I am trying to alter the Bash function below to execute each command argument. But when I run this script, the first echo works as intended, but the second echo that attempts to append to the scratch.txt file does not actually execute. It just gets echo'd into the prompt.
#!/bin/sh
clear
function each(){
while read line; do
for cmd in "$#"; do
cmd=${cmd//%/$line}
printf "%s\n" "$cmd"
$cmd
done
done
}
# pipe in the text file and run both commands
# on each line of the file
cat scratch.txt | each 'echo %' 'echo -e "%" >> "scratch.txt"'
exit 0
How do I get the $cmd variable to execute as a command?
I found the original code from answer 2 here:
Running multiple commands with xargs
You want eval. It's evil. Or at least, dangerous. Read all about it at BashFAQ #48.
Related
I have an unix shell script. I have put -x in shell to see all the execution step. Now I want to capture these in one log file on a daily basis.
Psb script.
#!/bin/ksh -x
Logfile= path.log.date
Print " copying file" | tee $logifle
Scp -i key source destination | tee -a $logfile.
Exit 0;
First line of the shell script is known as shebang , which indicates what interpreter has to be execute the below script.
Similarly first line is commented which denotes coming lines not related to that interpreted session.
To capture the output, run the script redirect your output while running the script.
ksh -x scriptname >> output_file
Note:it will output what your script's doing line by line
There are two cases, using ksh as your shell, then you need to do IO redirection accordingly, and using some other shell and executing a .ksh script, then IO redirection could be done based on that shell. Following method should work for most of the shells.
$ cat somescript.ksh
#!/bin/ksh -x
printf "Copy file \n";
printf "Do something else \n";
Run it:
$ ./somescript.ksh 1>some.log 2>&1
some.log will contain,
+ printf 'Copy file \n'
Copy file
+ printf 'Do something else \n'
Do something else
In your case, no need to specify logfile and/or tee. Script would look something like this,
#!/bin/ksh -x
printf "copying file\n"
scp -i key user#server /path/to/file
exit 0
Run it:
$ ./myscript 1>/path/to/logfile 2>&1
2>&1 captures both stderr and stdout into stdout and 1>logfile prints it out into logfile.
I would prefer to explicitly redirecting the output (including stderr 2> because set -x sends output to stderr).
This keeps the shebang short and you don't have to cram the redirecton and filename-building into it.
#!/bin/ksh
logfile=path.log.date
exec >> $logfile 2>&1 # redirecting all output to logfile (appending)
set -x # switch on debugging
# now start working
echo "print something"
Normally, we use
sh script.sh 1>t.log 2>t.err
to redirect log.
How can I use variant to log:
string="1>t.log 2>t.err"
sh script.sh $string
You need to use 'eval' shell builtin for this purpose. As per man page of bash command:
eval [arg ...]
The args are read and concatenated together into a single command. This command is then read and exe‐
cuted by the shell, and its exit status is returned as the value of eval. If there are no args, or
only null arguments, eval returns 0.
Run your command like below:
eval sh script.sh $string
However, do you really need to run script.sh through sh command? If you instead put sh interpreter line (using #!/bin/sh as the first line in your shell script) in your script itself and give it execute permission, that would let you access return code of ls command. Below is an example of using sh and not using sh. Notice the difference in exit codes.
Note: I had only one file try.sh in my current directory. So ls command was bound to exit with return code 2.
$ ls try1.sh try1.sh.backup 1>out.txt 2>err.txt
$ echo $?
2
$ eval sh ls try1.sh try1.sh.backup 1>out.txt 2>err.txt
$ echo $?
127
In the second case, the exit code is of sh shell. In first case, the exit code is of ls command. You need to make cautious choice depending on your needs.
I figure out one way but it's ugly:
echo script.sh $string | sh
I think you can just put the name into a string variable
and then use data redirection
file_name="file1"
outfile="$file_name"".log"
errorfile="$file_name"".err"
sh script.sh 1> $outfile 2> $errorfile
In a script.sh,
source a.sh
source b.sh
CMD1
CMD2
CMD3
how can I replace the source *.sh with their content (without executing the commands)?
I would like to see what the bash interpreter executes after sourcing the files and expanding all variables.
I know I can use set -n -v or run bash -n -v script.sh 2>output.sh, but that would not replace the source commands (and even less if a.sh or b.sh contain variables).
