Why is my command not found but works in the command line? - linux

This is a part of my script:
read -p "[q] What is the meteor project's location? "
PATH=$REPLY
cd $PATH
cd src
echo $(pwd)
mrt bundle bundle.tar.gz
It prints the correct working directory. Entering mrt bundle bundle.tar.gz in the console works.
Executing my script, I get meteor_bundle.sh: line 9: bash: command not found. What could be the reason?

You choose the variable name poorly: PATHhas a special meaning in Bash: it is used to indicate Bash where to look for executable programs. Choose another name (preferably lowercase, see #Gordon's comment) and it should be fine.

You should use the full path for mrt because you changed the value of the PATH. Or choose another name for the variable from the second line.
See the following post to understand exactly what is the purpose of the PATH environment variable:
http://www.linfo.org/path_env_var.html

Related

Running `cd` in `.sh` file in anaconda environment on Linux [duplicate]

I'm trying to write a small script to change the current directory to my project directory:
#!/bin/bash
cd /home/tree/projects/java
I saved this file as proj, added execute permission with chmod, and copied it to /usr/bin. When I call it by:
proj, it does nothing. What am I doing wrong?
Shell scripts are run inside a subshell, and each subshell has its own concept of what the current directory is. The cd succeeds, but as soon as the subshell exits, you're back in the interactive shell and nothing ever changed there.
One way to get around this is to use an alias instead:
alias proj="cd /home/tree/projects/java"
You're doing nothing wrong! You've changed the directory, but only within the subshell that runs the script.
You can run the script in your current process with the "dot" command:
. proj
But I'd prefer Greg's suggestion to use an alias in this simple case.
The cd in your script technically worked as it changed the directory of the shell that ran the script, but that was a separate process forked from your interactive shell.
A Posix-compatible way to solve this problem is to define a shell procedure rather than a shell-invoked command script.
jhome () {
cd /home/tree/projects/java
}
You can just type this in or put it in one of the various shell startup files.
The cd is done within the script's shell. When the script ends, that shell exits, and then you are left in the directory you were. "Source" the script, don't run it. Instead of:
./myscript.sh
do
. ./myscript.sh
(Notice the dot and space before the script name.)
To make a bash script that will cd to a select directory :
Create the script file
#!/bin/sh
# file : /scripts/cdjava
#
cd /home/askgelal/projects/java
Then create an alias in your startup file.
#!/bin/sh
# file /scripts/mastercode.sh
#
alias cdjava='. /scripts/cdjava'
I created a startup file where I dump all my aliases and custom functions.
Then I source this file into my .bashrc to have it set on each boot.
For example, create a master aliases/functions file: /scripts/mastercode.sh
(Put the alias in this file.)
Then at the end of your .bashrc file:
source /scripts/mastercode.sh
Now its easy to cd to your java directory, just type cdjava and you are there.
You can use . to execute a script in the current shell environment:
. script_name
or alternatively, its more readable but shell specific alias source:
source script_name
This avoids the subshell, and allows any variables or builtins (including cd) to affect the current shell instead.
Jeremy Ruten's idea of using a symlink triggered a thought that hasn't crossed any other answer. Use:
CDPATH=:$HOME/projects
The leading colon is important; it means that if there is a directory 'dir' in the current directory, then 'cd dir' will change to that, rather than hopping off somewhere else. With the value set as shown, you can do:
cd java
and, if there is no sub-directory called java in the current directory, then it will take you directly to $HOME/projects/java - no aliases, no scripts, no dubious execs or dot commands.
My $HOME is /Users/jleffler; my $CDPATH is:
:/Users/jleffler:/Users/jleffler/mail:/Users/jleffler/src:/Users/jleffler/src/perl:/Users/jleffler/src/sqltools:/Users/jleffler/lib:/Users/jleffler/doc:/Users/jleffler/work
Use exec bash at the end
A bash script operates on its current environment or on that of its
children, but never on its parent environment.
However, this question often gets asked because one wants to be left at a (new) bash prompt in a certain directory after execution of a bash script from within another directory.
