I have written a script that is shown below. However, when I run the script it just outputs the final command instead of running it.
What do I need to add at the end so that the command will be run?
Thank you
#!/bin/bash -l
#$ -cwd
# file name: singlecell.sh
rundir=$1
outdir=$2
samplesheet=$3
USAGE="Usage: $0 -i [path to input run directory] -o [path to output directory] -s [path to sample sheet]"
while getopts i:o:s:h opt
do
case "$opt" in
i) RunDir="$OPTARG";;
o) OutDir="$OPTARG";;
s) SampSheet="$OPTARG";;
h) echo $USAGE; exit 1
esac
done
if [[ $RunDir == "" || $OutDir == "" || $SampSheet == "" ]]
then
echo $USAGE
exit 1
fi
# asssume only one index type per run
index8=`grep "SI-GA\|SI-NA" $SampSheet | wc -l`
ignore=""
# check if single indexes.
if [ $index8 -gt 0 ]; then
ignore="--filter-single-index"
else
ignore="--filter-dual-index"
fi
cmd="cellranger mkfastq --run=$RunDir --output-dir=$OutDir --csv=$SampSheet --localcores=8 --localmem=40 $ignore --barcode-mismatches=0"
echo $cmd
$cmd
You can replace the last lines:
cmd="cellranger mkfastq --run=$RunDir --output-dir=$OutDir --csv=$SampSheet --localcores=8 --localmem=40 $ignore --barcode-mismatches=0"
echo $cmd
$cmd
With:
cellranger mkfastq --run=$RunDir --output-dir=$OutDir --csv=$SampSheet --localcores=8 --localmem=40 $ignore --barcode-mismatches=0
If you are in doubt the program actually ran as it may not give output in the terminal, the usual way is to check for its return value. This is the general status value returned by a program, such as the common return 0; at the end of main function in C programs.
cellranger mkfastq --run=$RunDir --output-dir=$OutDir --csv=$SampSheet --localcores=8 --localmem=40 $ignore --barcode-mismatches=0
cellranger_ret=$?
echo "cellranger_ret = ${cellranger_ret}"
This gives you a clue if the expected output from the program didn't show up and it is not reporting any error. Typically 0 means success and other values indicate error. Other useful tip is to look for debug related parameters (command line arguments) that will allow you to increase the verbosity level of the program.
Related
So I started today taking a look at scripting using vim and I'm just so very lost and was looking for some help in a few areas.
For my first project,I want to process a file as a command line argument, and if a file isn't included when the user executes this script, then a usage message should be displayed, followed by exiting the program.
I have no clue where to even start with that, will I need and if ... then statement, or what?
Save vim for later and try to learn one thing at a time. A simpler text editor is called nano.
Now, as far as checking for a file as an argument, and showing a usage message otherwise, this is a typical pattern:
PROGNAME="$0"
function show_usage()
{
echo "Usage: ${PROGNAME} <filename>" >&2
echo "..." >&2
exit 1
}
if [[ $# -lt 1 ]]; then
show_usage
fi
echo "Contents of ${1}:"
cat "$1"
Let's break this down.
PROGNAME="$0"
$0 is the name of the script, as it was called on the command line.
function show_usage()
{
echo "Usage: ${PROGNAME} <filename>" >&2
echo "..." >&2
exit 1
}
This is the function that prints the "usage" message and exits with a failure status code. 0 is success, anything other than 0 is a failure. Note that we redirect our echo to &2--this prints the usage message on Standard Error rather than Standard Output.
if [[ $# -lt 1 ]]; then
show_usage
fi
$# is the number of arguments passed to the script. If that number is less than 1, print the usage message and exit.
echo "Contents of ${1}:"
cat "$1"
$1 is out filename--the first argument of the script. We can do whatever processing we want to here, with $1 being the filename. Hope this helps!
i think you're asking how to write a bash script that requires a file as a command-line argument, and exits with a usage message if there's a problem with that:
#!/bin/bash
# check if user provided exactly one command-line argument:
if [ $# -ne 1 ]; then
echo "Usage: `basename "$0"` file"
exit 1
# now check if the provided argument corresponds to a real file
elif [ ! -f "$1" ]; then
echo "Error: couldn't find $1."
exit 1
fi
# do things with the file...
stat "$1"
head "$1"
tail "$1"
grep 'xyz' "$1"
I am trying to execute a hallo_word.sh that is stored at ~/bin from this script that is stored at my ~/Desktop. I have made both scripts executable. But all the time I get the problem message. Any ideas?
