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in bash, with
$ echo {1..10} 1 2 3 4 5 6 7 8 9 10 I can get a numbers sequence, but in some case I need
01 02 03 ... 10 how I can get this ?
and how I can get ?
001 002 ... 010 011 .. 100 
837
You can iterate the sequence of numbers in bash in two ways. One is by using the seq command, and another is by specifying the range in for loop. In the seq command, the sequence starts from one, the number increments by one in each step, and print each number in each line up to the upper limit by default.
$1 means an input argument and -z means non-defined or empty. You're testing whether an input argument to the script was defined when running the script. Follow this answer to receive notifications.
There are no Booleans in Bash. However, we can define the shell variable having value as 0 (“ False “) or 1 (“ True “) as per our needs.
A null string in Bash can be declared by equalizing a variable to “”. Then we have an “if” statement followed by the “-n” flag, which returns true if a string is not null. We have used this flag to test our string “name,” which is null.
This will work in any shell on a machine that has coreutils installed (thanks commenters for correcting me):
seq -w 1 10 and
seq -w 1 100 Explanation:
-w will:Equalize the widths of all numbers by padding with zeros as necessary.
seq [-w] [-f format] [-s string] [-t string] [first [incr]] lastprints a sequence of numbers, one per line (default), from first (default 1), to near last as possible, in increments of incr (default 1). When first is larger than last the default incr is -1
50
use seq command with -f parameter, try:
seq -f "%02g" 0 10
results: 00 01 02 03 04 05 06 07 08 09 10
seq -f "%03g" 0 10
results: 000 001 002 003 004 005 006 007 008 009 010
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