How I can check if the list in Python is empty
Question:
I have a list of users on the list and want to print for them some greetings but if I have in this list I should print some another text Hello Admin, would you like to see a status report?. Also I want to have some notification if the list is empty but I don’t know how to do this.
users_site = []
for site in users_site:
if site == 'Admin':
print ("Hello Admin, would you like to see a status report?")
if site == []:
print ("We need more user's")
else:
print (f"Hello, {site}")
Answers:
Normally, I’d put everything into a function and short-circuit it early with return
– you can detect if a list is empty with not
which will see if the right side is Falsey
def my_function(my_list):
if not my_list:
print("the list is empty")
return
for entry in my_list:
...
Does this help?
users_site = ['']
if not users_site:
print ("We need more user's")
else:
for site in users_site:
if site == 'Admin':
print ("Hello Admin, would you like to see a status report?")
else:
print (f"Hello, {site}")
I have a list of users on the list and want to print for them some greetings but if I have in this list I should print some another text Hello Admin, would you like to see a status report?. Also I want to have some notification if the list is empty but I don’t know how to do this.
users_site = []
for site in users_site:
if site == 'Admin':
print ("Hello Admin, would you like to see a status report?")
if site == []:
print ("We need more user's")
else:
print (f"Hello, {site}")
Normally, I’d put everything into a function and short-circuit it early with return
– you can detect if a list is empty with not
which will see if the right side is Falsey
def my_function(my_list):
if not my_list:
print("the list is empty")
return
for entry in my_list:
...
Does this help?
users_site = ['']
if not users_site:
print ("We need more user's")
else:
for site in users_site:
if site == 'Admin':
print ("Hello Admin, would you like to see a status report?")
else:
print (f"Hello, {site}")