Does the SoloLearn Python compiler have problems?

Question

Hello, In Python Intermediate course, Tuples, there is a code practice about contact searching. I wrote the code and worked properly in some external compilers, but Sololearn's compiler, throws an error message. Why?

Here is my solution code:

contacts = [
('James', 42),
('Amy', 24),
('John', 31),
('Amanda', 63),
('Bob', 18)
]

getListItem = int(input())

if getListItem < len(contacts):
name = contacts[getListItem][0]
age = contacts[getListItem][1]
contact = name + " is " + str(age) # concatenation does not work for Integers; Wrapped in str()
print(contact)
else:
print("Not found")

Answer 1

The input is string.but you convert it to integer. if i look on your code, you try to search the position of the contact. but you have to make the code to search the name in the contact list

Answer 2

Aram You're welcome!

Answer 3

Bertram Rayhan
Thank you for your reply.
I just thought we should access the elements using the indexes.

After your reply, I solved it by changing my code to this:

contacts = [
('James', 42),
('Amy', 24),
('John', 31),
('Amanda', 63),
('Bob', 18)
]

name = input()

for i in contacts: # for example, "i" now is "('Amy', 24)"
if name == i[0]: # "Amy" is the first element or index 0; So, inputed Amy == Amy element
age = i[1] # Sure, i[1] is the second element in "('Amy', 24)"; the "age" of Amy
contact = name + " is " + str(age) # concatenation does not work for Integers; Wrapped in str()
print(contact) # Print the contact as "Amy is 24"
break # Terminate the condition if the name is in the list and Don't run the else part
else:
print("Not Found") # Otherwise, if the name is not in the contact list, print "Not found"