Bashscripts.org

I have have a line of text which contains a 24hour time imbeded in it. Something like:

"It is now 2130 its time to begin"

The time can be anywhere in the string.

How can I strip out just the four digits that make up the time? I'm reading some stuff of regular expressions, and [0-9][0-9][0-9][0-9] is the proper pattern, but how can I use that to get the specific numbers out of the line? I'm just not getting how to use the expressions. Thanks.

doh, its grep -o I can't ever find anything in a man page until I read it 20 times...........

lol... i'm right there with you. I think that they should convert manpages to pdf... that way the flags can be indexed... and you could just jump to them in the order you want, and by pass all the extra crud

Hi, guys, I didn't find a solution with grep, but found workrounds with tr and sed. May these help.

First I want to make it clear that while [0-9][0-9][0-9][0-9] works quite well, [0-9]{4} do the same job.

As in the "tr" case, I tried

It works.
I have been learning the powerfull sed these days, so I gave it a try:

The sed command here replace whitespace with a ...... hmmm, line terminator ( Sorry, I am not sure about the right jargon. lol ), thus place every word in a seperate line.
BTW, I found the first line of grep manual is " grep , egrep, fgrep, rgrep, - print lines matching a pattern". So maybe it print "lines" instead of specified words in lines in my humble opion.

Slightly more simple than the above



irritatingly I cannot seem to persuade sed to use the [[:digit:]]{4} syntax

Why not just use this?

Anyway, I think the problem is that you haven't escaped {}.
http://www.ibm.com/developerworks/aix/l ... xp/#list16 there is a list of the simple things that needs to be escaped to use.