[SOLVED] Please help me with this sed command

towheedm · 01-22-2012, 11:58 AM

I'm trying to use sed to extract the URL portion of a line from a file.

The file contains the following line:

Code:

DOWNLOAD="http://ftp.a.b/c/d/file-to-download"

Using this command:

Code:

sed -n 's/^DOWNLOAD="//p' filename

returns:

Code:

http://ftp.a.b/c/d/file"

I can remove the last " character by piping the first command to another sed command:

Code:

sed -n 's/^DOWNLOAD="//p' filename | sed -n 's/"//p'

but there must be a better way.

The second part I have no idea where to even start. I would also like to return the "file-to-download" part of the URL.

Any help is greatly appreciated.

sycamorex · 01-22-2012, 12:04 PM

try:

Code:

sed 's/.*\(http.*\)\("\)/\1/' infile

druuna · 01-22-2012, 12:12 PM

Hi,

The URL part (alternative for sycamorex code):

Code:

sed 's/DOWNLOAD="\(.*\)"/\1/' infile

The file part:

Code:

sed 's%.*/\(.*\)"%\1%' infile

Hope this helps.

towheedm · 01-22-2012, 01:10 PM

druuna,

That's works. I understand the first part to return the URL. To print out only that line I did:

Code:

sed -n 's/DOWNLOAD="\(.*\)"/\1/p' infile

The part to return the file I don't get. I should have mentioned that 'infile' has several URLs, so the second part returns the file part of each URL. How can I get it to return (and print) only the file part of the line starting with DOWNLOAD=". I'm having a hard time understanding the second part. Is it too much to ask for a little explanation?

Thank you very much.

druuna · 01-22-2012, 01:39 PM

Hi,

Quote:

Originally Posted by towheedm

druuna,

That's works. I understand the first part to return the URL. To print out only that line I did:

Code:

sed -n 's/DOWNLOAD="\(.*\)"/\1/p' infile

The part to return the file I don't get. I should have mentioned that 'infile' has several URLs, so the second part returns the file part of each URL. How can I get it to return (and print) only the file part of the line starting with DOWNLOAD=". I'm having a hard time understanding the second part. Is it too much to ask for a little explanation?

Both my samples assume that DOWNLOAD="http://ftp.a.b/c/d/file-to-download" is all that's on that line and nothing else. If it isn't then both samples will probably work incorrectly.

Please provide us with a relevant sample, otherwise the answers given might not work.

About my second solution:

Code:

sed 's%.*/\(.*\)"%\1%' infile

- The brown part matches all up to and including the last forward slash,
- The green part matches the rest (minus the last ") and is used in the replace part as \1,
- I used % instead of / as separator because I need to match a forward slash.

The blue part will fail if there's more on a line. Sed is greedy and will grab all up to and including the last / it finds.

Hope this helps.

H_TeXMeX_H · 01-22-2012, 02:06 PM

It would help if you posted more of the file, I suspect it is an HTML file, and as such should have tags, which will make it much easier to deal with.

towheedm · 01-22-2012, 02:46 PM

Ah, I was wondering about the %. I usually use a comma so did not think about it as the separator. Thanks for the explanation. Now:

Code:

sed -n 's%DOWNLOAD=.*/\(.*\)"%\1%p' infile

returns the file part from the line starting with DOWNLOAD=.

BTW: Colors don't show in code or do I need to change some setting. Forget that...gotta get my eyes checked or get a new monitor. :-)

The file is actually the .info file from a SlackBuild archive. The format is:

Code:

PRGNAME="program-name"
VERSION="program-version"
HOMEPAGE="http://home/page/to/program/archive"
DOWNLOAD="http://link/to/program/archive/program-name"
MD5SUM="MD5 checksum value for program-name"
DOWNLOAD_xX86_64=""
MD5SUM_x86_64=""
MAINTAINER="Name of maintainer"
EMAIL="email address of maintainer"
APPROVED="Name of person who approved the SlackBuild script"

Thanks for all your help.

Cedrik · 01-22-2012, 06:55 PM

You can also use basename to extract the file name from the url

Code:

URL=$(sed -n 's/DOWNLOAD="\(.*\)"/\1/p' filename)
FILE=$(basename $URL)

echo "url: $URL"
echo "file: $FILE"