How to Print the Nth (5Th) Line of Every File Preceded by the Filename Using Any Linux Tool

How can I print the nth (5th) line of every file preceded by the FILENAME using any linux tool?

With GNU awk

gawk -v OFS=", " 'FNR == 5 {print "FILENAME: " FILENAME, $1, $2; nextfile}' *.gjf

Yes, FILENAME is the awk variable containing the current filename being processed.

Bash tool to get nth line from a file

head and pipe with tail will be slow for a huge file. I would suggest sed like this:

sed 'NUMq;d' file

Where NUM is the number of the line you want to print; so, for example, sed '10q;d' file to print the 10th line of file.

Explanation:

NUMq will quit immediately when the line number is NUM.

d will delete the line instead of printing it; this is inhibited on the last line because the q causes the rest of the script to be skipped when quitting.

If you have NUM in a variable, you will want to use double quotes instead of single:

sed "${NUM}q;d" file

Is there a way to filter a text file using grep (or any other tool), so that you can get a section of the file that's encased in bracers or brackets?

Yep, it's possible through grep which supports -P (Perl Regex) parameter.

$ grep -oPz '.*\[[^\[\]]*\]\s*=\s*\{[^{}]*\["life"\]\s*=\s*"No"[^{}]*}.*' file
        ["Alpha Centauri"] = {
            ["planets"] = "3",
            ["life"] = "No",
            ["asteroid"] = "20"
        },
        ["Rigel"] = {
            ["planets"] = "5",
            ["life"] = "No",
            ["asteroid"] = "11"
        }

DEMO

From grep --help

 -z, --null-data           a data line ends in 0 byte, not newline
 -o, --only-matching       show only the part of a line matching PATTERN

Update:

\[[^\n]*\]\h*=\h*\{(?!,\s*\[[^\[\]]*\]\h*=\h*{).*?\["fontSize"\]\h*=\h*20,.*?\}(?=,\s*\[[^\[\]]*\]\h*=\h*{|\s*})

DEMO

$ pcregrep -oM '(?s)[^\n]*\[[^\n]*\]\h*=\h*\{(?!,\s*\[[^\[\]]*\]\h*=\h*{).*?\["fontSize"\]\h*=\h*20,.*?\}(?=,\s*\[[^\[\]]*\]\h*=\h*{|\s*})' file
    ["frame 1"] = {
        ["fontSize"] = 20,
        ["displayStacks"] = "%p",
        ["xOffset"] = 251.000518798828,
        ["stacksPoint"] = "BOTTOM",
        ["regionType"] = "icon",
        ["yOffset"] = 416.000183105469,
        ["anchorPoint"] = "CENTER",
        ["parent"] = "Target Shit",
        ["numTriggers"] = 1,
        ["customTextUpdate"] = "update",
        ["id"] = "Invulnerabilities 2",
        ["icon"] = true,
        ["fontFlags"] = "OUTLINE",
        ["stacksContainment"] = "OUTSIDE",
        ["zoom"] = 0,
        ["auto"] = true,
        ["selfPoint"] = "CENTER",
        ["width"] = 60,
        ["frameStrata"] = 1,
        ["desaturate"] = false,
        ["stickyDuration"] = true,
        ["font"] = "Emblem",
        ["inverse"] = false,
        ["height"] = 60,
    }
    ["frame 2"] = {
        ["fontSize"] = 20,
        ["displayStacks"] = "%p",
        ["parent"] = "Target Shit",
        ["xOffset"] = 118.000427246094,
        ["stacksPoint"] = "BOTTOM",
        ["anchorPoint"] = "CENTER",
        ["untrigger"] = {
        },
        ["regionType"] = "icon",
        ["color"] = {
            1, -- [1]
            1, -- [2]
            1, -- [3]
            1, -- [4]
        },
        ["desaturate"] = false,
        ["frameStrata"] = 1,
        ["stickyDuration"] = true,
        ["width"] = 60,
        ["font"] = "Emblem",
        ["inverse"] = false,
        ["icon"] = true,
        ["height"] = 60,
        ["yOffset"] = 241
    }

(?s) DOTALL modifier which makes dots in your regex to match even line breaks.

Using awk to print all columns from the nth to the last

Print all columns:

awk '{print $0}' somefile

Print all but the first column:

awk '{$1=""; print $0}' somefile

Print all but the first two columns:

awk '{$1=$2=""; print $0}' somefile

Print a file, skipping the first X lines, in Bash

You'll need tail. Some examples:

$ tail great-big-file.log
< Last 10 lines of great-big-file.log >

If you really need to SKIP a particular number of "first" lines, use

$ tail -n +<N+1> <filename>
< filename, excluding first N lines. >

That is, if you want to skip N lines, you start printing line N+1. Example:

$ tail -n +11 /tmp/myfile
< /tmp/myfile, starting at line 11, or skipping the first 10 lines. >

If you want to just see the last so many lines, omit the "+":

$ tail -n <N> <filename>
< last N lines of file. >

How can I extract a predetermined range of lines from a text file on Unix?

sed -n '16224,16482p;16483q' filename > newfile

From the sed manual:

p -
Print out the pattern space (to the standard output). This command is usually only used in conjunction with the -n command-line option.

n -
If auto-print is not disabled, print the pattern space, then, regardless, replace the pattern space with the next line of input. If
there is no more input then sed exits without processing any more
commands.

q -
Exit sed without processing any more commands or input.
Note that the current pattern space is printed if auto-print is not disabled with the -n option.

and

Addresses in a sed script can be in any of the following forms:

number
Specifying a line number will match only that line in the input.

An address range can be specified by specifying two addresses
separated by a comma (,). An address range matches lines starting from
where the first address matches, and continues until the second
address matches (inclusively).

Quick unix command to display specific lines in the middle of a file?

with GNU-grep you could just say

grep --context=10 ...

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