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Old 09-05-2013, 12:17 PM   #1
casperdaghost
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Registered: Aug 2009
Posts: 349

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formatting output in bash


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

Last edited by casperdaghost; 09-05-2013 at 12:57 PM.
 
Old 09-05-2013, 12:49 PM   #2
Firerat
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Distribution: Debian Jessie / sid
Posts: 1,471

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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'

Last edited by Firerat; 09-05-2013 at 12:51 PM.
 
Old 09-05-2013, 12:56 PM   #3
casperdaghost
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Registered: Aug 2009
Posts: 349

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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.
 
Old 09-05-2013, 01:08 PM   #4
Firerat
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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
 
Old 09-05-2013, 01:28 PM   #5
grail
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Location: Perth
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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
 
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Old 09-05-2013, 01:57 PM   #6
casperdaghost
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Registered: Aug 2009
Posts: 349

Original Poster
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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.
 
Old 09-05-2013, 02:02 PM   #7
schneidz
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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
 
1 members found this post helpful.
  


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