I am trying to execute the following unix command from within perl
`git log --pretty=format:%H | grep $id | wc -l`;
I keep getting an error saying
sh: -c : syntax error near unexpected token '|'
sh -c : '| wc -l'
How can I fix this error
The problem is that you have 123<NEWLINE> instead of 123 in $id, so you're passing the following to the shell:
git log --pretty=format:%H | grep 123
| wc -l
instead of
git log --pretty=format:%H | grep 123 | wc -l
Fix the value in $id. The best way is probably through chomp.
You'll probably have a more robust script if you implement some of this functionality using Perl subroutines instead of external Unix commands.
Using backquotes is good for a quick script, but it's inherently fragile, because you are exposed to the full power and weirdness of whatever Unix shell is configured for the user that is running your script (with wildcard expansion, input/output redirection, piping, etc.). This really applies to any programming language, and most languages (including perl) have safer ways to run external commands without using the shell. But one of the reasons Perl was created was to be able to combine under one programming language the text-processing abilities of a slew of Unix commands.
Related
Hi I was wondering how to read the argument after "|" pipe from shell script.
For example, when I run ./tmp.sh ls -la | sort
I could only get 2 arguments, which is "ls" and "-la".
Is there any way to read "| sort" without modifying the command, and realize only with shell script?
Thanks a lot!!
One way would be to pass the entire command as a string to your script.
./tmp.sh -c "ls -la | sort"
...or without a flag...
./tmp.sh "ls -la | sort"
Afterward, you can split the string into an array in your script.
I guess you could check ps axf, but part of the beauty of pipes is the loose coupling they give, because bash knows what is in your pipeline, not the individual pieces of the pipeline. This makes writing filters simple.
with this grep it shows a comand I used:
echo `history | grep "ssh root" | head -1| cut -c6-`
with this output:
ssh root#107.170.70.100
I want the output to directly execute as the command instead of printed.
How can I do it?
In principle, this can be done by using the $() format, so
$(history | grep "ssh root" | head -1| cut -c6-)
should do what you ask for. However, I don't think that it is advisable to do so, as this will automatically execute the command that results from your grep, so if you did a mistake, a lot of bad things can happen. Instead I suggest reviewing your result before re-executing. bash history has a lot of nice shortcuts to deal with these kind of things. As an example, imagine:
> history | grep "ssh root"
756 ssh root#107.170.70.100
you can call this command on line 756 easily by typing
!756
It's definitely much safer. Hope this helps.
Ideally you'd be using the $(cmd) syntax rather than the `cmd` syntax. This makes it easier to nest subshells as well as keep track of what's going on.
That aside, if you remove the echo statement it will run the script:
# Prints out ls
echo $( echo ls )
# Runs the ls command
$( echo ls )
Use eval.
$ eval `history | grep "ssh root" | head -1| cut -c6-`
From eval command in Bash and its typical uses:
eval takes a string as its argument, and evaluates it as if you'd typed that string on a command line.
And the Bash Manual (https://www.gnu.org/software/bash/manual/html_node/Bourne-Shell-Builtins.html#Bourne-Shell-Builtins)
eval
eval [arguments]
The arguments are concatenated together into a single command, which is then read and executed, and its exit status returned as the exit status of eval. If there are no arguments or only empty arguments, the return status is zero.
I have a shell script of more than 1000 lines, i would like to check if all the commands used in the script are installed in my Linux operating system.
Is there any tool to get the list of Linux commands used in the shell script?
Or how can i write a small script which can do this for me?
The script runs successfully on the Ubuntu machine, it is invoked as a part of C++ application. we need to run the same on a device where a Linux with limited capability runs. I have identified manually, few commands which the script runs and not present on Device OS. before we try installing these commands i would like to check all other commands and install all at once.
Thanks in advance
I already tried this in the past and got to the conclusion that is very difficult to provide a solution which would work for all scripts. The reason is that each script with complex commands has a different approach in using the shells features.
In case of a simple linear script, it might be as easy as using debug mode.
For example: bash -x script.sh 2>&1 | grep ^+ | awk '{print $2}' | sort -u
In case the script has some decisions, then you might use the same approach an consider that for the "else" cases the commands would still be the same just with different arguments or would be something trivial (echo + exit).
In case of a complex script, I attempted to write a script that would just look for commands in the same place I would do it myself. The challenge is to create expressions that would help identify all used possibilities, I would say this is doable for about 80-90% of the script and the output should only be used as reference since it will contain invalid data (~20%).
Here is an example script that would parse itself using a very simple approach (separate commands on different lines, 1st word will be the command):
# 1. Eliminate all quoted text
# 2. Eliminate all comments
# 3. Replace all delimiters between commands with new lines ( ; | && || )
# 4. extract the command from 1st column and print it once
cat $0 \
| sed -e 's/\"/./g' -e "s/'[^']*'//g" -e 's/"[^"]*"//g' \
| sed -e "s/^[[:space:]]*#.*$//" -e "s/\([^\\]\)#[^\"']*$/\1/" \
| sed -e "s/&&/;/g" -e "s/||/;/g" | tr ";|" "\n\n" \
| awk '{print $1}' | sort -u
the output is:
.
