LogOn()
The else branch is called already for the first iteration (with x = 0)!
You should check after the while-loop if an account was found and otherwise print your error message.
Instead of recursing to LogOn() you could better use an outer (endless-)while-loop which calls LogOn(). In LogOn then just do a return in error case.

A few observations:
The else branch is called on the first iteration when x=0. You need to finish the for loop, then branch if the account has not been found.
if pin in accounts[x] is a security hole because it allows people to use their name or account number as their pin!
"if accnum in account[1]" should be "if accnum == account[1]"
Hope that helps!
accounts=[
["Luke",'00001','1234',1337],
["Louis",'00002','4321',420],
]
name = ""
x = []
def LogOn():
global name
global x
print ("Welcome to the Bank of Bailey")
accnum = raw_input("-please enter your account number-")
if len(accnum) != 5:
print("That number is not valid, please enter a valid number")
LogOn()
else:
found = False
for account in accounts:
if accnum == account[1]:
found = True
name = account[0]
print ("Account found")
pin = raw_input("-please enter your pin-")
if pin in account[2]:
print ("Pin found")
MainMenu()
else:
print("Pin not found")
print("Please try again")
LogOn()
break
if not found:
print("Account not found")
print("Please try again")
LogOn()

I like #TimSC's answer, but I'll add a few more observations:
LogOn is not a good function name; it should be log_on to match PEP8 naming conventions (LogOn suggests it is a class rather than a function).
You declare name and x as globals, then never use them. Globals are almost always a bad idea, and unused declarations just clutter your code.
As #CTX pointed out, you are using recursion (LogOn calling LogOn calling MainMenu calling LogOn etc) where iteration would make a lot more sense - you should back out on an error, not call again a level deeper.
Mixing input/output code and action code is usually a bad decision; it makes your code less flexible and reusable and often harder to debug as well. You should instead have an input/output function which calls an action function with the values it needs.
Giving separate error messages for bad account number, bad pin number is a security hole; it makes it much easier for would-be attackers to find valid account numbers to attack.
Looking things up in a list is relatively slow (O(n) ie time proportional to the number of items in the list). Use a dict instead (O(1) ie constant time - a quick hash calculation takes you right to the item you are looking for).
Here is a fancied-up version; may it give you things to think about:
import sys
# version compatibility shim
if sys.hexversion < 0x3000000:
inp = raw_input # Python 2.x
else:
inp = input # Python 3.x
class Account:
index = {}
__slots__ = ("accnum", "pin", "name", "balance")
#classmethod
def log_in(cls, accnum, pin):
key = (accnum, pin)
acc = Account.index.get(key, None)
if acc is None:
raise ValueError("No such login (bad account# or PIN)")
else:
return acc
def __init__(self, accnum, pin, name, balance):
# save values
self.accnum = accnum
self.pin = pin
self.name = name
self.balance = balance
# add self to account index
key = (accnum, pin)
Account.index[key] = self
# create accounts
Account('00001', '1234', "Luke", 1337)
Account('00002', '4321', "Louis", 420)
def main():
while True:
print("\nWelcome to the Bank of Bailey")
accnum = inp("Please enter your account number: ").strip()
pin = inp("Please enter your PIN: ").strip()
try:
acc = Account.log_in(accnum, pin)
print("\nYour account has a current balance of ${:0.2f}".format(acc.balance))
print("Thank you for banking with us!")
except ValueError as ve:
print(ve)
if __name__ == "__main__":
main()

Python Nested If

hey im making a simple little grocery list on Python. I know it's not the most eloquent... but I am just learning the syntax right now. I want to get into learning Django.
list = []
def makeList():
listing = True
while listing:
addTo = raw_input("Add to list: ")
if addTo == 'q':
listing = False
else:
list.append(addTo)
def checkList():
if check in list:
print "Yay there is " + check + " here"
else:
print "No you have not added that..."
addAnother = raw_input("Would you like to add it? ")
if str.lower(addAnother) == "yes":
list.append(check)
elif str.lower(addAnother) == "no":
print "Okay then here is your list."
print list
else:
print check
makeList()
check = raw_input("What item: ")
checkList()
I know its pretty complex and hard to understand O_o... but you can see that the nested if statement is not registering when you run it.
What is making it do this? I think that's the best way to ask this.

I've rewritten it a bit to make it cleaner and more Pythonic;
def get_list(prompt, halt):
lst = []
while True:
item = raw_input(prompt)
if item == halt:
return lst
else:
lst.append(item)
def check_list(lst, item):
if item in lst:
print('Yay there is {} here'.format(item))
return True
else:
print('No you have not added {}'.format(item))
return False
def get_yesno(prompt):
while True:
yesno = raw_input(prompt).lower()
if yesno in {'y', 'yes'}:
return True
elif yesno in {'n', 'no'}:
return False
def main():
mylist = get_list('Add to list:', 'q')
check = raw_input('Look for item:')
if not check_list(mylist, check):
if get_yesno('Would you like to add it?'):
mylist.append(check)
print(mylist)
if __name__=="__main__":
main()
Some style tips:
Don't use list as a variable name; it's a built-in function, and you don't want to overwrite it.
Global variables are almost always a bad idea; passing data around explicitly makes it much easier to figure out where bad data is coming from, and makes functions more reusable.
camelCase is generally denigrated; use_underscores for function names instead.

You probably intended to keep going rather than break when you append the new item (or at least print something to indicate success), but the nested if statement works just fine, appends the thing to the list as specified and then the function and program terminate.