I have wrote simple shell script test.sh as follows:
while getopts ":A:" OPTION
do
case $OPTION in
A)
echo $OPTARG
?)
echo "no option"
esac
done
And executed the scripts as follows
$ ./test.sh -A 1 2
Now if got argument 1 by $OPTARG but how can i access the second argument ( 2 in this case)?
Regards
Jayesh
There are several options.
(1) You can use shift and take $1
while -n "$1"
do
# do something with $1
shift
done
(2) You can iterate through the args:
for i
do
# do something with $i
done
There are other alternatives also.
Related
I have one shell script.
#
. ./shprofile
if [ "$1" != "INSERT_ALL_ITEMS" -a "$1" != "INSERT_ONE_ITEM" ]
then
echo "Usage: $0 [INSERT_ALL_ITEMS|INSERT_ONE_ITEM]"
exit
fi
#
echo "Start Date & Time is.." `date`
case "$1" in
INSERT_ALL_ITEMS)
echo "Executing all Items"
;;
INSERT_ONE_ITEM)
#
echo "Executing one Item"
;;
*)
echo "Invalid Options!!"
;;
esac
#
echo "Stop Date & Time is.." `date`
If I execute the above script with below command
./runItemsBatch.sh INSERT_ALL_ITEMS
Then it is showing echo message, instead of executing the script
Usage: ./runItemsBatch.sh [INSERT_ALL_ITEMS|INSERT_ONE_ITEM]
What is the wrong in script? Why it is not executing the script, even though I give correct option.
Need help.
Given that your script ran fine on my CentOS 7 virtual machine (commenting the reference to the other file), I think it may be that you didn't set up the interpreter properly.
The first line (#) should be a shabang (#!) followed by the interpreter. e.g.: #!/bin/sh.
You could also try to remove the quotes for the arguments in your script because the interpreter could ...interpret them as literals.
I mean that you should try to replace the line
if [ "$1" != "INSERT_ALL_ITEMS" -a "$1" != "INSERT_ONE_ITEM" ]
with
if [ $1 != "INSERT_ALL_ITEMS" -a $1 != "INSERT_ONE_ITEM" ]
Try with either this two things. In my case however, the script worked with every combination of "shabang / no shabang and quotes / no quotes"
Hope I helped.
EDIT: Try also to delete the spaces between variables and operators in the if (Again, these are rules depending by the interpreter you are using, or defaulted to use)
Due to my lack of thorough understanding using getopts, the title is definitely vague :0. I am currently writing a bash script and I would like to add an option that outputs the other options within the case statement in getopts. For the sake of scaling, I have shortened the program.
#!/bin/bash
while getopts :abc opt
do
case $opt in
a)
echo "Hello"
;;
b)
echo "Goodbye"
c)
:ab #****I WANT -c TO OUTPUT THE RESULTS OF a and b************
;;
esac
done
As you can see in option c, I would like this particular option (-c) to put out both the results of -a and -b. Is there a way to go about this by simply making c call on option a and b?
you can introduce functions to reduce duplications, something like this:
#!/bin/bash
do_a() {
echo "Hello"
}
do_b() {
echo "Goodbye"
}
while getopts :abc opt
do
case $opt in
a)
do_a
;;
b)
do_b
;;
c)
do_a
do_b
;;
esac
done
If you are using a recent version of Bash, instead of terminating case clauses with ;; you could use bash specific ;;& with multiple patterns:
#!/bin/bash
while getopts :abc opt
do
case $opt in
a|c)
echo "Hello"
;;&
b|c)
echo "Goodbye"
;;&
esac
done
And:
$ bash script.bash -a
Hello
$ bash script.bash -c
Hello
Goodbye
Using ‘;;&’ in place of ‘;;’ causes the shell to test the patterns in the next clause, if any, and execute any associated command-list on a successful match.
I am using a KSH script to execute a binary (program) that has the following syntax to execute correctly:
myprog [-v | --verbose (optional)] [input1] [input2]
The program prints nothing & returns exit code 0 (zero) on success. On failure it prints ERROR messages to STDERR & returns exit status > 0. If -v option is specified it prints verbose details to STDOUT both in case of success and failure.
To make this usable and reduce chances of argument swapping and user controlled logging I used a ksh shell script to invoke this binary. The syntax to run the ksh shell script is as:
myshell.sh [-v (optional)] [-a input1] [-b input2]
If -v option is specified, ksh redirects STDOUT to <execution_date_time>_out.log and STDERR to <execution_date_time>_err.log. My ksh script is as follows:
myshell.sh :
#! /bun/ksh
verbopt=""
log=""
arg1=""
arg2=""
dateTime=`date +%y-%m-%d_%H:%M:%S`
while getopts "va:b:" arg
do
case $arg in
v) # verbose output
verbopt="-v"
log="1>${dateTime}_out.log 2>${dateTime}_err.log"
;;
a) # Input 1
arg1=$OPTARG
;;
b) # Input 2
arg2=$OPTARG
;;
*) # usage
echo "USAGE: myshell.sh [-v] [-a input1] [-b input2]"
exit 2
;;
esac
done
if [[ -z $arg1|| -z $arg2]]
then
echo "Missing arguments"
exit 2
fi
myprog $verbopt $arg1 $arg2 $log
exit $?
