mjhjmtu

How to find a fixed strings with grep if the required output is not shown with grep -F? For example:

egrep -r 'string1|string2|string3' /var/www/http

with this what should we use?

Either use grep -E (the modern form for egrep) so | be treated as an extended regular expression alternation operator, but then you'd need to escape every other regular expression operator (like ., ?, *, ^, $, , [, (, ), , ) in your fixed strings.

Or use -F and pass the strings one per line or with several -es:

grep -Fe string1 -e string2 -e string3

Or:

grep -F 'string1
string2
string3'

Add the -x option if those fixed strings have to make up the entire matched line (as opposed to be found anywhere within the line without it). Most grep implementations also have a -w option for word match where for instance, string1 would match inside foo string1-2, but not inside foostring12.

For your specific example in comments, that would be:

grep -wE 'Allow from (8.8.2.5|192.3.0.4)|DenyAll'

Or:

grep -we 'Allow from 8.8.2.5' -e 'Allow from 192.3.0.4' -e 'DenyAll'

You'd want either -w or -x (assuming that's the whole lines, not even spaces around) here as otherwise it would also match on lines like Allow from 8.8.2.51.

You can also write it:

grep -xE '[[:space:]]*(Allow from (8.8.2.5|192.3.0.4)|DenyAll)[[:space:]]*'

To match on the full line (-x) but allowing leading and trailing spacing characters (here using [[:space:]] instead of [[:blank:]] to also allow the CR characters found at the end of lines coming from the MS-DOS world).

To look for those fixed strings anywhere in the line but not inside comments, you'd need to ensure that the part of the line leading to those fixed strings doesn't contain # characters. So something like:

grep -wE '^[^#]*(Allow from (8.8.2.5|192.3.0.4)|DenyAll)'

Note that the -w there doesn't apply to Allow as it applies to the whole matched string. So it would match on GAllow from for instance. Some grep implementations support <, > to match word boundaries explicitly (also known as b in some):

grep -E '^[^#]*<(Allow from (8.8.2.5|192.3.0.4)|DenyAll)>'

Also note that in apache2, configuration directives are case insensitive and any number of blanks are allowed between words. Also, I suspect you meant Deny from all instead of DenyAll, so maybe you'd want:

grep -iE '^[^#]*<(allow[[:blank:]]+from[[:blank:]]+(8.8.2.5|192.3.0.4)|deny[[:blank:]]+from[[:blank:]]+all)>'

The pipes | need to be escaped | like so:

grep 'string1|string2|string3' /var/www/http

The issue is that

grep -F -r 'string1|string2|string3' /var/www/http

would look for the literal string string1|string2|string3 in the files. Note, since you are looking for fixed strings, there is no reason to use egrep (or grep -E).

What you could do is to use

printf 'string1nstring2nstring3n' | grep -f /dev/stdin -F -r /var/www/http

This would make grep read the fixed strings to match against from standard input (provided by printf) and use them to get all lines that matched them in any file under the given directory.

Anything that generates a text document with the patterns on separate lines may replace the printf above, including a simple text file (which could be read directly by grep -f textfile -F ...).

Example from your comments:

Look for any of the strings Allow from 8.8.2.5, Allow from 192.3.0.4 and DenyAll in the files in $dirpath:

printf 'Allow from 8.8.2.5nAllow from 192.3.0.4nDenyAlln' |
grep -f /dev/stdin -F -r "$dirpath"