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

BashSed

Bash Problem Overview


From what I can find, when you use single quotes everything inside is considered literal. I want that for my substitution. But I also want to find a string that has single or double quotes.

For example,

sed -i 's/"http://www.fubar.com"/URL_FUBAR/g'

I want to replace "http://www.fubar.com" with URL_FUBAR. How is sed supposed to recognize my // or my double quotes?

Thanks for any help!

EDIT: Could I use s/\"http\:\/\/www\.fubar\.\com\"/URL_FUBAR/g ?

Does \ actually escape chars inside the single quotes?

Bash Solutions


Solution 1 - Bash

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.

Solution 2 - Bash

Regarding the single quote, see the code below used to replace the string let's with let us:

command:

echo "hello, let's go"|sed 's/let'"'"'s/let us/g'

result:

> hello, let us go

Solution 3 - Bash

My problem was that I needed to have the "" outside the expression since I have a dynamic variable inside the sed expression itself. So than the actual solution is that one from lenn jackman that you replace the " inside the sed regex with [\"].

So my complete bash is:

RELEASE_VERSION="0.6.6"

sed -i -e "s#value=[\"]trunk[\"]#value=\"tags/$RELEASE_VERSION\"#g" myfile.xml

Here is:

# is the sed separator

[\"] = " in regex

value = \"tags/$RELEASE_VERSION\" = my replacement string, important it has just the \" for the quotes

Solution 4 - Bash

It's hard to escape a single quote within single quotes. Try this:

sed "s@['\"]http://www.\([^.]\+).com['\"]@URL_\U\1@g" 

Example:

$ sed "s@['\"]http://www.\([^.]\+\).com['\"]@URL_\U\1@g" <<END
this is "http://www.fubar.com" and 'http://www.example.com' here
END

produces

this is URL_FUBAR and URL_EXAMPLE here

Solution 5 - Bash

Escaping a double quote can absolutely be necessary in sed: for instance, if you are using double quotes in the entire sed expression (as you need to do when you want to use a shell variable).

Here's an example that touches on escaping in sed but also captures some other quoting issues in bash:

# cat inventory
PURCHASED="2014-09-01"
SITE="Atlanta"
LOCATION="Room 154"

Let's say you wanted to change the room using a sed script that you can use over and over, so you variablize the input as follows:

# i="Room 101" (these quotes are there so the variable can contains spaces)

This script will add the whole line if it isn't there, or it will simply replace (using sed) the line that is there with the text plus the value of $i.

if grep -q LOCATION inventory; then 
## The sed expression is double quoted to allow for variable expansion; 
## the literal quotes are both escaped with \ 
    sed -i "/^LOCATION/c\LOCATION=\"$i\"" inventory
## Note the three layers of quotes to get echo to expand the variable
## AND insert the literal quotes
else 
    echo LOCATION='"'$i'"' >> inventory
fi

P.S. I wrote out the script above on multiple lines to make the comments parsable but I use it as a one-liner on the command line that looks like this:

i="your location"; if grep -q LOCATION inventory; then sed -i "/^LOCATION/c\LOCATION=\"$i\"" inventory; else echo LOCATION='"'$i'"' >> inventory; fi

Solution 6 - Bash

Aside: sed expressions containing BASH variables need to be double (")-quoted for the variable to be interpreted correctly.

If you also double-quote your $BASH variable (recommended practice)

... then you can escape the variable double quotes as shown:

sed -i "s/foo/bar ""$VARIABLE""/g" <file>

I.e., replace the $VARIABLE-associated " with "".

(Simply -escaping "$VAR" as \"$VAR\" results in a "-quoted output string.)


Examples

$ VAR='apples and bananas'
$ echo $VAR
apples and bananas

$ echo "$VAR"
apples and bananas

$ printf 'I like %s!\n' $VAR
I like apples!
I like and!
I like bananas!

$ printf 'I like %s!\n' "$VAR"
I like apples and bananas!

Here, $VAR is "-quoted before piping to sed (sed is either '- or "-quoted):

$ printf 'I like %s!\n' "$VAR" | sed 's/$VAR/cherries/g'
I like apples and bananas!

$ printf 'I like %s!\n' "$VAR" | sed 's/"$VAR"/cherries/g'
I like apples and bananas!

$ printf 'I like %s!\n' "$VAR" | sed 's/$VAR/cherries/g'
I like apples and bananas!

$ printf 'I like %s!\n' "$VAR" | sed 's/""$VAR""/cherries/g'
I like apples and bananas!

$ printf 'I like %s!\n' "$VAR" | sed "s/$VAR/cherries/g"
I like cherries!

$ printf 'I like %s!\n' "$VAR" | sed "s/""$VAR""/cherries/g"
I like cherries!

Compare that to:

$ printf 'I like %s!\n' $VAR | sed "s/$VAR/cherries/g"
I like apples!
I like and!
I like bananas!


$ printf 'I like %s!\n' $VAR | sed "s/""$VAR""/cherries/g"
I like apples!
I like and!
I like bananas!

... and so on ...

Conclusion

My recommendation, as standard practice, is to

  • "-quote BASH variables ("$VAR")
  • "-quote, again, those variables (""$VAR"") if they are used in a sed expression (which itself must be "-quoted, not '-quoted)
$ VAR='apples and bananas'

$ echo "$VAR"
apples and bananas

$ printf 'I like %s!\n' "$VAR" | sed "s/""$VAR""/cherries/g"
I like cherries!

Solution 7 - Bash

Prompt% cat t1
This is "Unix"
This is "Unix sed"
Prompt% sed -i 's/\"Unix\"/\"Linux\"/g' t1
Prompt% sed -i 's/\"Unix sed\"/\"Linux SED\"/g' t1
Prompt% cat t1
This is "Linux"
This is "Linux SED"
Prompt%

Solution 8 - Bash

You can use %

sed -i "s%http://www.fubar.com%URL_FUBAR%g"

Solution 9 - Bash

You need to use " for escaping " character (\ escape the following character

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

Solution 10 - Bash

May be the "" char, try this one:

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

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionKRBView Question on Stackoverflow
Solution 1 - BashKusalanandaView Answer on Stackoverflow
Solution 2 - Bashtony duanView Answer on Stackoverflow
Solution 3 - BashDenisView Answer on Stackoverflow
Solution 4 - Bashglenn jackmanView Answer on Stackoverflow
Solution 5 - Bashstarry_eyed_sysadminView Answer on Stackoverflow
Solution 6 - BashVictoria StuartView Answer on Stackoverflow
Solution 7 - BashSachinHView Answer on Stackoverflow
Solution 8 - BashAndreyView Answer on Stackoverflow
Solution 9 - BashGuilhem HoffmannView Answer on Stackoverflow
Solution 10 - BashGiuseppe La RosaView Answer on Stackoverflow