CODEWITHSUNDEEP
pythonpython-3.x
manos
manos

Reputation: 97

How to print the next few lines

So I want to make this script look for a word, which you can determine, and its line number. I want it to print out the text in this line and the two following ones. How do I code the last part (print the next three lines)?

def search_passwords():
    file = open("D:\\Libaries\\Documents\\resources_py\\pw.txt",  "r")
    print()
    search = input("Enter service: ")
    print()   
    for line in file:
        if search in line:  
           print(the next three lines)

Upvotes: 1

Views: 1326

Answers (3)

Alderven
Alderven

Reputation: 8260

Yet another approach with not found error handling:

with open('file.txt', 'r') as f:
    lines = [line.strip() for line in f.readlines()]
    try:
        i = lines.index(input('Enter service: '))
        for j in range(i, i+3):
            try:
                print(lines[j])
            except IndexError:
                pass
    except ValueError:
        print('Service not found!')

Upvotes: 1

statist32
statist32

Reputation: 56

There could be a better solution but this is a bit more flexible with a nigher amount of following rows.

def search_passwords():
    file = open("D:\\Libaries\\Documents\\resources_py\\pw.txt",  "r")
    print()
    search = input("Enter service: ")
    print()   
    counter = 3
    found = False
    for line in file:
        if search in line:  
           found = True
        if found:
            if counter > 0: 
                counter -= 1
                print(line, counter)
            if counter == 0:
                break

Upvotes: 1

minker
minker

Reputation: 680

There are some corner cases that you did not mention.

  • What if multiple lines have the searching word? What is your expected behavior?
  • What if the line right after the matching line has the searching word?
  • How large is the file and could you load it in memory?

If there's only one matching, or there won't be any matching in the lines you print right after, then you can use a single counter

def search_passwords():
    file = open("D:\\Libaries\\Documents\\resources_py\\pw.txt",  "r")
    print()
    search = input("Enter service: ")
    print()   
    counter = 0
    for line in file:
        if counter > 0:
           counter -= 1
           print(line)
        if search in line:  
           counter = 2
           print(line)

However, this will double print the lines if the following lines containing the searching word. You can of course make the second if elif, but then it will just ignore the following matching pattern.

And, if you only need the first or only appearance, you should break out of the loop once you print everything, not read the whole file.

Just like I said, you need to know the expected behavior and the requirement better, to write a solid program.

Upvotes: 1

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