Grep String Inside Double Quotes

Grep string inside 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 a string inside double quotes

You can write your loop a little more efficiently and use sed instead of multiple greps to get what you want:

for f in test.txt; do
  echo "Processing $f"
  while read line; do
    grep 'name="'$line'"' new1.xml 
  done < "$f" | sed -E 's/.+path="([^"]+)".+/\1/'
done

For your example, the above script gives this output:

Processing test.txt
abc/test/test

---

If you are just processing one file, you don't need the outer loop:

  while read line; do
    grep 'name="'$line'"' new1.xml
  done < "test.txt" | sed -E 's/.+path="([^"]+)".+/\1/'

Escape double quote in grep

The problem is that you aren't correctly escaping the input string, try:

echo "\"member\":\"time\"" | grep -e "member\""

Alternatively, you can use unescaped double quotes within single quotes:

echo '"member":"time"' | grep -e 'member"'

It's a matter of preference which you find clearer, although the second approach prevents you from nesting your command within another set of single quotes (e.g. ssh 'cmd').

grep between some characters (quotes, etc) of after (eg. hashtag) any content (text, numbers, emojis)

Thanks to @Aserre's pointings, I could come up with an answer.

In order for the "get every text when it appear AFTER a charater" and "get every text when it appear BETWEEN quotes" (grep) to work with any character, we have to replace [[:alpha:]] in the block to ...

So, it is:

echo '#first = "✅ Yes"' | grep -o '"...\+"' | tr -d '"' (get anything which is between double quotes)

and:

echo "Text and #hashtag" | grep -o '#...\+' | tr -d '"' (get anything which is after a hashtag)

Update:

If you want to support things with only 1 character (such as numbers ranging from 0 to 9), replace ... to . (single dot)

It works, as in the question, for: emojis, letters, numbers and other special characters.

How do I use GREP to match a string with quotation marks inside it?

With traditional regex, the double quotes are escaped by the single quotes. You only need to escape the quantifier + (one or more) and the literal dot in .mp4

grep -o '="http[^"]\+\.mp4"'

With PCRE, available in GNU Grep, you can actually match (without printing) the leading/trailing equals/doublequotes with Lookarounds:

grep -Po '(?<==")http[^"]+\.mp4(?=")'

• (?<= ... ) - lookbehind
• (?= ... ) - lookahead

returns:

https://www.something.com/file1.mp4
https://www.something.com/file2.mp4
https://www.something.com/file3.mp4
https://www.something.com/file4.mp4

How to search for a string with one double quote with git grep in Windows cmd

In Git CMD or Windows cmd console, you can use ^ to escape the " inside the terminal:

git grep ""^""Agent" > 1.txt

In Git Bash, you can simply use

git grep '"Agent' > 1.txt

See this text I made on my data with "Regex pattern:

How to match double quotes using grep

Use single quotes to consider literal meaning of every character.

grep 'and "Ben" is' trial.txt

How to find a double quoted string using grep and regex?

Your expression doesn't make sense to me. It seems like you're looking for:

grep -R '5\.1' .

Or if you want to include the " characters:

grep -R '"5\.1"' .

But I don't see anything related to those in your original post, so it's hard to say.

grep with double quotation in bash script

The simplest adaptation of the code you're using is:

for ip in $(cat /tmp/iplist)
do
    grep "$ip" /tmp/whitelist
done

Once you've got the value in the variable, its double quotes aren't damaged by further double-quoted variable expansion. If the /tmp/iplist file doesn't contain the double quotes but they're critical, then you can use:

grep "\"$ip\"" /tmp/whitelist

or you could use this:

grep \""$ip"\" /tmp/whitelist

(and there are two asymmetric permutations available too). It's a good idea to make sure you do know why that works. There are some ways to use single quotes if you want to, but the "$ip" part must be outside of the single quotes.

All your examples start the pattern argument to grep with a single quote. Single quotes suppress all expansions until the next single quote. So, for example, grep '"$ip"' /tmp/whitelist is looking for 5 characters — ", $, i, p, " — in the file. In none them is the variable ip ever expanded.

There will be problems if any IP address ever gets a space in it. Be cautious about using for ip in $(cat /tmp/iplist). Very often you'd do better with:

while read -r ip
do
    grep "$ip" /tmp/whitelist
done < /tmp/iplist

Another way to do this is with grep -F or fgrep:

grep -F -f /tmp/iplist /tmp/whitelist

This doesn't insist on double quotes around the IP addresses, but makes a single pass over the /tmp/whitelist file (and a single pass over the /tmp/iplist file too), which is about as efficient as it gets. This will produce lines in a slightly different order from before, which probably won't matter, but you should be aware of it.

If you must have the double quotes (to avoid selecting 11.2.3.44 when searching for 11.2.3.4, then:

grep -F -f <(sed 's/^/"/; s/$/"/' /tmp/iplist) /tmp/whitelist

This uses process substitution to pass an edited version of the /tmp/iplist file to the grep command. If you don't have process substitution in your shell, you can probably use:

sed 's/^/"/; s/$/"/' /tmp/iplist | grep -F -f - /tmp/whitelist

which makes grep read the list of patterns to match from standard input instead of a named file. If perchance -f - doesn't work (e.g. because you're working on a Mac using macOS, or probably a BSD machine), then -f /dev/stdin probably will, or -f /dev/fd/0.

You could also generate the /tmp/iplist file with the double quotes in place. You could generate the /tmp/whitelist file without the double quotes and then use grep -x to specify an exact match.

In case you haven't already gathered, there are quite a lot of different ways to do this.

how to grep everything between single quotes?

Try something like this: sed -n "s#version:\s*'\(.*\)'#\1#p" myfile.txt. This avoids the redundant cat and grep by finding the "version" line and extracting the contents between the single quotes.

Explanation:

the -n flag tells sed not to print lines automatically. We then use the p command at the end of our sed pattern to explicitly print when we've found the version line.

Search for pattern: version:\s*'\(.*\)'

• version:\s* Match "version:" followed by any amount of whitespace
• '\(.*\)' Match a single ', then capture everything until the next '

Replace with: \1; This is the first (and only) capture group above, containing contents between single quotes.

---

Related Topics