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Date:      Mon, 24 Dec 2001 19:29:19 +0100
From:      Nils Holland <nils@tisys.org>
To:        freebsd-questions@freebsd.org
Subject:   Questions to all script wizards out there...
Message-ID:  <20011224192919.A89314@tisys.org>

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Hi folks,

there are many things that I'd like to eventually learn, for example
writing advanced shell scripts. Right now, I'd like to set up a script for
a particular purpose, but I don't really know how to go about it.

Here's what this script should do:

I would like the script to use wget in order to fetch the program listing
of a radio station from the station's website. I would like to have seven
days fetched in advance.

Now, the URL always looks like this:
http://www.dradio.de/dlf/vorschau/<year>/<month>/<day>.html

So in order to make wget fetch today's program, I'd run it like this:

wget -options http://www.dradio.de/dlf/vorschau/2001/12/24.html

That's the theory, but how do I tell my shell script to do that? I have
two problems, one of which is probably easy to solve, while the other is
a little tough:

1) The shell script would have to have a look at today's date and construct
the appropriate URL. It would probably have to obtain the output of date
and assign the year, month and day values to some variables, and then
create the http://www.dradio.de/dlf/vorschau/<year>/<month>/<day>.html URL.
Now, any ideas how to get that done in a script?

2) As I said, I'd like to get seven days fetched in advance. An easy way to
do that would be to simply loop multiple times, always incrementing <day>
by one. However, not each month has the same number of days, so assuming
that after <day> has reached 31 it should start from 1 again is not always
accurate. I guess handling this in a shell script would be a fairly hard
thing to do, but if someone has any ideas if it's (easily) possible, I'd
like to know it.

Greetings
Nils

-- 
Nils Holland
Ti Systems - FreeBSD in Tiddische, Germany
http://www.tisys.org * nils@tisys.org

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