The problem here is, all the output STDERR & STDOUT is printed on the screen (i.e, No redirection took place) as well as no *.log files were created after successful or unsuccessful execution (i.e, exit status: 0 or >0 respectively).
Can anyone help me out on this?
Thanks.
Rather than trying to monkey patch redirections into the command line, just redirect the streams when you parse the flags. That is:
while getopts "va:b:" arg
do
case $arg in
v) # verbose output
verbopt="-v"
exec 1>${dateTime}_out.log 2>${dateTime}_err.log
;;
...
You need to be a little careful, since you do some error checking after this and you probably don't want your later error messages going to the *_err.log, but that's fairly trivial to fix. (eg, error check sooner, or do a test -n "$verbopt" && exec > ... after the error check, or similar)
The problem is that > is not expanded in the value of $log.
I'm afraid you will need to use a conditional for this, for example:
cmd="myprog $verbopt $arg1 $arg2"
if [ "$log" ]; then
$cmd 1>${dateTime}_out.log 2>${dateTime}_err.log
else
$cmd
fi
I would use the idiom exec redirection, which runs the rest of the script as if the given redirection had been supplied when it was run:
if need_to_log; then
exec >stdout_file 2>stderr_file
fi
this command will be logged if the above if statement was true
If you need to restore stdout and stderr afterward for the script to do more unlogged things, you can just run the logging part in a subshell:
(
if need_to_log; then
exec >stdout_file 2>stderr_file
fi
this command will be logged if the above if statement was true
)
this command will not be logged regardless
I would also build the command in an array, so you can add things like -v to it without having to have a separate variable for each possible parameter. If the order in which the -a and -b arguments are supplied to myprog doesn't matter, you can just add those to the array instead of having separate variables as well.
You can see my version below. Besides the above changes, I also don't bother getting the timestamp if not logging, since it's unneeded, and send error messages to standard error instead of standard out using the ksh builtin print.
Here's what I put together:
#!/usr/bin/env ksh
# new array syntax requires ksh93+; for older ksh, use this:
# set -A cmd myprog
cmd=(myprog) # build up the command to run in an array
log_flag=0 # nonzero if the command should be logged
input_a= # the two input filenames
input_b=
while getopts 'va:b:' arg; do
case $arg in
v) # verbose output
# older ksh: set -A cmd "${cmd[#]}" -v
cmd+=(-v)
log_flag=1
;;
a) # Input 1
input_a=$OPTARG
;;
b) # Input 2
input_b=$OPTARG
;;
*) # usage
print -u2 "USAGE: $0 [-v] [-a input1] [-b input2]"
exit 2
;;
esac
done
if [[ -z $input_a || -z $input_b ]]; then
print -u2 "$0: Missing arguments"
exit 2
fi
if (( log_flag )); then
timestamp=$(date +%y-%m-%d_%H:%M:%S)
exec >"${timestamp}_out.log" 2>"${timestamp}_err.log"
fi
"${cmd[#]}" "$input_a" "$input_b"
Your timestamp uses the two-digit year (%y); that and the underscore between the components are the only deviations from the ISO 8601 standard, so I would recommend you go ahead and adopt the standard format. That'd be %Y-%m-%dT%H:%M:%S, or, in C libraries with newer versions of strftime, %FT%T.
You could also be a little more clever and make log_flag a string that is either empty or -q, pass that to the command, and test it against the empty string to determine whether or not to open the log files, but I find the logic easier to follow with the simple 0/1 value treated as a Boolean.
Take a look at the eval command.
Replace ...
myprog $verbopt $arg1 $arg2 $log
with:
eval myprog $verbopt $arg1 $arg2 $log
I don't know what your myprog does but here's a simple example using eval to run date (valid command) and date xyz (invalid command), redirecting output to log.stdout/log.stderr accordingly:
$ cat logout
log='1>log.stdout 2>log.stderr'
'rm' -rf log.std* > /dev/null 2>&1
echo ""
echo 'eval date ${log}'
eval date ${log}
echo ""
echo "++++++++++++ log.stdout"
cat log.stdout
echo "++++++++++++ log.stderr"
cat log.stderr
echo "++++++++++++"
'rm' -rf log.std* > /dev/null 2>&1
echo ""
echo 'eval date xyz ${log}'
eval date xyz ${log}
echo ""
echo "++++++++++++ log.stdout"
cat log.stdout
echo "++++++++++++ log.stderr"
cat log.stderr
echo "++++++++++++"
Now run the script:
$ logout
eval date ${log}
++++++++++++ log.stdout
Sun Jul 23 15:56:01 CDT 2017
++++++++++++ log.stderr
++++++++++++
eval date xyz ${log}
++++++++++++ log.stdout
++++++++++++ log.stderr
date: invalid date `xyz'
++++++++++++
what is the best way to implement switches [ex: -m] for shell scripts?