/
/g.
awk
cat
sed
sort
tr
There are many more cases to consider (command substitutions, aliases etc.), 1, 2 and 3 are just beginning, but they would still cover 80% of most complex scripts.
The regular expressions used would need to be adjusted or extended to increase precision and special cases.
In conclusion if you really need something like this, then you can write a script as above, but don't trust the output until you verify it yourself.
Add export PATH='' to the second line of your script.
Execute your_script.sh 2>&1 > /dev/null | grep 'No such file or directory' | awk '{print $4;}' | grep -v '/' | sort | uniq | sed 's/.$//'.
If you have a fedora/redhat based system, bash has been patched with the --rpm-requires flag
--rpm-requires: Produce the list of files that are required for the shell script to run. This implies -n and is subject to the same limitations as compile time error checking checking; Command substitutions, Conditional expressions and eval builtin are not parsed so some dependencies may be missed.
So when you run the following:
$ bash --rpm-requires script.sh
executable(command1)
function(function1)
function(function2)
executable(command2)
function(function3)
There are some limitations here:
command and process substitutions and conditional expressions are not picked up. So the following are ignored:
$(command)
<(command)
>(command)
command1 && command2 || command3
commands as strings are not picked up. So the following line will be ignored
"/path/to/my/command"
commands that contain shell variables are not listed. This generally makes sense since
some might be the result of some script logic, but even the following is ignored
$HOME/bin/command
This point can however be bypassed by using envsubst and running it as
$ bash --rpm-requires <(<script envsubst)
However, if you use shellcheck, you most likely quoted this and it will still be ignored due to point 2
So if you want to use check if your scripts are all there, you can do something like:
while IFS='' read -r app; do
[ "${app%%(*}" == "executable" ] || continue
app="${app#*(}"; app="${app%)}";
if [ "$(type -t "${app}")" != "builtin" ] && \
! [ -x "$(command -v "${app}")" ]
then
echo "${app}: missing application"
fi
done < <(bash --rpm-requires <(<"$0" envsubst) )
If your script contains files that are sourced that might contain various functions and other important definitions, you might want to do something like
bash --rpm-requires <(cat source1 source2 ... <(<script.sh envsubst))
Based #czvtools’ answer, I added some extra checks to filter out bad values:
#!/usr/bin/fish
if test "$argv[1]" = ""
echo "Give path to command to be tested"
exit 1
end
set commands (cat $argv \
| sed -e 's/\"/./g' -e "s/'[^']*'//g" -e 's/"[^"]*"//g' \
| sed -e "s/^[[:space:]]*#.*\$//" -e "s/\([^\\]\)#[^\"']*\$/\1/" \
| sed -e "s/&&/;/g" -e "s/||/;/g" | tr ";|" "\n\n" \
| awk '{print $1}' | sort -u)
for command in $commands
if command -q -- $command
set -a resolved (realpath (which $command))
end
end
set resolved (string join0 $resolved | sort -z -u | string split0)
for command in $resolved
echo $command
end
In my program I need to output to user which shell he is using. So in file /etc/udate-motd.d/00-header I wrote printf "$SHELL" but the problem is that even when I switching my shell to zsh, $SHELL is still equal to /bin/bash. I searched through the internet and found that I can bo it by using MyShell='ps -hp $$', and here is again a problem. When I use it MyShell is a string with number of processes (/etc/update-motd.d/00-header is also there) but there no word zsh.
So how can I understand which shell use the logging in person.
"the internet" gave you one kind of ps syntax. You've tagged this linux, so don't use BSD syntax. Try this:
ps hp $$ -o cmd
no dash
The users shell is determined in /etc/passwd. Why not take the information from there? You could
grep $USER /etc/passwd | cut -f7 -d:
to get the shell.
I'm new to bash scripting and I've been learning as I go with a small project I'm taking on. However, I've run into a problem that I cannot seem to get past.
I have a variable that I need to include in a command. When ran directly in the shell (with the variable manually typed), the command returns the expected result. However, I can't get it to work when using a variable.
So, if I manually run this, it correctly returns 0 or 1, depending if it is running or not.
ps -ef | grep -v grep | grep -c ProcessName
However, when I try to embed that into this while clause, it always evaluates to 0 because it's not searching for the correct text.
while [ `ps -ef | grep -v grep | grep -c {$1}` -ne 0 ]
do
sleep 5
done
Is there a way I can accomplish this? I've tried a myriad of different things to no avail. I also tried using the $() syntax for command substitution, but I had no luck with that either.
Thanks!
I think that instead of {$1} you mean "$1". Also, you can just do pgrep -c "$1" instead of the two pipes.
In addition, there's also no need to compare the output of grep -c with 0, since you can just see if the command failed or not. So, a much simplified version might be:
while pgrep "$1" > /dev/null
do
sleep 4
done
You should really use -C with ps rather than the messy pipes if you're using the full process name. If you're interested in substring matching, then your way is the only thing I can think of.