#!/bin/sh
clear
dir="$PATH"
read -p "which file you want to execute" fl
echo ""
for fl in $dir
do
if [ -x "$fl" ]
then
echo "executing=====>"
./$fl
else
echo "Problem"
fi
done
This line has two problems:
for fl in $dir
$PATH is colon separated, but for expects whitespace separated values. You can change that by setting the IFS variable. This changes the FIELD SEPARATOR used by tools like for and awk.
$fl contains the name of the file you want to execute, but you overwrite its value with the contents of $dir.
Fixed:
#!/bin/sh
clear
read -p "which file you want to execute" file
echo
IFS=:
for dir in $PATH ; do
if [ -x "$dir/$file" ]
then
echo "executing $dir/$file"
exec "$dir/$file"
fi
done
echo "Problem"
You could also be lazy and let a subshell handle it.
PATH=(whatever) bash command -v my_command
if [ $? -ne 0 ]; then
# Problem, could not be found.
else
# No problem
fi
There is no need to over-complicate things.
command(1) is a builtin command that allows you to check if a command exists.
The PATH value contains all the directories in which executable files can be run without explicit qualification. So you can just call the command directly.
#!/bin/sh
clear
# r for raw input, e to use readline, add a space for clarity
read -rep "Which file you want to execute? " fl || exit 1
echo ""
"$fl" || { echo "Problem" ; exit 1 ; }
I quote the name as it could have spaces.
To test if the command exists before execution use type -p
#!/bin/sh
clear
# r for raw input, e to use readline, add a space for clarity
read -rep "Which file you want to execute? " fl || exit 1
echo ""
type -p "$fq" >/dev/null || exit 1
"$fl" || { echo "Problem" ; exit 1 ; }
How do I check if file exists in bash?
When I try to do it like this:
FILE1="${#:$OPTIND:1}"
if [ ! -e "$FILE1" ]
then
echo "requested file doesn't exist" >&2
exit 1
elif
<more code follows>
I always get following output:
requested file doesn't exist
The program is used like this:
script.sh [-g] [-p] [-r FUNCTION_ID|-d FUNCTION_ID] FILE
Any ideas please?
I will be glad for any help.
P.S. I wish I could show the entire file without the risk of being fired from school for having a duplicate. If there is a private method of communication I will happily oblige.
My mistake. Fas forcing a binary file into a wrong place. Thanks for everyone's help.
Little trick to debugging problems like this. Add these lines to the top of your script:
export PS4="\$LINENO: "
set -xv
The set -xv will print out each line before it is executed, and then the line once the shell interpolates variables, etc. The $PS4 is the prompt used by set -xv. This will print the line number of the shell script as it executes. You'll be able to follow what is going on and where you may have problems.
Here's an example of a test script:
#! /bin/bash
export PS4="\$LINENO: "
set -xv
FILE1="${#:$OPTIND:1}" # Line 6
if [ ! -e "$FILE1" ] # Line 7
then
echo "requested file doesn't exist" >&2
exit 1
else
echo "Found File $FILE1" # Line 12
fi
And here's what I get when I run it:
$ ./test.sh .profile
FILE1="${#:$OPTIND:1}"
6: FILE1=.profile
if [ ! -e "$FILE1" ]
then
echo "requested file doesn't exist" >&2
exit 1
else
echo "Found File $FILE1"
fi
7: [ ! -e .profile ]
12: echo 'Found File .profile'
Found File .profile
Here, I can see that I set $FILE1 to .profile, and that my script understood that ${#:$OPTIND:1}. The best thing about this is that it works on all shells down to the original Bourne shell. That means if you aren't running Bash as you think you might be, you'll see where your script is failing, and maybe fix the issue.