If this is the case, simply execute a child bash instance at the end of the script:
#!/usr/bin/env bash
cd /home/tree/projects/java
echo -e '\nHit [Ctrl]+[D] to exit this child shell.'
exec bash
To return to the previous, parental bash instance, use Ctrl+D.
Update
At least with newer versions of bash, the exec on the last line is no longer required. Furthermore, the script could be made to work with whatever preferred shell by using the $SHELL environment variable. This then gives:
#!/usr/bin/env bash
cd desired/directory
echo -e '\nHit [Ctrl]+[D] to exit this child shell.'
$SHELL
I got my code to work by using. <your file name>
./<your file name> dose not work because it doesn't change your directory in the terminal it just changes the directory specific to that script.
Here is my program
#!/bin/bash
echo "Taking you to eclipse's workspace."
cd /Developer/Java/workspace
Here is my terminal
nova:~ Kael$
nova:~ Kael$ . workspace.sh
Taking you to eclipe's workspace.
nova:workspace Kael$
simply run:
cd /home/xxx/yyy && command_you_want
When you fire a shell script, it runs a new instance of that shell (/bin/bash). Thus, your script just fires up a shell, changes the directory and exits. Put another way, cd (and other such commands) within a shell script do not affect nor have access to the shell from which they were launched.
You can do following:
#!/bin/bash
cd /your/project/directory
# start another shell and replacing the current
exec /bin/bash
EDIT: This could be 'dotted' as well, to prevent creation of subsequent shells.
Example:
. ./previous_script (with or without the first line)
On my particular case i needed too many times to change for the same directory.
So on my .bashrc (I use ubuntu) i've added the
1 -
$ nano ~./bashrc
function switchp
{
cd /home/tree/projects/$1
}
2-
$ source ~/.bashrc
3 -
$ switchp java
Directly it will do: cd /home/tree/projects/java
Hope that helps!
It only changes the directory for the script itself, while your current directory stays the same.
You might want to use a symbolic link instead. It allows you to make a "shortcut" to a file or directory, so you'd only have to type something like cd my-project.
You can combine Adam & Greg's alias and dot approaches to make something that can be more dynamic—
alias project=". project"
Now running the project alias will execute the project script in the current shell as opposed to the subshell.
You can combine an alias and a script,
alias proj="cd \`/usr/bin/proj !*\`"
provided that the script echos the destination path. Note that those are backticks surrounding the script name.
For example, your script could be
#!/bin/bash
echo /home/askgelal/projects/java/$1
The advantage with this technique is that the script could take any number of command line parameters and emit different destinations calculated by possibly complex logic.
to navigate directories quicky, there's $CDPATH, cdargs, and ways to generate aliases automatically
http://jackndempsey.blogspot.com/2008/07/cdargs.html
http://muness.blogspot.com/2008/06/lazy-bash-cd-aliaes.html
https://web.archive.org/web/1/http://articles.techrepublic%2ecom%2ecom/5100-10878_11-5827311.html
In your ~/.bash_profile file. add the next function
move_me() {
cd ~/path/to/dest
}
Restart terminal and you can type
move_me
and you will be moved to the destination folder.
You can use the operator && :
cd myDirectory && ls
While sourcing the script you want to run is one solution, you should be aware that this script then can directly modify the environment of your current shell. Also it is not possible to pass arguments anymore.
Another way to do, is to implement your script as a function in bash.
function cdbm() {
cd whereever_you_want_to_go
echo "Arguments to the functions were $1, $2, ..."
}
This technique is used by autojump: http://github.com/joelthelion/autojump/wiki to provide you with learning shell directory bookmarks.
You can create a function like below in your .bash_profile and it will work smoothly.
The following function takes an optional parameter which is a project.
For example, you can just run
cdproj
or
cdproj project_name
Here is the function definition.
cdproj(){
dir=/Users/yourname/projects
if [ "$1" ]; then
cd "${dir}/${1}"
else
cd "${dir}"
fi
}
Dont forget to source your .bash_profile
This should do what you want. Change to the directory of interest (from within the script), and then spawn a new bash shell.
#!/bin/bash
# saved as mov_dir.sh
cd ~/mt/v3/rt_linux-rt-tools/
bash
If you run this, it will take you to the directory of interest and when you exit it it will bring you back to the original place.
