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'which' vs 'command -v' in Bash [duplicate]
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Closed 11 months ago.
I dont understand what does this programme does :
if command -v apt-get > /dev/null; then
dpkg-query -W -f='...'
i dont understand how the command : "command" works
There are different ways to do things in the shell.
The default is to execute shell builtins.
Next commands (binaries, scripts, etc; basically something external to the shell) are looked up in the directories listed in the PATH variable.
This is followed by aliases. These are reusable abbreviations to commands, you can think of them as a way to save typing. They are usually defined in the shell's rc files. For example I have this defined in my ~/.bashrc: alias lt='ls --color=tty -lhtr'; it lets me type only lt to get a reverse sorted list of files (newest at the bottom).
More complex commands can be grouped into shell functions.
The command command is a shell builtin that allows you to bypass any functions that might also match a binary in your PATH, or any aliases. Say I use rsync frequently between specific hosts, so I wrote a function called rsync that includes the destination with all the options, allowing me to type less. However if I want to run rsync with a different destination, I could use command rsync <other rsync options> to bypass the function definition. Note that aliases are not bypassed; this is relevant if you are using command interactively. You can find the documentation in the builtin page linked above, or by typing help command
See man bash or help command in bash for help:
Runs COMMAND with ARGS suppressing shell function lookup, ...
The if checks whether there is a command apt-get; if there is a function apt-get, it's ignored.
Related
My default shell is bash in Ubuntu 14.04. I have a csh script file named clean.sh with the following make command:
#! /bin/csh -f
make -f commande.make del
And commande.make has
CKHOME=../CHEMKIN/DATA_BASES
LIN_DATA=${CKHOME}/LIN_FILES/
LINK_CKTP=${CKHOME}/LINK_CKTP_FILES/
#-----------------------------------------------------
include schema_cinetique.make
LINKFILE=${NAME}_LINK
LINKTPFILE=${NAME}_LINKTP
LINKFILE_OLD=${NAME_OLD}_LINK
LINKFILE_NEW=${NAME_NEW}_LINK
#-----------------------------------------------------
cplink :
${COPY} ${LINK_CKTP}${LINKFILE} LINK
cplink2 :
${COPY} ${LINK_CKTP}${LINKFILE} LINKZ1
tplink :
${COPY} ${LINK_CKTP}${LINKTPFILE} LINKTPZ1
calcul :
${COPY} jobtimp1 LJOBNZ1
${COPY} unsteadyf.dat1 DATZ1
del :
${DELETE} LINKZ1 LINKTPZ1 LJOBNZ1 DATZ1 SOLASUZ1
I opened the terminal and moved to the location and tried
./clean.sh
or
csh clean.sh &
or
csh -f clean.sh
Nothing worked.
I got the following line in the terminal,
LINKZ1 LINKTPZ1 LJOBNZ1 DATZ1 SOLASUZ1
make: LINKZ1: Command not found
make: *** [del] Error 127
So, how to run clean.sh file ?
You are confused. The Csh script contains a single command which actually runs identically in Bash.
#!/bin/bash
make -f commande.make del
Or, for that matter, the same with #!/bin/sh. Or, in this individual case, even sh clean.sh, since the shebang is then just a comment, and the commands in the file are available in sh just as well as in csh.
Once make runs, that is what parses and executes the commands in commande.make. make is not a "Fortran command", it is a utility for building projects (but the makefile named commande.make probably contains some instructions for how to compile Fortran code).
In the general case, Csh and Bash are incompatible, but the differences are in the shell's syntax itself (so, the syntax of loops and conditionals, etc, as well as variable assignments and various other shell builtins).
As an aside, Csh command files should probably not have a .sh extension, as that vaguely implies Bourne shell (sh) syntax. File extensions on Unix are just a hint to human readers, so not technically important; but please don't confuse them/us.
(As a further aside, nobody should be using Csh in 2022. There was a time when the C shell was attractive compared to its competition, but that was on the order of 40 years ago.)
The subsequent errors you are reporting seem to indicate that the makefile depends on some utilities which you have not installed. Figuring that out is a significant enough and separate enough question that you should probably ask a new question about that, probably with more debugging details. But in brief, it seems that make needs to be run with parameters to indicate what NAME and COPY (and probably some other variables) should be. Try with make -f commande.make COPY=cp DELETE=rm NAME=foobar for a start, but it's probably not yet anywhere near sufficient.
(I would actually assume that there will be a README file or similar to actually instruct you how to use commande.make since it seems to have some local conventions of its own.)
