Grep string inside double quotes

grep string between single quotes
grep escape single quote
grep string with double quotes
grep string between quotes
grep escape character
sed get string inside quotes
grep command
git grep double quote

Trying to grep a string inside double quotes at the moment I use this

grep user file | grep -e "[\'\"]"

This will get to the section of the file I need and highlight the double quotes but it will not give the sting in the double quotes

Try doing this :

$ cat aaaa
foo"bar"base
$ grep -oP '"\K[^"\047]+(?=["\047])' aaaa
bar

I use look around advanced regex techniques.

If you want the quotes too :

$ grep -Eo '["\047].*["\047]'
"bar"

Note :

\047

is the octal ascii representation of the single quote

Grep string inside double quotes & single quotes, Grep string inside double quotes & single quotes. The text file is multiline. It may have escaped double or single quotes inside quotes. The double quotes may have single quotes inside. The single quotes may have double quotes inside. The quotes must be paired match. how to use grep: finding a string with double quotes and multiple digits I have a file with a lot of lines (a lot!) that contain 10 digits between double quotes. ie "1726937489". The digits are random throughout, but always contain ten digits.

$ cat aaaa
foo"bar"base

$ grep -o '"[^"]\+"'
"bar"

Wildcards and double quotes with find -name command, When using the Find command Why should you enclose the pattern in single quotes? REGEX and GREP — Search string including optional and static words 1 how to use grep to extract the content of a string with multiple (") starting from the first double quote to the second one

Try:

grep "Test" /tmp/junk | tr '"' ' '

This will remove quotes from the output of grep

Or you could try the following:

grep "Test" /tmp/junk | cut -d '"' -f 2

This will use the quotes as a delimiter. Just specify the field you want to select. This lets you cherry-pick the information you want.

grep, –E, put a backslash ( \ ) in front of the character. With single quotes, the string is taken literally and no expansion takes place. With double quotes, variables are expanded. For example (with a Bourne-derived shell such as Bash or ZSH): VAR="serverfault" grep '$VAR' file1 grep "$VAR" file1. The first grep will look for the literal string "$VAR1" in file1.

This worked for me. This will print the data including the double quotes also.

 grep -o '"[^"]\+"' file.txt

If we want to print without double quotes, then we need to pipe the grep output to a sed command.

grep -o '"[^"]\+"' file.txt | sed 's/"//g'

how to use grep: finding a string with double quotes and multiple digits, I have a file with a lot of lines (a lot!) that contain 10 digits between double quotes​. ie "1726937489". The digits are random throughout, but always contain ten  Hello, I'm trying to use grep or egrep to exclude a whole range of characters but how do I exclude both a single and a double quote. It might be easier to say how do I use grep to find both single and double quotes. grep ' ' " ' file grep detects the first single quote within my expression as an end of expression marker.

When to use single-quote, double-quote in grep?, With single quotes, the string is taken literally and no expansion takes place. a singlequote inside a singlequoted string, and you have to use doublequotes. This will print the data including the double quotes also. grep -o '" [^"]\+"' file.txt If we want to print without double quotes, then we need to pipe the grep output to a sed command. grep -o '" [^"]\+"' file.txt | sed 's/"//g'

4.1 Searching for Patterns with grep, To search for a particular character string in a file, use the grep command. is more than one word long, enclose the string with single or double quotation marks: as such, even inside quotation marks, unless you escape it with a backslash. I have been fiddling around with the following pattern but it seems to only match strings on the same line. /\v"\zs(.*| *)\ze" The pattern should match something like this: "This is an example of a multi, line -- %123 string with some random 123 symbol's `inside` \\//><>< of it" What is wrong with the pattern I have?

Searching for Patterns With grep (Solaris Advanced User's Guide), To search for a particular character string in a file, use the grep command. is more than one word long, enclose the string with single or double quotation marks. as a metacharacter, even inside quotation marks, unless you escape it with a  I have already described in one of the earlier posts on how to print string within single quote in Awk print statement. Here is how we can print strings without "double quotes" in Awk. $ cat file.txt 6289693505455 Plan_DAIL_30D_AA 6289693505475 Plan_DAIL_30D_AA 6289693505462 Plan_DAIL_30D_AB Output required:

Comments
  • Thanks it nearly works - the output gets the string as well as ; outside the double quote
  • You want the quotes too ?
  • Nope just the sting but using the above grep gives me ; as well... try this for example user="bobhope";
  • What does the \K do?
  • Ironically, both of these yield a blank line when piped onto grep -o '"[^"]\+"', but grep -Eo '".*"' | tr -d '"' worked.
  • cut is not a good choice here because the information is not separated into fields. The number of quotes in a line could be 1, could be 5, could be none.