I suspect you might not be running your script in Bash. Did you put #! /bin/bash on the top?
script.sh [-g] [-p] [-r FUNCTION_ID|-d FUNCTION_ID] FILE
You may want to use getopts to parse your parameters:
#! /bin/bash
USAGE=" Usage:
script.sh [-g] [-p] [-r FUNCTION_ID|-d FUNCTION_ID] FILE
"
while getopts gpr:d: option
do
case $option in
g) g_opt=1;;
p) p_opt=1;;
r) rfunction_id="$OPTARG";;
d) dfunction_id="$OPTARG";;
[?])
echo "Invalid Usage" 1>&2
echo "$USAGE" 1>&2
exit 2
;;
esac
done
if [[ -n $rfunction_id && -n $dfunction_id ]]
then
echo "Invalid Usage: You can't specify both -r and -d" 1>&2
echo "$USAGE" >2&
exit 2
fi
shift $(($OPTIND - 1))
[[ -n $g_opt ]] && echo "-g was set"
[[ -n $p_opt ]] && echo "-p was set"
[[ -n $rfunction_id ]] && echo "-r was set to $rfunction_id"
[[ -n $dfunction_id ]] && echo "-d was set to $dfunction_id"
[[ -n $1 ]] && echo "File is $1"
To (recap) and add to #DavidW.'s excellent answer:
Check the shebang line (first line) of your script to ensure that it's executed by bash: is it #!/bin/bash or #!/usr/bin/env bash?
Inspect your script file for hidden control characters (such as \r) that can result in unexpected behavior; run cat -v scriptFile | fgrep ^ - it should produce NO output; if the file does contain \r chars., they would show as ^M.
To remove the \r instances (more accurately, to convert Windows-style \r\n newline sequences to Unix \n-only sequences), you can use dos2unix file to convert in place; if you don't have this utility, you can use sed 's/'$'\r''$//' file > outfile (CAVEAT: use a DIFFERENT output file, otherwise you'll destroy your input file); to remove all \r instances (even if not followed by \n), use tr -d '\r' < file > outfile (CAVEAT: use a DIFFERENT output file, otherwise you'll destroy your input file).
In addition to #DavidW.'s great debugging technique, you can add the following to visually inspect all arguments passed to your script:
i=0; for a; do echo "\$$((i+=1))=[$a]"; done
(The purpose of enclosing the value in [...] (for example), is to see the exact boundaries of the values.)
This will yield something like:
$1=[-g]
$2=[input.txt]
...
Note, though, that nothing at all is printed if no arguments were passed.
Try to print FILE1 to see if it has the value you want, if it is not the problem, here is a simple script (site below):
#!/bin/bash
file="${#:$OPTIND:1}"
if [ -f "$file" ]
then
echo "$file found."
else
echo "$file not found."
fi
http://www.cyberciti.biz/faq/unix-linux-test-existence-of-file-in-bash/
Instead of plucking an item out of "$#" in a tricky way, why don't you shift off the args you've processed with getopts:
while getopts ...
done
shift $(( OPTIND - 1 ))
FILE1=$1
I want to write code like this:
command="some command"
safeRunCommand $command
safeRunCommand() {
cmnd=$1
$($cmnd)
if [ $? != 0 ]; then
printf "Error when executing command: '$command'"
exit $ERROR_CODE
fi
}
But this code does not work the way I want. Where did I make the mistake?
Below is the fixed code:
#!/bin/ksh
safeRunCommand() {
typeset cmnd="$*"
typeset ret_code
echo cmnd=$cmnd
eval $cmnd
ret_code=$?
if [ $ret_code != 0 ]; then
printf "Error: [%d] when executing command: '$cmnd'" $ret_code
exit $ret_code
fi
}
command="ls -l | grep p"
safeRunCommand "$command"
Now if you look into this code, the few things that I changed are:
use of typeset is not necessary, but it is a good practice. It makes cmnd and ret_code local to safeRunCommand
use of ret_code is not necessary, but it is a good practice to store the return code in some variable (and store it ASAP), so that you can use it later like I did in printf "Error: [%d] when executing command: '$command'" $ret_code
pass the command with quotes surrounding the command like safeRunCommand "$command". If you don’t then cmnd will get only the value ls and not ls -l. And it is even more important if your command contains pipes.
you can use typeset cmnd="$*" instead of typeset cmnd="$1" if you want to keep the spaces. You can try with both depending upon how complex is your command argument.