I thought of using a subshell, but that still doesn't expand the source lines. I tried a combination of set +n +v and set -n -v before and after the source lines, but that still does not work.
I'm going to send that output to a remote machine using ssh.
I could use <<output.sh to pipe the content into the ssh command, but I can't log as root onto the remote machine, but I am however a sudoer.
Therefore, I thought I could create the script and send it as a base64-encoded string (using that clever trick )
base64 script | ssh remotehost 'base64 -d | sudo bash'
Is there a solution?
Or do you have a better idea?
You can do something like this:
inline.sh:
#!/usr/bin/env bash
while read line; do
if [[ "$line" =~ (\.|source)\s+.+ ]]; then
file="$(echo $line | cut -d' ' -f2)"
echo "$(cat $file)"
else
echo "$line"
fi
done < "$1"
Note this assumes the sourced files exist, and doesn't handle errors. You should also handle possible hashbangs. If the sourced files contain themselves source, you need to apply the script recursively, e.g. something like (not tested):
while egrep -q '^(source|\.)' main.sh; do
bash inline.sh main.sh > main.sh
done
Let's test it
main.sh:
source a.sh
. b.sh
echo cc
echo "$var_a $var_b"
a.sh:
echo aa
var_a="stack"
b.sh:
echo bb
var_b="overflow"
Result:
bash inline.sh main.sh
echo aa
var_a="stack"
echo bb
var_b="overflow"
echo cc
echo "$var_a $var_b"
bash inline.sh main.sh | bash
aa
bb
cc
stack overflow
BTW, if you just want to see what bash executes, you can run
bash -x [script]
or remotely
ssh user#host -t "bash -x [script]"
This must either be really simple or really complex, but I couldn't find anything about it... I am trying to open a new bash instance, then run a few commands inside it, and give the control back to the user inside that same instance.
I tried:
$ bash -lic "some_command"
but this executes some_command inside the new instance, then closes it. I want it to stay open.
One more detail which might affect answers: if I can get this to work I will use it in my .bashrc as alias(es), so bonus points for an alias implementation!
bash --rcfile <(echo '. ~/.bashrc; some_command')
dispenses the creation of temporary files. Question on other sites:
https://serverfault.com/questions/368054/run-an-interactive-bash-subshell-with-initial-commands-without-returning-to-the
https://unix.stackexchange.com/questions/123103/how-to-keep-bash-running-after-command-execution
This is a late answer, but I had the exact same problem and Google sent me to this page, so for completeness here is how I got around the problem.
As far as I can tell, bash does not have an option to do what the original poster wanted to do. The -c option will always return after the commands have been executed.
Broken solution: The simplest and obvious attempt around this is:
bash -c 'XXXX ; bash'
This partly works (albeit with an extra sub-shell layer). However, the problem is that while a sub-shell will inherit the exported environment variables, aliases and functions are not inherited. So this might work for some things but isn't a general solution.
Better: The way around this is to dynamically create a startup file and call bash with this new initialization file, making sure that your new init file calls your regular ~/.bashrc if necessary.
# Create a temporary file
TMPFILE=$(mktemp)
# Add stuff to the temporary file
echo "source ~/.bashrc" > $TMPFILE
echo "<other commands>" >> $TMPFILE
echo "rm -f $TMPFILE" >> $TMPFILE
# Start the new bash shell
bash --rcfile $TMPFILE
The nice thing is that the temporary init file will delete itself as soon as it is used, reducing the risk that it is not cleaned up correctly.
Note: I'm not sure if /etc/bashrc is usually called as part of a normal non-login shell. If so you might want to source /etc/bashrc as well as your ~/.bashrc.
You can pass --rcfile to Bash to cause it to read a file of your choice. This file will be read instead of your .bashrc. (If that's a problem, source ~/.bashrc from the other script.)
