I am quite new to Linux but am enjoying it. I am a level above a directory called dao_all. In the dao_all directory I have a script I would like to run called make_pyc_package.sh which will produce another folder (with output in a new directory called pkg).
I am trying to run the code as follows:
./dao_all/make_pyc_package.sh pkg
However when I run this I get the following output:
-bash: ./: Is a directory
I only can think of 2 situations that can create that error:
there is a ' ' (space) at start of dao_all and we cannot see it in the code you posted. In that case you get that error.
error is created by script. I mean script will run, but creates that error while running.
coud you use change directory command to switch directory to "dao_all" and then run the script ,try following commands to do so
```
cd dao_all - "change to dao_all directory"
ls - "just to check if you have make_pyc_package.sh"
if you see make_pyc_package.sh on ls command ,then run the script by
./make_pyc_package.sh
``
Related
I am building project source code in a SUSE server.
The project build.sh called "lzma" command to compress kernel.
The project build.sh need "sudo" to get access to some system command.
But I has tried to execute "sudo ./build.sh", and the shell always report error: "lzma: command not found."
I could execute "lzma" in shell with my user account. It works fine.
I also write a test shell script named "test.sh" which calls "lzma" command.
I found that it fails with same error message if I excute "test.sh" with "sudo" .
But if I execute "test.sh" without "sudo", it works fine.
Why ?
"Command not found" within sudo is almost invariably the result of an environment variable such as PATH, LD_LIBRARY_PATH (if what's missing is not the executable but a shared library it requires) or the like being altered.
You can pass working values through your environment variables through explicitly:
sudo PATH="$PATH" ./test.sh
Sudo uses a different Path then your user account.
EDIT (see comments)
Try and execute:
type lzma
Say the output reads something like '/usr/bin/lzma', then just copy that output into your sudo command like (for example):
sudo /usr/bin/lzma
That should do the trick. You should also write the full path of lzma into your shell script if you are to run it as root.
EDIT 2:
Or, as Charles Duffy mentioned in his answer, you could leave all things as is and simply use PATH="$PATH" in your command if you are trying to execute your file as SUDO or as a different user.
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.
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'm trying to run a shell script remotely. I'm having a resource(.txt file) in one machine and a shell script in another machine, but in the same network. The script resides in a particular directory and the script takes some binary files that resides in some other directory everything in the target machine. Now, I'm not able to run the script remotely. But I can run a simple shell program which does not have any local dependencies. Can you please suggest a solution to resolve this problem?
Here is the command that I used for running a shell-script-without-any-dependencies.
cat input.txt | ssh clrg#192.168.2.22 "sh path/to/shell/script/tokenize.sh"
On running a script with dependencies, I got the following error,
"ambiguous redirect"
../../deter-4/bin/determine: No such file or directory
../../deter-4/bin/determine1: No such file or directory
Now, how can I tell to accept dependencies while running the shell script
I think your problem is the working directory. Try something like this:
cat input.txt | ssh clrg#192.168.2.22 "sh -c 'cd /path/to/shell/script; ./tokenize.sh'"
Your script seems to expect to be run from a specific working directory only. The error messages you mentioned in your comment about the relative paths (starting with ../../) at least suggest that.
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.