Eureka
It mean's that i don't have to work on weekends and that i don't have to wakeup on 3.00AM to come to work on weekdays.
It's frequent the problems with that printer, and this will able them to repeat what they need without help.
Thank you very much once again, Ygor.
Hi there
i'm using this script from Ygor ,where i have made a small change to adapt it to my needs.
It just happens that i'm trying (not very successfuly) to change it a bit.
Right now i have a file , that the script loads 48 lines each time into an array and look for a match on the 6th entry, then prints those 48 lines where the match is found and print the next "xx" lines i indicate to the script.
here it is:
Code:
#!/usr/bin/ksh
echo ""
echo ""
printf "file :"
read sFile
echo ""
echo ""
printf "Client :"
read clientID
echo ""
echo ""
printf "How many more?"
read Qta
qt=$(($Qta * 48))
if [ $clientID ]
then
awk -v cid="$clientID" -v xx="$qt" '
{
arr[++cnt]=$0
if (cnt==48) {
if (arr[6] ~ cid) {
for (idx=1;idx<=48;idx++) {
print arr[idx]
}
for (idx=1;idx<=xx;idx++) {
getline
print $0
}
}
cnt=0
}
}' /usr/imp/$sFile > /usr/imp/rep.txt
fi
this works wonderfuly.
My problem now, is that some times the user don't know how many more pages in the file he has to print. What he allways know is the first page and last page to repeat.
So, i need to change the script indicating not HOW MANY PAGES MORE, but a second match to be found in the file and then print from the 48 lines where the first match is found to the 48 lines where the second match is found.
awk -v cid="$clientID" -v eid="$endingID" '
{
if (NR%48==6 && $0~cid) {
for(x=NR-5;x<NR;x++)
print a[x%5]
cnt=1000000
}
if (NR%48==6 && $0~eid) cnt=40 --------------------- this should be 43 in my understanding, otherwise it will make the last page 3 lines short
a[NR%5]=$0
}
cnt-->0' /usr/imp/$sFile > /usr/imp/rep.txt
thanks Ygor, i've been learning a lot with you. Thanks for that
OS version: RHEL 6.7
Shell : Bash
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