I can do it via the switch case statement. But i am curious to know is there any other standard way to get all the arguments into a variable via a switch.
Ex:
-m A1 A2 -c c1 c2
So that,
M[] can take -m
and C[] can all take -c
The best known way is to use getopts, see http://wiki.bash-hackers.org/howto/getopts_tutorial
An example :
#!/bin/bash
while getopts ":a" opt; do
case $opt in
a)
echo "-a was triggered!" >&2
;;
\?)
echo "Invalid option: -$OPTARG" >&2
;;
esac
done
I have a script, which is essentially a wrapper around an executable by the same name on a different machine. For the sake of example, i'll wrap printf here. My current script looks like this:
#!/bin/bash
ssh user#hostname.tld. printf "$#"
Unfortunately, this breaks when one of the arguments contains a space, e.g. i'd expect the following commands to give identical outputs.:
~$ ./wrap_printf "%s_%s" "hello world" "1"
hello_world1_
~$ printf "%s_%s" "hello world" "1"
hello world_1
The problem gets even worse when (escaped) newlines are involved. How would I properly escape my arguments here?
Based on the answer from Peter Lyons, but also allow quotes inside arguments:
#!/bin/bash
QUOTE_ARGS=''
for ARG in "$#"
do
ARG=$(printf "%q" "$ARG")
QUOTE_ARGS="${QUOTE_ARGS} $ARG"
done
ssh user#hostname.tld. "printf ${QUOTE_ARGS}"
This works for everything i've tested so far, except newlines:
$ /tmp/wrap_printf "[-%s-]" "hello'\$t\""
[-hello'$t"-]
#!/bin/sh
QUOTE_ARGS=''
for ARG in "$#"
do
QUOTE_ARGS="${QUOTE_ARGS} '${ARG}'"
done
ssh user#hostname.tld. "${QUOTE_ARGS}"
This works for spaces. It doesn't work if the argument has an embedded single quote.
Getting quoting right is pretty hard and doing it in bash (in a general and robust way) almost impossible.
Use Perl:
#!/usr/bin/perl
use Net::OpenSSH;
my $ssh = Net::OpenSSH->new('user#hostname');
$ssh->system('printf', #ARGV);
Based on the answers from Koert and Peter Lyons, here a wrapper for ssh; i call it "sshsystem". (also available at https://gist.github.com/4672115)
#!/bin/bash
# quote command in ssh call to prevent remote side from expanding any arguments
# uses bash printf %q for quoting - no idea how compatible this is with other shells.
# http://stackoverflow.com/questions/6592376/prevent-ssh-from-breaking-up-shell-script-parameters
sshargs=()
while (( $# > 0 )); do
case "$1" in
-[1246AaCfgKkMNnqsTtVvXxYy])
# simple argument
sshargs+=("$1")
shift
;;
-[bcDeFIiLlmOopRSWw])
# argument with parameter
sshargs+=("$1")
shift
if (( $# == 0 )); then
echo "missing second part of long argument" >&2
exit 99
fi
sshargs+=("$1")
shift
;;
-[bcDeFIiLlmOopRSWw]*)
# argument with parameter appended without space
sshargs+=("$1")
shift
;;
--)
# end of arguments
sshargs+=("$1")
shift
break
;;
-*)
echo "unrecognized argument: '$1'" >&2
exit 99
;;
*)
# end of arguments
break
;;
esac
done
# user#host
sshargs+=("$1")
shift
# command - quote
if (( $# > 0 )); then
# no need to make COMMAND an array - ssh will merge it anyway
COMMAND=
while (( $# > 0 )); do
arg=$(printf "%q" "$1")
COMMAND="${COMMAND} ${arg}"
shift
done
sshargs+=("${COMMAND}")
fi
exec ssh "${sshargs[#]}"
The easiest and quickest is to just use Bash's Quoting Parameter Transformation: ${parameter#Q}. This can automatically applied during array expansion with ${array[#]#Q}, but when using the builtin argument array, the name and the brackets are dropped, so it becomes ${##Q}. Therefore the original script only needs 4 characters added to it to work.
#!/bin/bash
ssh user#hostname.tld. printf "${##Q}"
Now any escaping will work, even terminal colors like this:
./wrap_printf "%s\e[39m\e[49m\n" $'\e[30m\e[42mBlack on Green' "Just Normal Text"