root#intel-corei7-64:~# ./mov_dir.sh
root#intel-corei7-64:~/mt/v3/rt_linux-rt-tools# exit
root#intel-corei7-64:~#
This will even take you to back to your original directory when you exit (CTRL+d)
I did the following:
create a file called case
paste the following in the file:
#!/bin/sh
cd /home/"$1"
save it and then:
chmod +x case
I also created an alias in my .bashrc:
alias disk='cd /home/; . case'
now when I type:
case 12345
essentially I am typing:
cd /home/12345
You can type any folder after 'case':
case 12
case 15
case 17
which is like typing:
cd /home/12
cd /home/15
cd /home/17
respectively
In my case the path is much longer - these guys summed it up with the ~ info earlier.
As explained on the other answers, you have changed the directory, but only within the sub-shell that runs the script. this does not impact the parent shell.
One solution is to use bash functions instead of a bash script (sh); by placing your bash script code into a function. That makes the function available as a command and then, this will be executed without a child process and thus any cd command will impact the caller shell.
Bash functions :
One feature of the bash profile is to store custom functions that can be run in the terminal or in bash scripts the same way you run application/commands this also could be used as a shortcut for long commands.
To make your function efficient system widely you will need to copy your function at the end of several files
/home/user/.bashrc
/home/user/.bash_profile
/root/.bashrc
/root/.bash_profile
You can sudo kwrite /home/user/.bashrc /home/user/.bash_profile /root/.bashrc /root/.bash_profile to edit/create those files quickly
Howto :
Copy your bash script code inside a new function at the end of your bash's profile file and restart your terminal, you can then run cdd or whatever the function you wrote.
Script Example
Making shortcut to cd .. with cdd
cdd() {
cd ..
}
ls shortcut
ll() {
ls -l -h
}
ls shortcut
lll() {
ls -l -h -a
}
If you are using fish as your shell, the best solution is to create a function. As an example, given the original question, you could copy the 4 lines below and paste them into your fish command line:
function proj
cd /home/tree/projects/java
end
funcsave proj
This will create the function and save it for use later. If your project changes, just repeat the process using the new path.
If you prefer, you can manually add the function file by doing the following:
nano ~/.config/fish/functions/proj.fish
and enter the text:
function proj
cd /home/tree/projects/java
end
and finally press ctrl+x to exit and y followed by return to save your changes.
(NOTE: the first method of using funcsave creates the proj.fish file for you).
You need no script, only set the correct option and create an environment variable.
shopt -s cdable_vars
in your ~/.bashrc allows to cd to the content of environment variables.
Create such an environment variable:
export myjava="/home/tree/projects/java"
and you can use:
cd myjava
Other alternatives.
Note the discussion How do I set the working directory of the parent process?
It contains some hackish answers, e.g.
https://stackoverflow.com/a/2375174/755804 (changing the parent process directory via gdb, don't do this) and https://stackoverflow.com/a/51985735/755804 (the command tailcd that injects cd dirname to the input stream of the parent process; well, ideally it should be a part of bash rather than a hack)
It is an old question, but I am really surprised I don't see this trick here
Instead of using cd you can use
export PWD=the/path/you/want
No need to create subshells or use aliases.
Note that it is your responsibility to make sure the/path/you/want exists.
I have to work in tcsh, and I know this is not an elegant solution, but for example, if I had to change folders to a path where one word is different, the whole thing can be done in the alias
a alias_name 'set a = `pwd`; set b = `echo $a | replace "Trees" "Tests"` ; cd $b'
If the path is always fixed, the just
a alias_name2 'cd path/you/always/need'
should work
In the line above, the new folder path is set
This combines the answer by Serge with an unrelated answer by David. It changes the directory, and then instead of forcing a bash shell, it launches the user's default shell. It however requires both getent and /etc/passwd to detect the default shell.
#!/usr/bin/env bash
cd desired/directory
USER_SHELL=$(getent passwd <USER> | cut -d : -f 7)
$USER_SHELL
Of course this still has the same deficiency of creating a nested shell.

