grep and 'Pattern'

[mvdberg112]

I try to find with Grep lines that contain both of two characters. For example the letter 'a' and also 'p' regardless of place on the line.

grep(1): print lines matching pattern - Linux man page
grep [options] PATTERN [FILE...]
grep, egrep, fgrep - print lines matching a pattern

So what qualifies as PATERN? Where does grep define what qualifies as 'pattern' in case the -E is not specified? Can we use BOOLEAN logic? Does it depend on the shell that is used? Or does grep evaluate the expression all by itself?

Example of what I want:
e.g. four files in the current directory:
pap1.txt
foo2.bar
foo3.png
foo4.test
The two lines below do not work:

ls  | grep -E "a & p"
ls | grep "a & p"

I want only to use grep once.
How can I make a logical AND?

I have read:

Tao of Regular Expressions (Linux Reviews)...
All about Linux: Learn how to use regular expressions the easy way...
Linux Command Directory: expr...
Tao of Regular Expressions (Linux Reviews)

[khafa]

you can use grep as follows

ls | grep "a*p"

but here it will match words like apple (where a comes before p) and not words like parent (where p comes first).
to solve this you can do

ls | grep a | grep p

[mvdberg112]

Thank you! I've still this question:
Is logical OR possible with normal grep? (without the -E option) What is the definition of pattern?

In the mean time, this answer "a*p" pushed me in the right direction. Because the logical OR between "a*p" and "p*a" is the same as a logical AND betwen 'a' and 'p'.
For -E option the syntax is: a+p, the '+' that the preceding charcter occurs at least once. See examples below.

root@erver:/home/user# ps -A |  grep -E "i+f"
   28 ?        00:00:00 kacpi_notify
12544 ?        00:00:00 update-notifier
12659 ?        00:00:02 notification-da
root@erver:/home/user# ps -A |  grep -E "f+i"
12544 ?        00:00:00 update-notifier
12659 ?        00:00:02 notification-da
12983 ?        00:00:00 soffice
12999 ?        00:00:30 soffice.bin
13252 ?        00:00:00 firefox
13268 ?        00:42:37 firefox-bin 
root@erver:/home/user# ps -A |  grep -E "i+f|f+i"
   28 ?        00:00:00 kacpi_notify
12544 ?        00:00:00 update-notifier         
12659 ?        00:00:02 notification-da
12983 ?        00:00:00 soffice
12999 ?        00:00:30 soffice.bin
13252 ?        00:00:00 firefox
13268 ?        00:42:37 firefox-bin

[kakariko81280]

You can specify multiple patterns to match with the -e switch.

chris@angua:~/dev/scratch$ cat test.txt
a
b
c
abc
chris@angua:~/dev/scratch$ grep -e a -e b test.txt
a
b
abc

Let us know how you get on,

Chris...

[vsemaska]

you can use grep as follows

ls | grep "a*p"

but here it will match words like apple (where a comes before p) and not words like parent (where p comes first).

It would also match 'pple' because the "a*" means zero or more occurances of "a". The correct pattern would be "a.*p" which means "a" followed by zero or more occurances of any character followed by "p".

to solve this you can do

ls | grep a | grep p

Using extended expression you can do it this was:

ls | grep -E "(a.*p|p.*a)"

[khafa]

vsemaska,

thanx for the remarks.
for the first one , my bad . you are totally right.

for the second one the author of the thread said

in case the -E is not specified