If I run the following in a bash shell:
./script /path/to/file.txt
echo !$:t
it outputs file.txt and all is good.
If in my script I have:
echo $1:t
it outputs /path/to/file.txt:t
How can I get it to output file.txt as per the behaviour I see in a shell? Thanks in advance.
Use the parameter expansion syntax:
echo ${1##*/}
Modifier only work on word designators
In bash you can use the ${1##*/} expansion to get the basename of the file with all leading path components removed:
$ set -- /path/to/file
$ echo "$1"
/path/to/file
$ echo "${1##*/}"
file
You can use this in a script as well:
#!/bin/sh
echo "${1##*/}"
While ${1##*/} will work when Bash is called as /bin/sh, other Bash features require that you use #!/bin/bash at the start of your script. This notation may also not be available in other shells.
A more portable solution is this:
#!/bin/sh
echo `basename "$1"`
Related
Having executed:
cd ~ && mkdir mytmp && cd mytmp/
echo > somefile
and doing this in a bash script mytmp/myscript.sh:
#!/bin/bash
# version 1
cmd="find . -type f -printf %f\n"
$cmd
renders the wanted result, i.e.:
somefile
myscript.sh
Notice that for some reason I don't need to surround %f\n with quotes as I'd do it if I were to write the command in the terminal. Doing so would render a bad result:
#!/bin/bash
# version 2
cmd="find . -type f -printf '%f\n'"
$cmd
results in:
'somefile
''myscript.sh
'
I need to execute $cmd and at the same time print it as I'd write it in the terminal.
Adding echo $cmd in version 1 executes the command properly but echoes the command without quotes.
Adding echo $cmd in version 2 echoes the command with quotes, like I want, but the result of command execution is bad.
How can I achieve this?
Use set -v.
Example Script
I used some overly complicated script, to test the output for quoting and so on.
#! /bin/bash
set -v
myVariable='test'
# a comment
echo "$(echo "$myVariable") two" | cat -
Output When Running The Script
$ ./myscript
myVariable='test'
# a comment
echo "$(echo "$myVariable") two" | cat -
echo "$myVariable"
test two
As we can see, quotes, variable names, and comments are retained, but commands from subshells will appear twice. Since you don't use any subshells, that shouldn't be a problem.
I actually had the answer to this question, but since I "accidentally" got what I wanted and haven't found the answer online after searching for it, I thought I'd post it here anyway.
The solution is including the quotes as in version 2 and getting rid of them when executing the command by using eval:
#!/bin/bash
cmd="find . -type f -printf '%f\n'"
eval $cmd
echo $cmd
In user's console I have bash:
$ echo $SHELL
/bin/bash
$ bash --version
GNU bash, version 4.2.46(1)-release (x86_64-redhat-linux-gnu)
I have code in file test.sh:
$ cat test.sh
aaa='---aa-aa---'
echo "${aaa}"
echo 'does not work...'
# trim "-"
echo ${aaa/+(-)}
echo ${aaa%%+(-)}
echo 'works for one symbol...'
echo ${aaa%-}
echo ${aaa/-}
The last two rows work fine but previous ones.
$ bash test.sh
---aa-aa---
does not work...
---aa-aa---
---aa-aa---
works for one symbol...
---aa-aa--
--aa-aa---
In the same time, if you would try to make this console it works:
$ aaa='---aa-aa---'
$ echo ${aaa/+(-)}
aa-aa---
$ echo ${aaa%%+(-)}
---aa-aa
So, why it doesn't work in a script?
You seem to have shopt -s extglob enabled in your interactive shell, which turns on extended globbing. This is not the default behavior, and needs to be explicitly enabled in your script. See extended pattern matching in the bash hackers wiki for details.
How can a script determine it's path when it is sourced by ksh? i.e.
$ ksh ". foo.sh"
I've seen very nice ways of doing this in BASH posted on stackoverflow and elsewhere but haven't yet found a ksh method.
Using "$0" doesn't work. This simply refers to "ksh".