I'm learning about shebangs. How do I make it work with node.js in a Mac terminal?

I have:
#!/usr/bin/env node
console.log("It works!");
I learned that env finds the node program and interprets it with node. I checked that env exists in /usr/bin.
When I call node itworks.js it works and outputs It works!. However, from what I understand, I should just be able to call itworks.js without node due to the shebang. But when I make this command it says -bash: itworks.js: command not found.
Could someone help me get the shebang to work?
First of all you need to make the file executable:
chmod +x itworks.js
Then you need to call it by specifying the path as well. Either:
/where/it/is/on/disk/itworks.js
or:
./itworks.js
The reason for :
-bash: itworks.js: command not found
is because bash looks for programs in directories in the PATH environment variable when you do not say where the file is - it does not look in the current directory unless you tell it.
You could update the PATH variable with the current directory shortcut ., but that can be a security risk, so most run the program like this:
./itworks.js
Of course if you put your scripts all in one directory then you could add that to PATH in one of your start-up files. For example, if you had a directory called bin in your home directory that held your scripts:
PATH=$PATH:"$HOME/bin"
You also need to add the execute permissions to the script:
chmod u+x itworks.js
The u indicates that we only give permission for the current user to execute this file. If we omit the u then anyone can run it.

Program runs fine from the terminal, but does not work from a shell script

I am on Linux Mint.
I have program called samtools, and it is stored in a folder on my desktop. I've added path to the executable into $PATH variable. In other words my local ~/.bashrc file has a line:
export PATH="~/Desktop/samtools/samtools-1.1:$PATH"
Executable file named samtools is in this folder.
So when I try to launch it from a command line like a simple command, by just typing "samtools" it works. It also works when I type direct path to the executable in the command line.
However when I try to launch it from a shell script, it does not launch and says either No such file or directory or not found.
Actually, I am trying to use another software which uses some shell scripts to preprocess some data. The error I am getting looks like this:
Indexing...
./RD_capture//process_one_capture.sh: 17: ./RD_capture//process_one_capture.sh: samtools: not found
Sorting...
./RD_capture//process_one_capture.sh: 20: ./RD_capture//process_one_capture.sh: samtools: not found
Piling up...
./RD_capture//process_one_capture.sh: 23: ./RD_capture//process_one_capture.sh: samtools: not found
Shell code looks like this:
echo "Indexing..."
samtools index $INPUTDIR/$sample.bam
#Then we sort them
echo "Sorting..."
samtools sort $INPUTDIR/$sample.bam $TMPDIR/$sample.sorted
#Finally we pile them up
echo "Piling up..."
samtools mpileup $TMPDIR/$sample.sorted.bam | cut -f 1-4 > $OUTPUT/$(basename $sample .bam).pile
Can anyone help me to solve this problem?
It looks like Linux is not able to find samtools.
To solve this use:
Declare this in the beginning (change pwd with the path that leads to executable from pwd)
export set CURRENT_DIR=`pwd`
and while calling samtools use
$CURRENT_DIR/samtools
Note 1: you are responsible to tell the exact path to shell script as shell scripts execute from /usr/bin/sh. try echoing the path on the console with the command, if this does not help, as follows:
echo `pwd`
Note 2: the use of back quote above that is located on left hand side top corner of your keyboard.
Note 3: the export set is used to store global variables in a shell script so that commands that appear after this command can use this variable and update it when ever required.

What does this shell script line of code mean

I need some help understanding following shell script line,
apphome = "`cd \`dirname $0\` && pwd && cd - >/dev/null`"
All I understand is, this is creating a variable called apphome.
This is not a valid shell code.
The shell don't allow spaces around =
For the rest, while this seems broken, it try to cd to the dir of the script itself, display the current dir & finally cd back to the latest cd place redirecting his standard output STDOUT to the /dev/null trash-bin (that's makes not any sense, cd display only on standard error STDERR when it fails, never on STDOUT)
If you want to do this in a proper a simple way :
apphome="$(dirname $0)"
That's all you need.
NOTE
The backquote
`
is used in the old-style command substitution, e.g.
foo=`command`
The
foo=$(command)
syntax is recommended instead. Backslash handling inside $() is less surprising, and $() is easier to nest. See http://mywiki.wooledge.org/BashFAQ/082
It seems to assign a command to the "apphome" variable. This command can be executed later.
dirname returns a directory portion of a file name. $0 is the name of the script this line contains (if I am not mistaken).
Now, executing dirname <name> will return a directory, and cd will use the value.
So, what it would do is execute three command in the row assuming that each one of them succeeds. The commands are:
cd `dirname [name of the script]`
pwd
cd -
First command will change directory to the directory containing your script; second will print current directory; third will take yo back to the original directory. Output of the third command will not be printed out.
In summary, it will print out a name of a directory containing the script that contains the line in question.
At least, this is how I understand it.

Creating a command in linux

I have created a simple script:
echo "the path of the current directory is `pwd`"
and saved it by the name pathinfo
then i have created a bin directory at my home page with path as
/home/vpnsadmin/bin
and copied my script(pathinfo) to that bin directory.
Now i want run this script as a command but it is showing error
-bash: /usr/bin/test2: No such file or directory
but if copy my script(pathinfo) to "/usr/bin/" then it runs as a command.
the PATH environment variable is set as-
PATH=/usr/kerberos/sbin:/usr/kerberos/bin:/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin:/home/vpnsadmin/bin
My question is why does the shell not run it as a command when it is present in /home/vpnsadmin/bin.
or else
why does it only check for the binary at /usr/bin and not at /home/vpnsadmin/bin or at /bin
The shell that is to execute your command needs to have the correct PATH variable set at the time of execution and, depending on shell, might need to have created its own internal (hash)map of the available commands.
Assuming you are using bash, try the following with your script saved in /usr/bin:
$ PATH=/ test2
$ PATH=/usr/bin test2
In the first case you should get an expected "not found" error, in the second it should work. The third test to perform is left as an exercise...
And I have to say that the supplied error message looks a bit odd if you actually tried to do
$ test2
and not
$ /usr/bin/test2
before copying the command to /usr/bin.
Edit:
Also, avoid naming your scripts test, in any way shape or form. This causes so much confusion for beginners.
Hint:
man test
Did you have the path to bash at the top of your script and did you use backticks around pwd?
#!/bin/bash
echo "the path of the current directory is `pwd`"
Did you make the file executable?
chmod +x pathinfo
There is another script pathinfo somewhere in your path which contains a call to /usr/bin/test2
Try whereis pathinfo to see how many there are and which pathinfo to see which one your shell currently prefers.

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