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  1. Prepare
  2. Linux Shell
  3. Text Processing
  4. Cut #7
  5. Discussions

Cut #7

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98 Discussions

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  • jesiel1990arthas
     + 1 comment

    Okay guys, I understand that at first it seems nonsensical, but yes, it prints the first two because the delimiter ' ' space doesn't exist. However, as we saw earlier, before it was tabs and now it's spaces. The sample doesn't have four words, but if you add them, it prints the fourth. The error is the sample; the command works.

  • farouk_m_alim
     + 0 comments
    cut -d ' ' -f4

  • achraf_crispy
     + 0 comments
    while ISF=' ' read -r -a line; do

    if  [ ${#line[@]} -ne 1 ]
    then
        echo ${line[3]}
    else
        echo ${line[0]}
    fi
done

-r Option
-r: When the -r option is used, read treats backslashes literally. This means that the backslashes will not be used as escape characters. Without -r, read would interpret backslashes as escape characters, which could be used to escape special characters (like newline).
-a Option
-a: The -a option allows you to read the input into an array. This means that each word of the input will be stored as an element in the array. You need to specify the name of the array immediately after the -a option.


e

  • devanshmudgal96
     + 0 comments
    while read lines 
do 
    echo $lines | cut -d " " -f 4
done
    • -d flag (delimiter flag) for specifying space (" ") as the delimiter
    • -f flag (field flag) has a default delimiter of tab

  • anuradhahadkar30
     + 0 comments

    cut -d ' ' -f4