It seems the script is written having portability in mind, i.e. the name of the cp and rm binaries is kept in variables rather than hard-coding it. My best guess is that this has been done to make it possible to run the script on non UNIX systems, like Windows.
To make it work, export the respective variables before running the script. For the del action you are calling, only the DELETE variable is needed. It should be set to rm which is the command used to remove files on Linux:
export DELETE=rm
./clean.sh
Note: exporting the variable can also be done in one line when invoking the script, by prepending it to the command line:
DELETE=rm ./clean.sh
This behaviour is described in the bash manual:
The environment for any simple command or function may be augmented temporarily by prefixing it with parameter assignments, as described in Shell Parameters. These assignment statements affect only the environment seen by that command.
Original Title: Indirect parameter substitution breaks when the script is sourced (zsh)
zsh 5.7.1 (x86_64-apple-darwin19.0)
GNU bash, version 4.4.20(1)-release (x86_64-pc-linux-gnu)
I’m developing a shell script on a Mac and I’m trying to keep it portable between bash & zsh, so array indexing is a consideration. I know that I can set KSH_ARRAYS to get indexing to start at 0, but I decided to query the OS for the shell that’s in use and set the start index accordingly, which led to the issue described below.
It made sense (to me anyway!) to use indirect expansion, which is what led to the problem. Consider the script indirect.sh:
#! /bin/bash
declare -r ARRAY_START_BASH=0
declare -r ARRAY_START_ZSH=1
declare -r SHELL_BASH=0
declare -r SHELL_ZSH=1
# Indirect expansion is used to reference the values of the variables declared
# in this case statement e.g. ${!ARRAY_START}
case $(basename $SHELL) in
"bash" )
declare -r SHELL_ID=SHELL_BASH
declare -r ARRAY_START=ARRAY_START_BASH
;;
"zsh" )
declare -r SHELL_ID=SHELL_ZSH
declare -r ARRAY_START=ARRAY_START_ZSH
;;
* )
return 1
;;
esac
echo "Shell ID: ${!SHELL_ID} Index arrays from: ${!ARRAY_START}"
It works fine when run from the command line while in the same directory:
<my home> ~ % echo "$(./indirect.sh)"
Shell ID: 1 Index arrays from: 1
Problems arise when I source the script:
<my home> ~ % echo "$(. ~/indirect.sh)"
/Users/<me>/indirect.sh:28: bad substitution
I don’t understand why sourcing the script changes the behavior of the parameter expansion.
Is this expected behavior? If so, I’d be grateful if someone could explain it and hopefully, offer a work around.
The problem described in the original post has nothing to do with indirect expansion. The difference in behavior is a result of different shells being invoked depending on whether the script is “executed” or “sourced”. These differences reveal the basic flaw in deriving the shell from the $SHELL variable that underpins the script's design. If the shell defined in $SHELL does not match the shebang, the script will fail either when sourced or executed. An explanation follows.
Indirect expansion doesn’t offer value in the given scenario because values could just as easily be assigned directly. They’ll have to be assigned that way regardless given the different syntax used for indirect expansion between shells. In fact, other syntax differences between shells makes the entire premise for detecting the shell moot! However, putting that aside, the difference in behavior is a result of different shells being invoked based on whether the script is “executed” or “sourced”. The behavior of sourcing is well documented with numerous explanations on the web, but for context here’s how it works:
Executing a Script
Use the “./“ syntax to execute a script.
When run this way, the script executes in a sub-shell. Any changes the
script makes to it’s shell are applied to the sub-shell, not the shell
in which the script was launched, so those changes are lost when the
shell exits because the sub-shell in which it executed is destroyed as
well. For example, if the script changes the working directory, it
does so in the sub-shell. The working directory of the main shell that
launched the script is unchanged when the script terminates. If you
want to make changes to the shell in which the script was launched, it
must be sourced.
Sourcing a Script
Use the “source “ syntax to source a
script. When run this way, the script essentially becomes an argument
for the source command, which handles invoking the appropriate
execution. Some shells (e.g. ksh) use a single period “.” instead of
“source”.
When a script is executed with the “./“ syntax, the shebang at the top of the file is used to determine which shell to use. When a script is sourced, the shebang is ignored and the shell in which the script is launched is used instead. Also note that the period that appears in the “./“ command syntax used to execute a script, is not related to the period that’s occasionally used as an alias for the source command.
