formatting output in bash

casperdaghost · 09-05-2013, 12:17 PM

So i have a list -
cat /tmp/foo

Code:

InvSellSide42
onSellSide42
CapSellSide42
gerSellSide42
FortSellSide42
M42BuySide
MeSellSide42
PSPSSide40

and if you cat out just one of these files it looks something like this -

Code:

[timers/plugin_timer_0_0]
action = doStart
trigger = 0 35 18 ? * 1,2,3,4,5;

[timers/plugin_timer_1_0]
action = doStop; doReset
trigger = 0 00 18 ? * 2,3,4,5,6;

All i need to get is the name, start and stop times.
which i can do - but it gets printed out like this.

Code:

ASellSide40,
0 35 18,
0 00 18
BSellSide42,
0 00 18,
0 50 17
CSellSide42,
0 30 19,
0 30 16
ESellSide42,
0 30 07,
0 30 17

I need it printed out like this:

Code:

ASellSide40, 0 35 18, 0 00 18
BSellSide42, 0 00 18, 0 50 17
CSellSide42, 0 30 19, 0 30 16
ESellSide42, 0 30 07, 0 30 17

here is the code - the code runs ok, i just do not know how to format the print out.

Code:

#!/bin/bash
for i in $(cat /tmp/foohoo)
do
dostart=$(cat /tmp/conf/$i.ini  | egrep -A2 "action = doStart")
parse_dostart=$(echo "$dostart" | perl -nle 'print /trigger\s=\s(\w{1,2}\s\w{1,2}\s\w{1,2})/')
dostop=$(cat /opt/ullink/service/ulbridge/conf/$i.ini  | egrep -A2 "action = doStop")
parse_dostop=$(echo "$dostop" | perl -nle 'print /trigger\s=\s(\w{1,2}\s\w{1,2}\s\w{1,2})/')
#printf "%10s\t%10d\t%10d \n"  $i, $parse_dostart, $parse_dostop
printf "$i,\t$parse_dostart,\t $parse_dostop\n"
done

Firerat · 09-05-2013, 12:49 PM

I would use awk,

Code:

for file in $(cat /tmp/foo); do
    awk -v File=$file 'BEGIN{printf File}
                    /trigger/{printf ", "$3" "$4" "$5}
                    END{printf"\n"}' $file
done

It does make a number of assumptions,
one being foo has full path
another is that each file in foo has the correct 'format'

casperdaghost · 09-05-2013, 12:56 PM

Is there a way to do it with printf formatting?
It can't be that hard. I have tried google - cant really find a explanation.

Firerat · 09-05-2013, 01:08 PM

I don't know perl.. but try removing the -l flag ( which processes line ends ) that is where your 'extra' end of lines are coming from

grail · 09-05-2013, 01:28 PM

I agree with Firerat that awk, perl or ruby would possibly be better choices for the entire code.

My personal choice if using bash though would be to go all bash:

Code:

while read -r file_name
do
    echo -n "$file_name"

    while read -r line
    do
        [[ $line =~ trigger ]] && echo -n ", ${line:10:7}"
    done<"/tmp/conf/$file_name.ini"
    echo
done</tmp/foohoo

casperdaghost · 09-05-2013, 01:57 PM

removing the newline - l from the perl statement worked.

how do i mark this as solved - and how do i make certain that everybody who answered this thread gets some points.

schneidz · 09-05-2013, 02:02 PM

to mark as solved ... select the dropdown under thread tools (above the first post) and select mark as solved.

to give points click on 'Did you find this post helpful? yes