Output everything before the first slash '/' in a line

Question

I have a set of URLs and I'm really only interested in anything up to the first /.

How can I capture this info to a text file?

Input (foo.txt):

apple.com/nothing.js  
t1.msn.com/cookie=22  
happy.net/whatever

Output (redirected to file: foo_filter.txt):

apple.com/  
t1.msn.com/  
happy.net/  

Answer 1

If you don't want the trailing slash, it's very straightforward

cut -d/ -f1 foo.txt
awk -F/ '{print $1}' foo.txt
sed 's!/.*!!' foo.txt

If you do want that trailing slash, then

awk -F/ '{print $1 "/"}' foo.txt
sed 's!/.*!/!' foo.txt

All of these will write to stdout (your screen) so you can see the result immediately. To redirect them to your target file, use >foo_filter.txt on the end of the command. For example,

awk -F/ '{print $1 "/"}' foo.txt >foo_filter.txt

Answer 2

Using Miller:

mlr --nidx --ifs '/' -N cut -f 1 file

or using GNU datamash:

datamash dirname 1 <file

Answer 3

$ awk 'sub("/.*","/")' foo.txt
apple.com/
t1.msn.com/
happy.net/

Answer 4

Your better option here is actually sed, because it edits the stream on a line by line basis.

Try the following:

sed 's/\/.*//' foo.txt > foo_filter.txt

This tells sed that - per line - replace anything after the / with nothing. You then redirect the output to the new file with the >. You can read more from the sed manual here.

Note: because sed is greedy, you might need to specify the first slash with a 1 at the end of the command:

sed 's/\/.*//1' foo.txt > foo_filter.txt

You definitely can use awk if you have strings with multiple slashes:

awk -F"/" '{print $1"/"}' foo.txt > foo_filter.txt

The -F"/" sets the field delimiter to forward slash, and '{print $1"/"}' prints the first field followed by a slash (since it's the field delimiter, it gets removed on print and has to be re-included).

Answer 5

Here's an awk and a cut solution.

$ cut -f1 -d/ foo.txt
apple.com
t1.msn.com
happy.net
$ awk -F/ '{ print $1 }' foo.txt
apple.com
t1.msn.com
happy.net
$ awk -F/ '{ print $1"/" }' foo.txt
apple.com/
t1.msn.com/
happy.net/
$

Answer 6

With just grep:

$ grep -oE '^[^/]+/' foo.txt

Output:

apple.com/
t1.msn.com/
happy.net/

To fulfill all the requirements:

grep -oE '^[^/]+/' foo.txt | tee foo_filter.txt