The script in the post uses bash in the shebang statement, so it works when executed because it’s run using bash. When it’s sourced from zsh, it encounters the incorrect indirect expansion syntax:
“${!A_VAR}"
The correct syntax is:
"${(P)A_VAR}"
However, correcting the syntax won’t help because it will then fail when executed. The shebang will invoke bash and the syntax will be wrong again. That renders indirection useless for accessing a variable designed to indicate the shell in use. More importantly, a design based on querying an environment variable for the shell is flawed due to differences in the shell that’s ultimately used depending on whether the script is executed or sourced.
To add to your answer (what I'm going to say is too long for a comment), I can not think of any application, why your script could be useful if not sourced. Actually, I came accross the need of such a script by myself in exactly one occasion:
Since I use as interactive shell not only zsh, but also sometimes bash, so I have written my .zshrc and .bashrc to set up everything (including defining variables and shell functions for interactive use). In order to safe work,
I try to put code which works under both bash and zsh into a single file (say: .commonrc), and my .zshrc and .bashrc have inside them a
source .commonrc
While many things are so different in bash and zsh, that I can't put them into .commonrc, some can, provided I do some tweaking. One reason for headache is obviously the different indexing of arrays, which you seemingly try to solve. So I have also a similar feature. However, I don't nee ca case construct for this. Instead, my .bashrc looks like this (using your naming of the variables):
...
declare -r ARRAY_START=0
source .commonrc
...
and my .zshrc looks like this:
...
declare -r ARRAY_START=1
source .commonrc
...
Since it does not happen that the .bashrc is run from a zsh and vice versa, I don't need to query what kind of shell I have.
I have a shell script on a mac (OSX 10.9) named msii810161816_TMP_CMD with the following content.
matlab
When I execute it, I get
./msii810161816_TMP_CMD: line 1: matlab: command not found
However, when I type matlab into the shell directly it starts as normal. How can it be that the same command works inside the shell but not inside a shell script? I copy-pasted the command directly from the script into the shell and it worked ...
PS: When I replace the content of the script with
echo matlab
I get the desired result, so I can definitely execute the shell script (I use ./msii810161816_TMP_CMD)
Thanks guys!
By default, aliases are not expanded in non-interactive shells, which is what shell scripts are. Aliases are intended to be used by a person at the keyboard as a typing aid.
If your goal is to not have to type the full path to matlab, instead of creating an alias you should modify your $PATH. Add /Applications/MATLAB_R2014a.app/bin to your $PATH environment variable and then both you and your shell scripts will be able to simply say
matlab
This is because, as commenters have stated, the PATH variable inside of the shell executing the script does not include the directory containing the matlab executable.
When a command name is used, like "matlab", your shell looks at every directory in the PATH in order, searching for one containing an executable file with the name "matlab".
Without going into too much detail, the PATH is determined by the shell being invoked.
When you execute bash, it combines a global setting for basic directories that must be in the PATH with any settings in your ~/.bashrc which alter the PATH.
Most likely, you are not running your script in a shell where the PATH includes matlab's directory.
To verify this, you can take the following steps:
Run which matlab. This will show you the path to the matlab executable.
Run echo "$PATH". This will show you your current PATH settings. Note that the directory from which matlab is included in the colon-separated list.
Add a line to the beginning of your script that does echo "$PATH". Note that the directory from which matlab is not included.
To resolve this, ensure that your script is executed in a shell that has the needed directory in the PATH.
You can do this a few ways, but the two most highly recommended ones would be
Add a shebang line to the start of your script. Assuming that you want to run it with bash, do #!/bin/bash or whatever the path to your bash interpreter is.
The shebang line is not actually fully standardized by POSIX, so BSD-derived systems like OSX will happily handle multiple arguments to the shebanged executable, while Linux systems pass at most one argument.
In spite of this, the shebang is an easy and simple way to document what should be used to execute the script, so it's a good solution.
Explicitly invoke your script with a shell as its interpreter, as in bash myscript.sh or tcsh myscript.sh or even sh myscript.sh
This is not incompatible with using a shebang line, and using both is a common practice.
I believe that the default shell on OSX is always bash, so you should start by trying with that.
If these instructions don't help, then you'll have to dig deeper to find out why or how the PATH is being altered between the calling context and the script's internal context.
Ultimately, this is almost certainly the source of your issue.
I am trying to create an AppleScript with commands below. An issue I am having is there is an error at the third line. I have no problem using the lame command in the terminal directly. In addition, lame is not a native Mac utility; I installed it on my own. Does anybody have a solution?
do shell script "cd ~/Downloads"
do shell script "say -f ~/Downloads/RE.txt -o ~/Downloads/recording.aiff"
do shell script "lame -m m ~/Downloads/recording.aiff ~/Downloads/recording.mp3"
-- error "sh: lame: command not found" number 127
do shell script "rm recording.aiff RE.txt"
To complement Paul R's helpful answer:
The thing to note is that do shell script - regrettably - does NOT see the same $PATH as shells created by Terminal.app - a notable absence is /usr/local/bin.
