How to Escape Single Quotes in Bash/Grep

how to escape single quotes in grep

Just wrap the string within double quotes:

grep "'true' format" your_file

Test

$ cat a
this is 'true' format
and this is true formatting
test

$ grep "'true' format" a
this is 'true' format

How to escape single quotes in Bash/Grep?

grep -i "something ~\* '[[:alnum:]]*'" /var/log/syslog

works for me.

• escape the first * to match a literal * instead of making it the zero-or-more-matches character:
  
  ~* would match zero or more occurrences of ~ while
  
  ~\* matches the expression ~* after something
• use double brackets around :alnum: (see example here)
• use a * after [[:alnum::]] to match not only one character between your single quotes but several of them
• the single quotes don't have to be escaped at all because they are contained in an expression that is limited by double quotes.

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.

How to do a grep regex search for single-quotes?

You are using a PCRE. So, you need the -P flag. So, use this:

grep -rnwi some/path -P "ABC\s*=\s*[\'\"][^\'\"]+[\'\"]"

We don't need a \\ for single quotes inside the character classes. So, your regex can also be written as:

"ABC\s*=\s*['\"][^'\"]+['\"]" 

Input file:

ABC="123"
ABC='123'

Run grep with your PCRE:

grep -P "ABC\s*=\s*['\"][^'\"]+['\"]" input.txt

Output:

ABC="123"
ABC='123'

extract path between single quotes using grep

Since the question asks for a solution in grep, a single GNU grep command to extract the specified path could be:

grep -Po "^Saving to: .\\K[^']*"

provided the Perl Regular Expressions are implemented in the grep (not all greps implement those).

Of course, it can be used in a pipe also:

wget_command | grep -Po "^Saving to: .\\K[^']*" | tee log.txt

Note that I used a single quote (') character to anchor the end of path in the pattern match expression, but in the question, Unicode Character Left Single Quotation Mark (U+2018) (‘) and Unicode Character Right Single Quotation Mark (U+2019) (’) are used in the sample input. If this is really intended then just replace the [^'] with the [^’] in the pattern match expression above.

bash string expansion for grep -E expression in single quotes

Why are you adding the single quote marks? Just remove this line:

 EXCLUDE_PATTERN="'""$EXCLUDE_PATTERN""'"

I'm getting the following without that line:

 Excluding files that match the string alice|bar|bob|foo
 Keeping alice.tmp
 Keeping bob.tmp
 Deleting crump.tmp
 Deleting dammitall.tmp
 Keeping foo.tmp

How do I escape slashes and double and single quotes in sed?

The s/// command in sed allows you to use other characters instead of / as the delimiter, as in

sed 's#"http://www\.fubar\.com"#URL_FUBAR#g'

or

sed 's,"http://www\.fubar\.com",URL_FUBAR,g'

The double quotes are not a problem. For matching single quotes, switch the two types of quotes around. Note that a single quoted string may not contain single quotes (not even escaped ones).

The dots need to be escaped if sed is to interpret them as literal dots and not as the regular expression pattern . which matches any one character.

bash: matching single and double quotes together by grep

I am probably mis-interpreting the question, but it seems like you are asking how to deal with properly escaping the quotes. Just do:

grep '\(["'"']\).*\1.*\1"

Search for single quoted grep strings?

The main problem here is that you are not quoting the argument being passed to grep. The only thing that needs to be escaped is \$ (if double quoted) and []. If you want the exact string (not using regex), just use fgrep (grep -F) which does exact string matching:

grep -F "\$server['fish_stick']"

Works on my system:

$ foo="\$server['fish_stick']"
$ echo "$foo" | grep -F "\$server['fish_stick']"
$server['fish_stick']

Using regex:

$ echo "$foo" | grep "\$server\['fish_stick'\]"
$server['fish_stick']

Using regex and handling nested single quotes:

$ echo "$foo" | grep '\$server\['\''fish_stick'\''\]'
$server['fish_stick']

Inside of single quotes, nested single quotes can not be not be escaped. You have to close the quotes, and then reopen it to "escape" the single quotes.

http://mywiki.wooledge.org/Quotes

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