Filter lines with special symbols using grep -E?

Question

file.txt:

vwive
wbvo$nivw
nowev^VWen
wev$
qnwe^

Filter lines that constains '^' in it using command :

1. grep -E "\^" file.txt

Output :

nowev^VWen
qnwe^

Filter lines that constains '$' in it using command :

2. grep -E "\$" file.txt

Output :

vwive
wbvo$nivw
nowev^VWen
wev$
qnwe^

Command 1 gives correct output by removing the special meaning of '^' but not command 2.

Why the behaviour of both command are different?

Try single quotes, `grep -E '\$' file.txt`, see [demo](https://ideone.com/OYolEe). — Wiktor Stribiżew, Aug 07 '20 at 09:50
Yeah, Single quotes work, but how double quotes worked in command 1 and not in command 2? — tusharRawat, Aug 07 '20 at 09:51
The `"\$"` is equal to `'$'`. `[[ '$' == "\$" ]] && echo "yes" || echo "no";` shows `yes`. — Wiktor Stribiżew, Aug 07 '20 at 09:54
shell expands variable names inside double-quoted strings where `$` is a special symbol which is used in variable names but `^` isn't — anubhava, Aug 07 '20 at 09:55
Thanks @anubhava. Why single backslash '\' doesn't remove special meaning of '$' and we have to use backslash 2 times to search for '$' in double quotes? — tusharRawat, Aug 07 '20 at 10:05
Also, see https://stackoverflow.com/questions/19147498/egrep-and-grep-difference-with-dollar and https://stackoverflow.com/questions/59081624/whats-the-differences-between-grep-and-grep — Wiktor Stribiżew, Aug 07 '20 at 10:07

score 0 · Answer 1 · answered Aug 07 '20 at 10:03

0

Why the behaviour of both command are different?

Because the \ character is an escape character inside " quotes, and \$ is one of escape sequences. Some parts from bash manual double qoutes:

Enclosing characters in double quotes (‘"’) preserves the literal value of all characters within the quotes, with the exception of ‘$’, ‘`’, ‘\’, and, when history expansion is enabled, ‘!‘. [...]
The backslash retains its special meaning only when followed by one of the following characters: ‘$’, ‘`’, ‘"’, ‘\’, or newline.
Within double quotes, backslashes that are followed by one of these characters are removed.
Backslashes preceding characters without a special meaning are left unmodified.

The "special meaning" of \ character is meant to be the escaping thing:

A non-quoted backslash ‘\’ is the Bash escape character. It preserves the literal value of the next character that follows

Because ^ is not one of $ ` " \, then "\^" preserves both \ and ^ so it's is equal to '\^'.

Because $ is one of them, then "\$" is equal to '$'. The slash get's removed and \$ is interpreted as $.

Such typos are easy to find out with set -x.

$ ( set -x; grep -E "\$" /dev/null; )
+ grep -E '$' /dev/null
$ ( set -x; grep -E '\$' /dev/null; )
+ grep -E '\$' /dev/null

answered Aug 07 '20 at 10:03

KamilCuk

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what if I want to use variable in regex like searching lines with 2 dot characters, would this work ? dot="." grep -E "\.\$dot" file.txt – tusharRawat Aug 07 '20 at 11:34
1

No it will not. Please take your time to learn about quoting. I would `grep -E '\.\'"$dot"` or `grep -E "\\.\\$dot"` – KamilCuk Aug 07 '20 at 11:36
grep -E "\.\\$dot" also works ? – tusharRawat Aug 07 '20 at 12:11
Yes, because `\.` is preserved – KamilCuk Aug 07 '20 at 12:20
But If variable expansion happens first than don't you think expression "\.\$dot" should get convert to "\.\." first and then regex matching? – tusharRawat Aug 07 '20 at 12:30
I do not understand. `"\$dot"` does not trigger expansion, because `\$` disables the special meaning of `$`. The expansion does not "happen first" then. Or maybe, the expansion happen only if there is a `$` has "special meaning" (ie. it triggers the expansion). Within double quotes and when preceded by "\", then `$` looses it's special meaning. – KamilCuk Aug 07 '20 at 12:32
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/219406/discussion-between-tusharrawat-and-kamilcuk). – tusharRawat Aug 07 '20 at 12:34

Filter lines with special symbols using grep -E?

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