In bash i have a string like this, please notice the first line and last line is blank.
How can i get the 300 number and the 20 number?
#first blank line
Found 300 modems:
/org/freedesktop/ModemManager1/Modem/20 [huawei] E3131
/org/freedesktop/ModemManager1/Modem/1 [huawei] E3131
/org/freedesktop/ModemManager1/Modem/0 [huawei] E303
#last blank line
Here's a Perl version...
perl -ne 'if(m[Found (\d+)] || m[Modem/(\d+)]) {print "$1\n"}' < file_name
Note. You can use the following script top play with the regex...
#!/usr/bin/perl
use strict;
use warnings;
while(<DATA>) {
if(m[Found (\d+)] || m[Modem/(\d+)]) {
print "$1 \n";
}
}
__DATA__
#first blank line
Found 300 modems:
/org/freedesktop/ModemManager1/Modem/20 [huawei] E3131
/org/freedesktop/ModemManager1/Modem/1 [huawei] E3131
/org/freedesktop/ModemManager1/Modem/0 [huawei] E303
#last blank line
If you save that script as test.pl you can use it like this...
perl -ne 'if(m[Found (\d+)] || m[Modem/(\d+)]) {print "$1\n"}' < test.pl
and I got the following output...
300
20
1
0
If you have that output stored in a variable called ${var}, you can do
sed -n 's/Found \([0-9]*\) modems:/\1/p' <<< "${var}"
sed loops over all lines in the string, -n suppresses the output. If we find the line with Found n modems:, we substitute it with the number only (s command) and print the line (p flag).
This will print 300, 20, 1, and 0:
sed -ne 's/Found \(.*\) modems:/\1/p' -e 's:.*Modem/\([^ ]*\) .*:\1:p' FILENAME
-n says not to print the output by default
-e COMMAND executes COMMAND. You can use it more than once.
Ending the s command with p says to print if any substitution happens.
Run man sed for more information.
If you really only want 300 and 20 as output, append |head -2 to limit the output to the first two lines.