Edit: So a function to start a new shell with the stuff from ~/.more.sh would look something like:
more() { bash --rcfile ~/.more.sh ; }
... and in .more.sh you would have the commands you want to execute when the shell starts. (I suppose it would be elegant to avoid a separate startup file -- you cannot use standard input because then the shell will not be interactive, but you could create a startup file from a here document in a temporary location, then read it.)
bash -c '<some command> ; exec /bin/bash'
will avoid additional shell sublayer
You can get the functionality you want by sourcing the script instead of running it. eg:
$cat script
cmd1
cmd2
$ . script
$ at this point cmd1 and cmd2 have been run inside this shell
Append to ~/.bashrc a section like this:
if [ "$subshell" = 'true' ]
then
# commands to execute only on a subshell
date
fi
alias sub='subshell=true bash'
Then you can start the subshell with sub.
The accepted answer is really helpful! Just to add that process substitution (i.e., <(COMMAND)) is not supported in some shells (e.g., dash).
In my case, I was trying to create a custom action (basically a one-line shell script) in Thunar file manager to start a shell and activate the selected Python virtual environment. My first attempt was:
urxvt -e bash --rcfile <(echo ". $HOME/.bashrc; . %f/bin/activate;")
where %f is the path to the virtual environment handled by Thunar.
I got an error (by running Thunar from command line):
/bin/sh: 1: Syntax error: "(" unexpected
Then I realized that my sh (essentially dash) does not support process substitution.
My solution was to invoke bash at the top level to interpret the process substitution, at the expense of an extra level of shell:
bash -c 'urxvt -e bash --rcfile <(echo "source $HOME/.bashrc; source %f/bin/activate;")'
Alternatively, I tried to use here-document for dash but with no success. Something like:
echo -e " <<EOF\n. $HOME/.bashrc; . %f/bin/activate;\nEOF\n" | xargs -0 urxvt -e bash --rcfile
P.S.: I do not have enough reputation to post comments, moderators please feel free to move it to comments or remove it if not helpful with this question.
With accordance with the answer by daveraja, here is a bash script which will solve the purpose.
Consider a situation if you are using C-shell and you want to execute a command
without leaving the C-shell context/window as follows,
Command to be executed: Search exact word 'Testing' in current directory recursively only in *.h, *.c files
grep -nrs --color -w --include="*.{h,c}" Testing ./
Solution 1: Enter into bash from C-shell and execute the command
bash
grep -nrs --color -w --include="*.{h,c}" Testing ./
exit
Solution 2: Write the intended command into a text file and execute it using bash
echo 'grep -nrs --color -w --include="*.{h,c}" Testing ./' > tmp_file.txt
bash tmp_file.txt
Solution 3: Run command on the same line using bash
bash -c 'grep -nrs --color -w --include="*.{h,c}" Testing ./'
Solution 4: Create a sciprt (one-time) and use it for all future commands
alias ebash './execute_command_on_bash.sh'
ebash grep -nrs --color -w --include="*.{h,c}" Testing ./
The script is as follows,
#!/bin/bash
# =========================================================================
# References:
# https://stackoverflow.com/a/13343457/5409274
# https://stackoverflow.com/a/26733366/5409274
# https://stackoverflow.com/a/2853811/5409274
# https://stackoverflow.com/a/2853811/5409274
# https://www.linuxquestions.org/questions/other-%2Anix-55/how-can-i-run-a-command-on-another-shell-without-changing-the-current-shell-794580/
# https://www.tldp.org/LDP/abs/html/internalvariables.html
# https://stackoverflow.com/a/4277753/5409274
# =========================================================================
# Enable following line to see the script commands
# getting printing along with their execution. This will help for debugging.
#set -o verbose
E_BADARGS=85
if [ ! -n "$1" ]
then
echo "Usage: `basename $0` grep -nrs --color -w --include=\"*.{h,c}\" Testing ."
echo "Usage: `basename $0` find . -name \"*.txt\""
exit $E_BADARGS
fi
# Create a temporary file
TMPFILE=$(mktemp)
# Add stuff to the temporary file
#echo "echo Hello World...." >> $TMPFILE
#initialize the variable that will contain the whole argument string
argList=""
#iterate on each argument
for arg in "$#"
do
#if an argument contains a white space, enclose it in double quotes and append to the list
#otherwise simply append the argument to the list
if echo $arg | grep -q " "; then
argList="$argList \"$arg\""
else
argList="$argList $arg"
fi
done
#remove a possible trailing space at the beginning of the list
argList=$(echo $argList | sed 's/^ *//')
# Echoing the command to be executed to tmp file
echo "$argList" >> $TMPFILE
# Note: This should be your last command
# Important last command which deletes the tmp file
last_command="rm -f $TMPFILE"
echo "$last_command" >> $TMPFILE
#echo "---------------------------------------------"
#echo "TMPFILE is $TMPFILE as follows"
#cat $TMPFILE
#echo "---------------------------------------------"
check_for_last_line=$(tail -n 1 $TMPFILE | grep -o "$last_command")
#echo $check_for_last_line
#if tail -n 1 $TMPFILE | grep -o "$last_command"
if [ "$check_for_last_line" == "$last_command" ]
then
#echo "Okay..."