'eval' is used to evaluate so that a command containing pipes can work fine
Note: Do remember some commands give 1 as the return code even though there isn't any error like grep. If grep found something it will return 0, else 1.
I had tested with KornShell and Bash. And it worked fine. Let me know if you face issues running this.
Try
safeRunCommand() {
"$#"
if [ $? != 0 ]; then
printf "Error when executing command: '$1'"
exit $ERROR_CODE
fi
}
It should be $cmd instead of $($cmd). It works fine with that on my box.
Your script works only for one-word commands, like ls. It will not work for "ls cpp". For this to work, replace cmd="$1"; $cmd with "$#". And, do not run your script as command="some cmd"; safeRun command. Run it as safeRun some cmd.
Also, when you have to debug your Bash scripts, execute with '-x' flag. [bash -x s.sh].
There are several things wrong with your script.
Functions (subroutines) should be declared before attempting to call them. You probably want to return() but not exit() from your subroutine to allow the calling block to test the success or failure of a particular command. That aside, you don't capture 'ERROR_CODE' so that is always zero (undefined).
It's good practice to surround your variable references with curly braces, too. Your code might look like:
#!/bin/sh
command="/bin/date -u" #...Example Only
safeRunCommand() {
cmnd="$#" #...insure whitespace passed and preserved
$cmnd
ERROR_CODE=$? #...so we have it for the command we want
if [ ${ERROR_CODE} != 0 ]; then
printf "Error when executing command: '${command}'\n"
exit ${ERROR_CODE} #...consider 'return()' here
fi
}
safeRunCommand $command
command="cp"
safeRunCommand $command
The normal idea would be to run the command and then use $? to get the exit code. However, sometimes you have multiple cases in which you need to get the exit code. For example, you might need to hide its output, but still return the exit code, or print both the exit code and the output.
ec() { [[ "$1" == "-h" ]] && { shift && eval $* > /dev/null 2>&1; ec=$?; echo $ec; } || eval $*; ec=$?; }
This will give you the option to suppress the output of the command you want the exit code for. When the output is suppressed for the command, the exit code will directly be returned by the function.
I personally like to put this function in my .bashrc file.
Below I demonstrate a few ways in which you can use this:
# In this example, the output for the command will be
# normally displayed, and the exit code will be stored
# in the variable $ec.
$ ec echo test
test
$ echo $ec
0
# In this example, the exit code is output
# and the output of the command passed
# to the `ec` function is suppressed.
$ echo "Exit Code: $(ec -h echo test)"
Exit Code: 0
# In this example, the output of the command
# passed to the `ec` function is suppressed
# and the exit code is stored in `$ec`
$ ec -h echo test
$ echo $ec
0
Solution to your code using this function
#!/bin/bash
if [[ "$(ec -h 'ls -l | grep p')" != "0" ]]; then
echo "Error when executing command: 'grep p' [$ec]"
exit $ec;
fi
You should also note that the exit code you will be seeing will be for the grep command that's being run, as it is the last command being executed. Not the ls.
#!/bin/bash
#!/bin/sh
# Need help
__help() { echo "$0 [ stop|start ]" 1>&2; exit 1; }
# Not enough args to run properly
[ $# -ne 1 ] && __help
# See what we're called with
case "$1" in
start) # Start sniffer as root, under a different argv[0] and make it drop rights
s=$(/usr/local/sbin/tcpdump -n -nn -f -q -i lo | awk 'END {print NR}')
echo "$s" > eppps_$(/bin/date +'%Y%m%d%H%M')
;;
stop) # End run, first "friendly", then strict:
/usr/bin/pkill -15 -f /usr/local/sbin/tcpdump >/dev/null 2>&1|| { sleep 3s; /usr/bin/pkill -9 -f /usr/local/sbin/tc$
;;
*) # Superfluous but show we only accept these args
__help
;;
esac
exit 0
This code runs perfectly on manual testing. But when i couple it with cron it just doesn't do anything. No output file is created.
My cron entries for the script looks like
http://postimage.org/image/1pztgd6xw/
It looks like you are not setting the working directory, so you may need to give an absolute path for the output file