How to Grep '---' in Linux? Grep: Unrecognized Option '---'

How to grep '---' in Linux? grep: unrecognized option '---'

This happens because grep interprets --- as an option instead of a text to look for. Instead, use --:

grep -- "---" your_file

This way, you tell grep that the rest is not a command line option.

Other options:

• use grep -e (see Kent's solution, as I added it when he had already posted it - didn't notice it until now):

• use awk (see anubhava's solution) or sed:

  sed -n '/---/p' file
  

-n prevents sed from printing the lines (its default action). Then /--- matches those lines containing --- and /p makes them be printed.

Grep unrecognized option '--' while parsion content of html element

The line

response=$(curl ...)

puts the output of the curl command in a variable named response.

In your grep command you try to pass the expansion of the variable as an argument.

output=$(grep -o '<div class="error-icon">[^<]*' "$response" | ...)

grep tries to interpret the value as command line arguments which may result in various errors depending on the actual output. In my test I got a message grep: <some html code>: File name too long because it tries to interpret this as a file name argument.

You should save the data in a file and pass this to grep. Adapt the name and location of the temporary file as necessary. Example:

#!/bin/sh

verifyCard=$1

if [ -z "${verifyCard}" ]; then echo "No argument supplied"; exit 1; fi

curl -o response-file 'https://www.isic.org/verify/' -H 'User-Agent: Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:80.0) Gecko/20100101 Firefox/80.0' -H 'Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8' -H 'Accept-Language: en-US,en;q=0.5' --compressed -H 'Content-Type: application/x-www-form-urlencoded' -H 'Origin: https://www.isic.org' -H 'Connection: keep-alive' -H 'Referer: https://www.isic.org/verify/' -H 'Cookie: PHPSESSID=46plnh6b31e2pusv1do6thfbm7; AWSELB=AF73034F18B38D7DCED6DEDC728D31BA3F3A73F96747FEE7FA7C4F7A74BC9954E5928CBDDD5C053FFB2A37CE37136C4BA008B15163192B34CFA04D35BEC4ED0D0D2913A2FB; AWSELBCORS=AF73034F18B38D7DCED6DEDC728D31BA3F3A73F96747FEE7FA7C4F7A74BC9954E5928CBDDD5C053FFB2A37CE37136C4BA008B15163192B34CFA04D35BEC4ED0D0D2913A2FB; _ga=GA1.2.650910486.1600495658; _gid=GA1.2.731428038.1600495658; _gat=1' -H 'Upgrade-Insecure-Requests: 1' --data-raw 'verify_card_number=${$verifyCard}'

output=$(grep -o '<div class="error-icon">[^<]*' response-file | grep -o '[^>]*$')

rm response-file

echo "$output"

If you use bash or zsh instead of sh, there are ways to substitute some variable value as an input file, see e.g. the answers in using a Bash variable in place of a file as input for an executable.

grep -r string * error: unrecognized option `--DIRAC3LE--'?

Your shell is probably expanding * and there is a '--DIRAC3LE--' in the working directory. grep is then confusing the leading -- with command line options.

Try using

grep -r "string" .

This will recursively search for everything in the working directory.

Also try using "--" before the *. POSIX commands use this to indicate the end of the command line options to prevent such ambiguity (see comments for a reference).

Match from beginning to word as long as there are no . in between: Convert grep -Po command to sed

You can use

awk -F'.' '$2 == "enabled"{print $1}' file
sed -n 's/^\([^.]*\)\.enabled.*/\1/p' file

See the online demo.

Details:

awk:

• -F'.' - the field separator is set to a .
• $2 == "enabled" - if Group 2 value is enabled, then
• {print $1} - print Field 1 value
  
  sed:
  
• -n - suppresses default line output in the sed command
• s/^\([^.]*\)\.enabled.*/\1/p - finds any zero or more chars other than . at the start of string (placing them into Group 1, \1), then a .enabled and then the rest of the string and replaces with the Group 1 value, and prints the resulting value.

grep command not accepting a string with multiple options in my script

Do not double quote multiple options/flags into as one option. When you quote a string grep will treat it as single string.

For example:

Valid:

grep -ins --color=auto "foo" inputfile

Valid:

 grep "-ins" "--color=auto" "foo" inputfile

Invalid:

 grep "-ins --color=auto" "foo" inputfile

Because there is no single option as "-ins --color=auto"

grep search for -- or -- gives unrecognised or invalid option

Well, dashes start options, so it's only natural grep tries to interpret that as one. You'd have the same problem if searching for -R.

Use -e:

-e PATTERN, --regexp=PATTERN

Use PATTERN as the pattern. If this option is used multiple times or is combined with the -f (--file) option, search for all patterns given. This
option can be used to protect a pattern beginning with “-”.

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