LinuxQuestions.org - simple parameter expansion

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- - simple parameter expansion (https://www.linuxquestions.org/questions/linux-newbie-8/simple-parameter-expansion-4175476875/)

casperdaghost

09-12-2013 10:41 AM

simple parameter expansion

how would you guys go about adding a zero to the beginning if the integer in the parameter expansion is one digit, and ignoring it if the parameter expansion is 2 digits.

Code:

[casper@casper]$  for i in {6..11}; do  echo "201309${i}.txt.bz2"; done

2013096.txt.bz2

2013097.txt.bz2

2013098.txt.bz2

2013099.txt.bz2

20130910.txt.bz2

20130911.txt.bz2

i need to expand the range to :

Code:

20130906.txt.bz2

20130907.txt.bz2

20130908.txt.bz2

20130909.txt.bz2

20130910.txt.bz2

20130911.txt.bz2

colucix

09-12-2013 10:50 AM

Recent versions of bash accept zeroes in front of the first number:

Code:

$ echo {06..11}

06 07 08 09 10 11

Otherwise you can try printf and use the format specifiers at your pleasure:

Code:

for i in {6..11}

do

  echo 201309$(printf "%02d" $i).txt.bz2

done

In this specific case, you can manage dates using the date command and its format specifications:

Code:

day=20130906

while [[ $day -le 20130911 ]]

do

  echo ${day}.txt.bz2

  day=$(date -d "$day 1 day" +%Y%m%d)

done

pan64

09-12-2013 10:52 AM

see: http://wiki.bash-hackers.org/commands/builtin/printf
for i in {6..11}; do printf "201309%02d.txt.bz2\n" $i; done

casperdaghost

09-12-2013 10:53 AM

amazing - thanks

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