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Top Forums Shell Programming and Scripting Print arguments with the help of variable Post 302498220 by hakermania on Sunday 20th of February 2011 03:04:59 PM
Old 02-20-2011
Question Print arguments with the help of variable

Let's say I want to print the arguments $4 till $#, how can I do this?
$# contains the number of arguments
$@ contain all the arguments as string
What i need is something like
Code:
for i in $4_till_$#; do
#do something with $i
convert $i ~/$output
done

The first 3 arguments are used as options for other things....
Thanks in advance!
 

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print(1)							   User Commands							  print(1)

NAME
print - shell built-in function to output characters to the screen or window SYNOPSIS
ksh print [ -Rnprsu [n]] [arg...] DESCRIPTION
ksh The shell output mechanism. With no flags or with flag - or -, the arguments are printed on standard output as described by echo(1). OPTIONS
The following options are supported: -n suppresses new-line from being added to the output. -R -r (raw mode) ignore the escape conventions of echo. The -R option will print all subsequent arguments and options other than -n. -p causes the arguments to be written onto the pipe of the process spawned with |& instead of standard output. -s causes the arguments to be written onto the history file instead of standard output. -u [ n ] flag can be used to specify a one digit file descriptor unit number n on which the output will be placed. The default is 1. EXIT STATUS
The following exit values are returned: 0 Successful operation. >0 Output file is not open for writing. ATTRIBUTES
See attributes(5) for descriptions of the following attributes: +-----------------------------+-----------------------------+ | ATTRIBUTE TYPE | ATTRIBUTE VALUE | +-----------------------------+-----------------------------+ |Availability |SUNWcsu | +-----------------------------+-----------------------------+ SEE ALSO
echo(1), ksh(1), attributes(5) SunOS 5.10 15 Apr 1994 print(1)
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