On my OS X 10.9.3 system, running do shell script "echo $PATH" yields merely:
/usr/bin:/bin:/usr/sbin:/sbin
There are various ways around this:
Use the full path to executables, as in Paul's solution.
Manually prepend/append /usr/local/bin, where many non-system executables live, to the $PATH - worth considering if you invoke multiple executables in a single do shell script command; e.g.:
do shell script "export PATH=\"/usr/local/bin:$PATH\"
cd ~/Downloads
say -f ~/Downloads/RE.txt -o ~/Downloads/recording.aiff
lame -m m ~/Downloads/recording.aiff ~/Downloads/recording.mp3
rm recording.aiff RE.txt"
Note how the above use a single do shell script command with multiple commands in a single string - commands can be separated by newlines or, if on the same line, with ;.
This is more efficient than multiple invocations, though adding error handling both inside the script code and around the do shell script command is advisable.
To get the same $PATH that interactive shells see (except additions made in your bash profile), you can invoke eval $(/usr/libexec/path_helper -s); as the first statement in your command string.
Other important considerations with do shell script:
bash is invoked as sh, which results in changes in behavior, most notably:
process substitution (<(...)) is not available
echo by default accepts no options and interprets escape sequences such as \n.
other, subtle changes in behavior; see http://www.gnu.org/software/bash/manual/html_node/Bash-POSIX-Mode.html
You could address these issues manually by prepending shopt -uo posix; shopt -u xpg_echo; to your command string.
The locale is set to the generic "C" locale instead of to your system's; to fix that, manually prepend export LANG='" & user locale of (system info) & ".UTF-8' to your command string.
No startup files (profiles) are read; this is not surprising, because the shell created is a noninteractive (non-login) shell, but sometimes it's handy to load one's profile by manually by prepending . ~/.bash_profile to the command string; note, however, that this makes your AppleScript less portable.
do shell script command reference: http://developer.apple.com/library/mac/#technotes/tn2065/_index.html
Probably a PATH problem - use the full path for lame, e.g.
do shell script "/usr/local/bin/lame -m m ~/Downloads/recording.aiff ~/Downloads/recording.mp3"
I have been struggling to get the path of an installed BASH command via Applescript for a long time. Using the information here, I finally succeeded.
tell me to set sox_path to (do shell script "eval $(/usr/libexec/path_helper -s); which sox")
Thanks.
Url：http://sourceforge.net/project/showfiles.php?group_id=290&package_id=309
./configure
make install
I'm a newbie to scripting languages trying to learn bash programming.
I have very basic question. Suppose I want to create three folders like $HOME/folder/
with two child folders folder1 and folder2.
If I execute command in shell like
mkdir -p $HOME/folder/{folder1,folder2}
folder will be created along with child folder.
If the same thing is executed through script I'm not able get expected result. If sample.sh contains
#!/bin/sh
mkdir -p $HOME/folder/{folder1,folder2}
and I execute sh ./sample.sh, the first folder will be created then in that a single {folder1,folder2} directory is created. The separate child folders are not created.
My query is
How the script file works when we compared to as terminal command? i.e., why is it not the same?
How to make it work?
bash behaves differently when invoked as sh, to more closely mimic the POSIX standard. One of the things that changes is that brace expansion (which is absent from POSIX) is no longer recognized. You have several options:
Run your script using bash ./sample.sh. This ignores the hashbang and explicitly uses bash to run the script.
Change the hashbang to read #!/bin/bash, which allows you to run the script by itself (assuming you set its execute bit with chmod +x sample.sh).
Note that running it as sh ./sample.sh would still fail, since the hashbang is only used when running the file itself as the executable.
Don't use brace expansion in your script. You could still use as a longer method for avoiding duplicate code:
for d in folder1 folder2; do
mkdir -p "$HOME/folder/$d"
done
Brace expansion doesn't happen in sh.
In sh:
$ echo {1,2}
produces
{1,2}
In bash:
$ echo {1,2}
produces
1 2
Execute your script using bash instead of using sh and you should see expected results.
This is probably happening because while your tags indicate you think you are using Bash, you may not be. This is because of the very first line:
#/bin/sh
That says "use the system default shell." That may not be bash. Try this instead:
#!/usr/bin/env bash
Oh, and note that you were missing the ! after #. I'm not sure if that's just a copy-paste error here, but you need the !.