You return four variables s1,s2,s3,s4 and receive them using a single variable obj. This is what is called a tuple, obj is associated with 4 values, the values of s1,s2,s3,s4. So, use index as you use in a list to get the value you want, in order.

Copyobj=list_benefits()
print obj[0] + " is a benefit of functions!"
print obj[1] + " is a benefit of functions!"
print obj[2] + " is a benefit of functions!"
print obj[3] + " is a benefit of functions!"

p1=(input("Name: "),input("Age: "))

As of right now, you're making a 2-tuple, not a Person. This should be

p1 = Person(input("Name: "), input("Age: "))

Please post the entire error message with all the information.

Are you initializing guessed_num as a tuple verses an empty list? See the difference in the ipython session below: ```python In [43]: guessed_num = () # declared as a tuple In [44]: guessed_num.append(4) # this will fail AttributeError Traceback (most recent call last) in () ----> 1 guessed_num.append(4) AttributeError: 'tuple' object has no attribute 'append' In [45]: guessed_num = [] # declared as a list In [46]: guessed_num.append(4) # append at will! In [47]: guessed_num Out[47]: [4] ```

Remove the * in front of *arg. That is converting the argument into a tuple.

In the line:

Jobs = ()

you create a tuple. A tuple is immutable and has no methods to add, remove or alter elements. You probably wanted to create a list (lists have an .append-method). To create a list use the square brackets instead of round ones:

Jobs = []

or use the list-"constructor":

Jobs = list()

---

However some suggestions for your code:

opening a file requires that you close it again. Otherwise Python will keep the file handle as long as it is running. To make it easier there is a context manager for this:

with open('Jobs.txt') as openFile:
    x = 1
    while x != 0:
        Stuff = openFile.readline(x)
        if Stuff != '':
            Jobs.append(Stuff)
        else:
            x = 0

As soon as the context manager finishes the file will be closed automatically, even if an exception occurs.

---

It's used very rarely but iter accepts two arguments. If you give it two arguments, then it will call the first each iteration and stop as soon as the second argument is encountered. That seems like a perfect fit here:

with open('Jobs.txt') as openFile:
    for Stuff in iter(openFile.readline, ''):
        Jobs.append(Stuff)

I'm not sure if that's actually working like expected because openFile.readline keeps trailing newline characters (\n) so if you want to stop at the first empty line you need for Stuff in iter(openFile.readline, '\n'). (Could also be a windows thingy on my computer, ignore this if you don't have problems!)

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This can also be done in two lines, without creating the Jobs before you start the loop:

with open('Jobs.txt') as openFile:
    # you could also use "tuple" instead of "list" here.
    Jobs = list(iter(openFile.readline, ''))  

---

Besides iter with two arguments you could also use itertools.takewhile:

import itertools
with open('Jobs.txt') as openFile:
    Jobs = list(itertools.takewhile(lambda x: x != '', openFile))

The lambda is a bit slow, if you need it faster you could also use ''.__ne__ or bool (the latter one works because an empty string is considered False):

import itertools
with open('Jobs.txt') as openFile:
    Jobs = list(itertools.takewhile(''.__ne__, openFile))

items() is a method specifically for dictionaries, not tuples. Tuples are like lists, except they are immutable.

This error indicates that value is a tuple, and not a string as you might expect. This indicates a problem with your application.

Here the problem is that fetchone() returns a one-tuple. You should change from this line:

value = results.fetchone()

to this (notice the comma after value):

value, = results.fetchone()

or this (not recommended):

value = results.fetchone()[0]

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But why is fetchone() returning a tuple instead of a string? Because you can SELECT multiple columns. Consider for example the following SQL statement:

SELECT a, b, c FROM my_table;

In this case, fetchone() will return a three-tuple.

Writing value, = fetchone() you are telling Python that you are expecting a one-tuple and you want that single item placed into value. If you were expecting the three-tuple, you'd have used column_a, column_b, column_c = resulsts.fetchone().

This is the reason why you should prefer value, over fetchone()[0].

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Bonus tip: I noticed you are using Python 3. In this case, you can write:

translation_table = dict.fromkeys(b'+-(),', None)

Speeding up your code and making it more clean.

What does your imports look like?