UNIX for Dummies Questions & Answers

I have the following code:

Code:

ls -al /bin | tr -s ' ' | grep 'x[ ]'

ls -al: Lists all the files in a given director such as /bin
tr -s ' ': removes additional spaces between characters so that there is only one space
grep 'x[ ]': match all "x" characters that are followed by a whitespace.

I was able to repalce [ ] with [[:space:]] but it wouldn't seem to allow for [/s] or [/t]. I'm just seeing what other ways I can represent a whitespace. I also tried to implenment free-space mode but to no avail. When I use /s or /t, I just get no output.

On a side note does [ ] equal the exact samething as [[:space:]]?

Hi, MykC,

[:space:]=space, tab, newline, formfeed, verticaltab, carriage-return
[:blank:] =space, tab

so [[:blank:]]=[\t ]

I noticed you used a / instead of a \ .

BTW

Code:

grep 'x '

will get the same result. I don't think anything other than " " is being used so that should do.
If you just want the world executable entries a safer bet would be:

Code:

grep '^.\{9\}x '

(a line starting with 9 characters followed by an x)
If you also want to know if the items that symbolic links point to are in fact world executable you could use the -L (dereference) option if your ls supports it:

Code:

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

Hey, thanks for the help.

I did implement the repetion for safety which prevents sequences else where matching 'x ' from being flagged.

Ok, so when I use ls -aLl, it will replace all the files that were represented by a symbolic link (link a windows shortcut) with the information from the actual file. Which is probably better since symbolic links all have the same permissions were as the actual file may not.

So I now have:

Code:

 
ls -aLl /bin | grep '[[:graph:]]\{9\}x '

I don't know if I should use [:graph:] but most tutorials reccommend the "." whenever I can and I though [:graph:] would do the trick.

You can use [[:graph:]] to mean any non-space-or-tab character or you can use :

Code:

[^ \t]

(neither space nor tab )

If you use the "^" as the first character in your grep string, it means "at the beginning of the line", then I think "." would suffice since the -l output is always going to produce 10 characters in the first column. So:

Code:

grep '^[^ \t]\{9\}x '

or

Code:

grep '^.\{9\}x '