Ubuntu – grep pattern with leading spaces

bashgrepregex

I need some help in setting the correct pattern for grep. I need to find all occurrences of pattern where line may have leading space(s). For example: In the following file:

 1. No pattern recognized.
 2. Pattern to be recognized
 3.          Pattern to be recognized here also
 4.  pattern with only one leading space

I would like to grep only lines 2,3 and 4. The line numbers are just for the reference.

So far I have tried the following:

grep -i '^ [[:blank:]]pattern', but it returns only line 4.

grep -i '[[:blank:]]pattern' returns 1,3 and 4.

grep -i '^[[:blank:]]pattern' returns 1,3 and 4.

— Mike
P.S.
If this is not the appropriate forum, then please guide me accordingly.

Best Answer

Your line 2 and 3 has upper case P and requires zero or more spaces, so specify exactly that:

$ grep '[[:blank:]]*Pattern'  input.txt                                  
 Pattern to be recognized
         Pattern to be recognized here also

Personally, I'd recommend extending your pattern with something else, like '[[:blank:]]Pattern.*recognized'

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Ubuntu – grep recursively for specific files

You are getting these Permission denied errors because there are some python files that your user does not have permission to read, so grep is not working for those files (you can't search them).

You want to "get rid of" the Permission denied errors.

You can add 2>/dev/null to the end of your command like this:
```
grep -rn --include="*.py" "search-pattern" 2>/dev/null
```
This suppresses all error messages, so you won't see the Permission denied errors popping up. But this obviously still means that grep is not searching those files, so if they do contain your search-pattern, you won't see them in the output.

Explanation:
In Linux, there are three things called file descriptors: stdin, stdout, stderr. If you run a command and it gives you an error, the error is written to the file descriptor stderr. By default, stderr outputs the errors to the terminal. So what we can do is redirect the errors, i.e. we redirect stderr somewhere else instead of the terminal. We redirect stuff in Linux by using the > sign. So where are we going to redirect stderr? To /dev/null. /dev/null is a special file; think of it as a black hole. Anything you redirect to it gets thrown away. Finally, we refer to stderr as simply the number 2. So 2>/dev/null tells Linux to redirect 2 to /dev/null. 1 means stdout (which is the normal output of a command, i.e. not errors). So, for example, if you wanted to save the actual output of your grep command to a file rather than have it displayed in the terminal, you can use 1>/path/to/filename, which will redirect stdout to that file.
Use sudo with your command, assuming that you have the privileges:
```
sudo grep -rn --include="*.py" "search-pattern"
```
If your username has root privileges (i.e. if you're an administrator on the machine), using sudo before a command runs it as the root user, so you will never get the Permission denied errors and grep will search those files and output the search-pattern if it finds it inside them.

Ubuntu – How to grep lines, based on a certain pattern

Try the below grep command which uses -P (Perl-regexp) parameter.

grep -P '(?<!\S)/aa/bbbb/cccccc(?!\S)' file

(?<!\S) This negative lookbehind asserts that the character which preceeds the string /aa/bbbb/cccccc would be any but not a non-space character.
(?!\S) Negative lookahead asserts that the character following the match would be any but not a non-space character.

Another grep,

 grep -E '(^|\s)/aa/bbbb/cccccc(\s|$)' file

Through python,

script.py

#!/usr/bin/python3
import re
import sys
file = sys.argv[1]
with open(file, 'r') as f:
    for line in f:
        for i in line.split():
            if i == "/aa/bbbb/cccccc":
                print(line, end='')

Save the above code in a file and name it as script.py. Then execute the above script by

python3 script.py /path/to/the/file/you/want/to/work/with

Best Answer

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