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.
Related
I always have this doubt and I don't know why the file inside the bin folder which create a simple http server doesn't have a .js extension.
Is there a reason behind this?
It's tradition on unix that executables don't have extension.
For example, on Linux and MacOS to list a directory you type:
ls
you don't type
ls.exe
Another example, to launch the Dropbox service on Linux you can type
dropbox
you don't type
dropbox.py
even though dropbox is just a text file containing Python code.
Unix (and also bash terminal on Windows) have a feature where if a file is marked as executable (using the chmod command) and the first line contains:
#!
.. then the shell (the program controlling the command line) will remove the first two characters (#!) and execute the rest of that first line. This is often called the shbang line (sh = shell, ! = bang).
Therefore, if you want to develop a command-line program in node.js all you need to do is start the file with #! /usr/bin/env node:
#! /usr/bin/env node
// ^
// |
// the 'env' command will find the correct install path of node
console.log('hello world');
Then use chmod to make the file executable:
chmod +x my-script.js
Of course, creating a program that ends in .js does not look "professional". For example you don't type gulp.js when you run gulp and you don't type npm.js when you run npm. So people follow tradition and make their executable scripts have no extension - it makes it harder for people to realise that you didn't write the program in C or assembly language.
Because www file is executable.
Open a console, go to the bin folder, and type ./www . The server will run.
Say I have a programme that's run from a shell script:
cd /path/to/file/
./programme
How would I bind that to a single command so I only have to type one thing? ie for most installed programmes I can just type the name of the programme and it's running
programme
you must add your script path in $PATH variable in ~/.bashrc file ,like this:
export PATH=$PATH:/path/to/file
or put your script in linux binary directory like /usr/local/bin , /usr/bin
For this you need to create a script with a name and give execute permission to it. Then Copy that script to /usr/bin directory . Now you can run your scrpt as a command in terminal. For details please refer the link https://devopsmanual.in/2018/04/17/create-our-own-script-in-linux/
I need to run a Make script that invokes a shell script.
I can run the shell script directly as root but when running make on the makefile (still as root) make is denied permission to run the same shell script?
The offending line in the Makefile is that one:
PLATFORM=$(shell $(ROOT)/systype.sh)
I could go in and hardcode the value of every PLATFORM variable of every Makefile scrip on the system but that would be pointless fix, I'd like to understand why there is that Permission Denied error:
make[1]: execvp: ../systype.sh: Permission denied
PS: The content of the shell script is not the issue even if the shell script only contain ls or echo linux the Permission is Denied to the Make utility to run the shell script.
PS: I am not a make expert by an mean so if the explanation is related to Make please be as specific as you can.
In your comments above you say when you "run it manually" you use . scriptname.sh, is that correct? You use . followed by scriptname.sh?
That does not run the script, that sources the script. Your statement that scriptname.sh will execute with and without the x permission since it is a shell script is wrong. You can source the script if you have read permissions. But you cannot execute the script unless you have execute permissions.
"Sourcing" means that a new shell is not started: instead your current shell (where you type that command) reads the contents of the script and runs them just as if you'd typed them in by hand, in the current shell. At the end all the side-effects (directory changes, variable assignments, etc.) that were performed in that script are still available in your current script.
"Executing" means that the script is treated like a program, but the program is a new shell that's started, which then reads the contents of the script and executes it. Once the script ends the shell exits and all side-effects are lost.
The $(shell ...) function in make will not source your script (unless you also use . there, which you did not). It will try to run your script. The error you show implies that either systype.sh did not have the execution bit set, or else that it had an invalid #! line. There's no other explanation I can think of.
If sourcing the file really does what you want then why not just use the same method in $(shell ...) that you use in your own personal use:
PLATFORM=$(shell . $(ROOT)/systype.sh)
If changing the user permission didn't work, are you sure that whatever user owns the script is the same user you're using to invoke make? You say you're "running as root"; is the script owned by root? Or is it owned by you and you're running sudo make or similar?
I don't know why you don't just use:
chmod +x systype.sh
and call it a day.
Adding execution permission to the file Group rather that the file User fixed the issue.
PS: I wonder why? It seems the Make utility run shell scripts not with the same user that started Make...
I am using a tool called Droidbox for experiment.
The tool has a shell script droidbox.sh which I can invoke through terminal.
droidbox.sh takes one argument i.e path of the apk
Usage: ./droidbox.sh APK
I want to call the droidbox.sh through a shell script.
I wrote a shell script like
#!/bin/bash
ARG1="/home/xxx/a.apk"
/home/xxx/DroidBox_4.1.1/droidbox.sh "$ARG1"
I am getting error which says
python: can't open file 'scripts/droidbox.py': [Errno 2] No such file or directory
Can anybody point out what am I doing wrong?
Your error message does not come from your script, but from the called one.
It sounds like the droidbox.sh script is not very smart and requires you to set the current working directory before you can call it.
I would typically use also some more variables, so you better see what belongs together:
#!/bin/sh
set -e
BASEDIR="/home/xxx"
DROIDDIR="$BASEDIR/DrroidBox_4.1.1"
APKFILE="$BASEDIR/a.apk"
cd "$DROIDDIR"
"$DROIDDIR/droidbox.sh" "$APKFILE"
If you dont use set -e you better combine commands which need to succeed together:
cd "$DROIDDIR" && "$DROIDDIR/droidbox.sh" "$APKFILE"
Otherwise the cd might fail when the directory is missing and all following commands execute in the wrong directory.
This error is because you're running the script in a different folder than the folder that houses your "scripts/droidbox.py" file. You can fix this in the following way(s):
Move the "scripts/" folder to the directory that you're running this script out of using "mv /path/to/scripts/ ."
Move your customer script to the folder that contains scripts/droidbox.py and run the script from there
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.