bash $TMPFILE
exit 0
else
echo "Something is wrong"
echo "Last command in your tmp file should be removing itself"
echo "Aborting the process"
exit 1
fi
$ bash --init-file <(echo 'some_command')
$ bash --rcfile <(echo 'some_command')
In case you can't or don't want to use process substitution:
$ cat script
some_command
$ bash --init-file script
Another way:
$ bash -c 'some_command; exec bash'
$ sh -c 'some_command; exec sh'
sh-only way (dash, busybox):
$ ENV=script sh
Here is yet another (working) variant:
This opens a new gnome terminal, then in the new terminal it runs bash. The user's rc file is read first, then a command ls -la is sent for execution to the new shell before it turns interactive.
The last echo adds an extra newline that is needed to finish execution.
gnome-terminal -- bash -c 'bash --rcfile <( cat ~/.bashrc; echo ls -la ; echo)'
I also find it useful sometimes to decorate the terminal, e.g. with colorfor better orientation.
gnome-terminal --profile green -- bash -c 'bash --rcfile <( cat ~/.bashrc; echo ls -la ; echo)'
I have a file named cmd that contains a list of Unix commands as follows:
hostname
pwd
ls /tmp
cat /etc/hostname
ls -la
ps -ef | grep java
cat cmd
I have another script that executes the commands in cmd as:
IFS=$'\n'
clear
for cmds in `cat cmd`
do
if [ $cmds ] ; then
$cmds;
echo "****************************";
fi
done
The problem is that commands in cmd without spaces run fine, but those with spaces are not correctly interpreted by the script. Following is the output:
patrick-laptop
****************************
/home/patrick/bashFiles
****************************
./prog.sh: line 6: ls /tmp: No such file or directory
****************************
./prog.sh: line 6: cat /etc/hostname: No such file or directory
****************************
./prog.sh: line 6: ls -la: command not found
****************************
./prog.sh: line 6: ps -ef | grep java: command not found
****************************
./prog.sh: line 6: cat cmd: command not found
****************************
What am I missing here?
Try changing the one line to eval $cmds rather than just $cmds
You can replace your script with the command
sh cmd
The shell’s job is to read commands and run them! If you want output/progress indicators, run the shell in verbose mode
sh -v cmd
I personally like this approach better - I don't want to munge the IFS if I don't have to do so. You do need to use an eval if you are going to use pipes in your commands. The pipe needs to be processed by the shell not the command. I believe the shell parses out pipes before the expanding strings.
Note that if your cmd file contains commands that take input there will be an issue. (But you can always create a new fd for the read command to read from.)
clear
while read cmds
do
if [ -n "$cmds" ] ; then
eval $cmds
echo "****************************";
fi
done < cmd
Edit: Turns out this fails on pipes and redirection. Thanks, Andomar.
You need to change IFS back inside the loop so that bash knows where to split the arguments:
IFS=$'\n'
clear
for cmds in `cat cmd`
do
if [ $cmds ] ; then
IFS=$' \t\n' # the default
$cmds;
echo "****************************";
IFS=$'\n'
fi
done
EDIT: The comment by Ben Blank pointed out that my old answer was wrong, thanks.
Looks like you're executing commands as a single string, so bash sees them as the script/executable name.
One way to avoid that would be to invoke bash on the command. Change:
if [ $cmds ] ; then
$cmds;
echo "****************************";
fi
to
if [ $cmds ] ; then
bash -c $cmds
echo "****************************";
fi
sed 'aecho "-----------"' cmd > cmd.sh; bash cmd.sh
sed 'aXX' appends XX to every line. This will not work for multiline-commands like:
for f in *
do
something $f
fi
but for single-line commands in most cases, it should do.