'How to get multiline input from user [duplicate]

I want to write a program that gets multiple line input and work with it line by line. Why there is no function like raw_input in Python 3?

input does not allow user to put lines separated by newline (Enter), it prints back only the first line.

Can it be stored in variable or even read it to a list?



Solution 1:[1]

raw_input can correctly handle the EOF, so we can write a loop, read till we have received an EOF (Ctrl-D) from user:

Python 3

print("Enter/Paste your content. Ctrl-D or Ctrl-Z ( windows ) to save it.")
contents = []
while True:
    try:
        line = input()
    except EOFError:
        break
    contents.append(line)

Python 2

print "Enter/Paste your content. Ctrl-D or Ctrl-Z ( windows ) to save it."
contents = []
while True:
    try:
        line = raw_input("")
    except EOFError:
        break
    contents.append(line)

Solution 2:[2]

input(prompt) is basically equivalent to

def input(prompt):
    print(prompt, end='', file=sys.stderr, flush=True)
    return sys.stdin.readline()

You can read directly from sys.stdin if you like.

lines = sys.stdin.readlines()

lines = [line for line in sys.stdin]

five_lines = list(itertools.islice(sys.stdin, 5))
    

The first two require that the input end somehow, either by reaching the end of a file or by the user typing Control-D (or Control-Z in Windows) to signal the end. The last one will return after five lines have been read, whether from a file or from the terminal/keyboard.

Solution 3:[3]

Use the input() built-in function to get a input line from the user.

You can read the help here.

You can use the following code to get several line at once (finishing by an empty one):

while input() != '':
    do_thing

Solution 4:[4]

no_of_lines = 5
lines = ""
for i in xrange(5):
    lines+=input()+"\n"
    a=raw_input("if u want to continue (Y/n)")
    ""
    if(a=='y'):
        continue
    else:
        break
    print lines

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Raj
Solution 2 Community
Solution 3 maggick
Solution 4 Michael Dorner