UNIX for Beginners Questions & Answers

I'm trying to learn about regular expressions. Let's say I want to list all the files in /usr/bin beginning with "p", ending with "x", and containing an "a".
I know this works:

Code:

ls | grep ^p | grep x$ | grep a

but I'm thinking there must be a way to do it without typing grep three times. Some of my attempts:

Code:

ls | grep '^p' 'x$' 'a'
ls | grep ^p | x$ | [a]
ls | grep -E "[[ ^p ]]+[[ x$ ]]+[[ a ]]"

Ubuntu 18.04.2; Xfce 4.12.3; kernel 4.15.0-45-generic; bash 4.4.19(1); Dell Inspiron-518

Try this...

Code:

ls | grep "^p.*a.*x$"

.* is for any character, 0 to any number of occurrances.

So the task is to create one combined regular expression that matches all criteria you want.

You are right and there is - in fact there are two ways:

First, you can use several expressions in grep at once by using the -e-switch:

Code:

grep -e "one" -e "two" /some/file

will list all lines containing "one" AND all lines containing "two" from that file. It is in fact a logical OR for these two expressions.

The second possibility (and this is probably what you wanted) is to combine regular expressions. i.e. to get all lines containing "one" and "two" in that order you would write:

Code:

grep 'one.*two' /some/file

The regexp means: "one", followed by anything (".*"), followed by "two". If the order of the two words should not matter you need two regexps, which you can combine with the method above:

Code:

grep -e 'one.*two' -e 'two.*one' /some/file

Search for "'one', something, then 'two'" or for "'two', something, then "one"'.

You can also use sed (the stream editor) for such (or even more complex) purposes where grep might get a bit unwieldy: sed -n will only print the lines you explicitly print (the default is to print every line after it is processed, which includes passing it through unchanged) therefore:

Code:

sed -n '/one/p' /some/file

basically is the same as

Code:

grep 'one' /some/file

But you can nest certain rules in sed which grep cannot do. i.e.

Code:

sed -n '/one/ {
             /two/ {
                   /three/p
             }
         }' /some/file

Would be similar to the example above: print only the lines containing "one", "two" and "three" but in any order.

I hope this helps.

bakunin

@stomp--ls | grep "^p.*a.*x$" works great, with either single or double quotes. I'm used to thinking of the asterisk as a wildcard, which was confusing me at first.
@bakunin--this works great

Code:

sed -n '/^p/ {
             /x$/ {
                  /a/p
                  }
             }' catalogue

Not really related to the question, but:

Code:

ls -d p*a*x

@Scrutinizer--it does relate to the question, as it performs the task. It found the target file, partx, in /usr/bin.


I don't understand why, however, as the man page for ls says the -d option is to "list directories themselves, not their contents".

Normally ls lists the contents of a given directory.
Say there would be a pax/ subdirecory beside the partx file,
ls p*a*x would list partx and the contents of the pax/
While ls -d p*a*x would list the two items partx and pax.