Update: I've tried using the "history" command but that isn't aware of the history outside the current script.
$ cat k.ksh
#!/bin/ksh
. j.ksh
$ cat j.ksh
#!/bin/ksh
a=$(history | tail -1)
echo $a
$ ./k.ksh
270 ./k.ksh
I would want it echo "* ./j.ksh".
If it's the AT&T ksh93, this information is stored in the .sh namespace, in the variable .sh.file.
Example
sourced.sh:
(
echo "Sourced: ${.sh.file}"
)
Invocation:
$ ksh -c '. ./sourced.sh'
Result:
Sourced: /var/tmp/sourced.sh
The .sh.file variable is distinct from $0. While $0 can be ksh or /usr/bin/ksh, or the name of the currently running script, .sh.file will always refer to the file for the current scope.
In an interactive shell, this variable won't even exist:
$ echo ${.sh.file:?}
-ksh: .sh.file: parameter not set
I believe the only portable solution is to override the source command:
source() {
sourced=$1
. "$1"
}
And then use source instead of . (the script name will be in $sourced).
The difference of course between sourcing and forking is that sourcing results in the invoked script being executed within the calling process. Henk showed an elegant solution in ksh93, but if, like me, you're stuck with ksh88 then you need an alternative. I'd rather not change the default ksh method of sourcing by using C-shell syntax, and at work it would be against our coding standards, so creating and using a source() function would be unworkable for me. ps, $0 and $_ are unreliable, so here's an alternative:
$ cat b.sh ; cat c.sh ; ./b.sh
#!/bin/ksh
export SCRIPT=c.sh
. $SCRIPT
echo "PPID: $$"
echo "FORKING c.sh"
./c.sh
If we set the invoked script in a variable, and source it using the variable, that variable will be available to the invoked script, since they are in the same process space.
#!/bin/ksh
arguments=$_
pid=$$
echo "PID:$pid"
command=`ps -o args -p $pid | tail -1`
echo "COMMAND (from ps -o args of the PID): $command"
echo "COMMAND (from c.sh's \$_ ) : $arguments"
echo "\$SCRIPT variable: $SCRIPT"
echo dirname: `dirname $0`
echo ; echo
Output is as follows:
PID:21665
COMMAND (from ps -o args of the PID): /bin/ksh ./b.sh
COMMAND (from c.sh's $_ ) : SCRIPT=c.sh
$SCRIPT variable: c.sh
dirname: .
PPID: 21665
FORKING c.sh
PID:21669
COMMAND (from ps -o args of the PID): /bin/ksh ./c.sh
COMMAND (from c.sh's $_ ) : ./c.sh
$SCRIPT variable: c.sh
dirname: .
So when we set the SCRIPT variable in the caller script, the variable is either accessible from the sourced script's operands, or, in the case of a forked process, the variable along with all other environment variables of the parent process are copied for the child process. In either case, the SCRIPT variable can contain your command and arguments, and will be accessible in the case of both sourcing and forking.
You should find it as last command in the history.
I need to write a script to test if the command blablabla exists in the classpath. So I wrote the following code:
if ! hash blablabla >/dev/null 2>&1; then
echo not found
fi
This works fine when the script is executed in the bash. But if I try it in KSH, then it doesn't work:
#! /usr/bin/ksh
if ! hash blablabla >/dev/null 2>&1; then
echo not found
fi
I expect the echo not found to be executed but instead I get nothing. What's the problem?
I believe command is portable (if that matters):
command -v -- some_command >/dev/null 2>&1 ||
printf '%s\n' "not found"
In bash hash is a builtin command. In ksh it's an alias; aliases aren't active in shell scripts.
alias hash='alias -t --'
Try the which command, which is an external command and therefore shell-independent:
if ! which -s blablabla; then
echo not found >&2
fi
Tha hash command is a shell built-in command in bash, but not in ksh. You might want to use whence instead.
if ! whence blah; then print urgh; fi
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)'