UNIX for Dummies Questions & Answers

Hello,

I looking to use grep to return a string with exactly n matches.

I'm building off this:

Code:

 
ls -aLl /bin | grep '^.\{9\}x' | tr -s ' '

Code:

-rwxr-xr-x 1 root root 632816 Nov 25 2008 vi
-rwxr-xr-x 1 root root 632816 Nov 25 2008 view
-rwxr-xr-x 1 root root 16008 May 25 2008 ypdomainname
-rwxr-xr-x 3 root root 62864 May 28 2008 zcat
-rwxr-xr-x 1 root root 594848 Jan 7 2007 zsh

Which prints a list of files in long format that are have an 'x' in the last file permissions position, then remove spaces between characters so that there is at most one. I now want to print just the file name. The following code does exactly that:

Code:

ls -aLl /bin | grep '^.\{9\}x' | tr -s ' ' | cut -d' ' -f9

Code:

vi
view
ypdomainname
zcat
zsh

What I want to see if I can do is use grep instead of cut.
I figure I would need grep to be able to match a pattern exactly n times then match .*/n$.
I would want to match [[:blank:]] 8 times then match any number of characters followed by an end of line character at the end of the string.

Code:

ls -aLl /bin | grep '^.\{9\}x' | tr -s ' ' | grep 'match[[:blank:]] 8 sequential or non-sequential times then .*/n$'

Any ideas if I can do this with grep and the power of regular expression?
I tried various expressions on the built in dictionary to see if I could return words with say exactly 4 'a' anywhere in the word without successs.

This should work:

Code:

grep -o "[^ ]*$"

It is probably better to do a pattern match anyway instead of a column number match, since the ls -l listing may take up more or fewer fields, depending on your locale.

Or use find:

Code:

find -L /bin/* -maxdepth 1 -perm /001

If you want to filter out the leading path you can use sed:

Code:

find -L /bin/* -maxdepth 1 -perm /001|sed 's|.*/||'

or a find option:

Code:

find -L /bin/* -maxdepth 1 -perm /001 -printf "%P\n"

or if you only want the files

Code:

find -L /bin -maxdepth 1 -type f -perm /001 -printf "%P\n"

The thread id changed because I merged this thread with another by mistake, sorry.
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Ok, cool.

grep -o prints just the matching sting on its own line.
[^ ]*$ matches any number of non-space, non-tab characters in a sequence that are at the of a line.

So, this seems good for grabbing text at the end of any line.

However, I'm still trying to match lines that can exactly n matches. MY current thoughts are find a line with only two 'x':

Code:

ls -aLl -R /bin | grep -o 'x.*x[^x]*'

x.*x[^x]* = find a 'x', followed by any number of any character, find a 'x', followed by any number of characters not 'x'
This just list highlights the entire line.

Code:

ls -aLl -R /bin | grep 'x'

This highlights each 'x' in a line without highlighting any other characters. Match any number of 'x'.

Code:

ls -aLl -R /bin | grep 'x\+'

This does the same as the above. Match any number of 'x'.

Code:

ls -aLl -R /bin | grep 'x\{n\}'

n = 1, matches like the above two examples. Match any number of 'x'.
n = 2, matches only when 'x' in a sequence of two. Sequence of four match because 'xxxx' is 'xx' 'xx', 'xxx' is only matched 'xx'. (I think).

Code:

ls -aLl -R /bin | grep 'x.*x'

This highlights all the characters between the first 'x' and last 'x'. Match at least two 'x'.

Code:

ls -aLl -R /bin | grep 'x.*x\{1\}'

This, appears to match the same sections in a line as the above code. Match at least two 'x'.

Code:

ls -aLl -R /bin | grep  '[^x]*'

This only highlights the line up until the first 'x'. Match all characters up until the first 'x'.

I think the issue may relate to the "greedy" regex engine, so to control it is the issue.
I think I may need to specify hard limit with \{1\}.
Also I'm not sure if I can use look ahead/behind/around or conidtional with grep's regular expressions, but these commands may solve the problem as well. x\(?!x\)
I may possibly need to use \? before the "." for cases that are xxx and I want to match exactly two 'x'.

Side question:
[^ ] == [^[:blank:]] == horizontal space, tab == \t
Horizontal single space == " " but it has no numerical representation for an amount like " " is the literal equivalent to " "?

you could just do:

Code:

 ls -aLl /bin | awk '/x /{print $9}'

which should be close ;-)

Thanks, that definitely does it.

I haven't gotten around to awk and sed so I'm still unsure on how they operate exactly. This is mainly an exercise for myself to understand the basics on regular expressions though and limitations of some of the simplier programs that take regular expressions.

So, I'm still trying to use grep and my short term goal is to return lines from:

Code:

 
ls -aLl -R /bin

That contain exactly two 'x' characters. No more, no less. Preferably using grep.

..contain exactly two 'x' characters. No more, no less. Preferably using grep...

one way:

Code:

#  ls -aLl -R /bin | grep "^[^x]*x[^x]*x[^x]*$"

HTH