Show Special Characters in Unix While Using 'Less' Command

Show special characters in Unix while using 'less' Command

less will look in its environment to see if there is a variable named LESS

You can set LESS in one of your ~/.profile (.bash_rc, etc, etc) and then anytime you run less from the comand line, it will find the LESS.

Try adding this

export LESS="-CQaix4"

This is the setup I use, there are some behaviors embedded in that may confuse you, so you can find out about what all of these mean from the help function in less, just tap the 'h' key and nose around, or run less --help.

Edit:

I looked at the help, and noticed there is also an -r option

-r  -R  ....  --raw-control-chars  --RAW-CONTROL-CHARS
                Output "raw" control characters.

I agree that cat may be the most exact match to your stated needs.

cat -vet file | less

Will add '$' at end of each line and convert tab char to visual '^I'.

cat --help
   (edited)
    -e                       equivalent to -vE
    -E, --show-ends          display $ at end of each line
    -t                       equivalent to -vT
    -T, --show-tabs          display TAB characters as ^I
    -v, --show-nonprinting   use ^ and M- notation, except for LFD and TAB

I hope this helps.

linux less command search invisible character

CtrlE means something special at the beginning of the search field regardless of how it is entered. Type another non-"special" character first, then press the sequence, then remove the first character.

grep for special characters in Unix

Tell grep to treat your input as fixed string using -F option.

grep -F '*^%Q&$*&^@$&*!^@$*&^&^*&^&' application.log

Option -n is required to get the line number,

grep -Fn '*^%Q&$*&^@$&*!^@$*&^&^*&^&' application.log

How to find /* in LESS?

When you enter / in less, the following expression is a regex. As such, * has special meaning there; /* thus searches for zero-or-more instances of /.

To prevent any character from having special meaning, you can enter it as a character class:

/[/][*]

...ensures that both of the characters you're searching for are treated literally. This works in other regular-expression contexts (grep, etc) as well.

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That said, the above is somewhat more paranoid than necessary. In my tests, the below work as well:

//[*]
//\*

How to remove all special characters in Linux text

Remove everything except the printable characters (character class [:print:]), with sed:

sed $'s/[^[:print:]\t]//g' file.txt

[:print:] includes:

• [:alnum:] (alpha-numerics)
• [:punct:] (punctuations)
• space

The ANSI C quoting ($'') is used for interpreting \t as literal tab inside $'' (in bash and alike).

UNIX: Strange output if piped to less

I'm guessing that you have the --color=always option to ls set, either through an alias, functions or the LS_COLORS environment variable and ls is sending color directives to a non-terminal (that is, your pipe to less).

Handling special characters in bash script

Please do not post so much in one question. Please one problem per question. One script per question, etc.

Make sure to check your scripts with shellcheck. It will catch your mistakes. See https://mywiki.wooledge.org/Quotes .

1. When run without arguments:

filename=("'" name "'") inside awk script is a invalid way to pass anything with ' quotes to system() call, so you are getting unterminated ' error, as expected, because there will be 3 ' characters. Fix the AWS script, or better rewrite it in Bash, no need for awk. Maybe rewrite it all in Python or Perl.

Moreover, tmp=(name " " $i); deletes tabs and multiple spaces from filenames. It's all meant to work with only nice filenames.

The script will break on newlines in filenames anyway.

2. When run with arguments:

$@ undergoes word splitting and filename expansion (topics you should research). Word splitting splits the input into words on spaces. Use "$@". Quote the expansions.

3. What I want:

You'll be doing that with "$@"

4. What I tried:

The variable content is irrelevant. You have to change the way you use the variable, not it's content. I.e. use quotes around the use of the variable. Not the content.

5. Extra:

You did not quote the expansion. Use "$i" not $i. It's "$i"/*. $1 undergoes word splitting.

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And finally, after that all, your script may look like, with GNU tools:

if (($# == 0)); then
   set -- *
fi
du -hs0 "$@" |
sort -zh |
sed -z 's/\t/\x00/' |
while IFS= read -r -d '' size && IFS= read -r -d '' file; do
   printf "%s " "$size";
   ls -d "$file"
done

Also see How can I find and safely handle file names containing newlines, spaces or both? https://mywiki.wooledge.org/BashFAQ/001 .

Also, you can chain any statements:

if stuff; then
   stuff1
else
   stuff2
fi | 
sort -h |
awk -f yourscriptrt 

And also don't repeat yourself - use bash arrays:

args=()
if stuff; then
  args=(*)
else
  args=("$@")
fi
du -hs "${args[@]}" | stuff...

And so that sort has less work to do, I would put it right after du, not after parsing.

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