I'm still new to Unix. Is it possible to run multiple commands of Unix in one time? Such as write all those commands that I want to run in a file, then after I call that file, it will run all the commands inside that file? or is there any way(or better) which i do not know?
Thanks for giving all the comments and suggestions, I will appreciate it.
Short answer is, yes. The concept is known as shell scripting, or bash scripts (a common shell). In order to create a simple bash script, create a text file with this at the top:
#!/bin/bash
Then paste your commands inside of it, one to a line.
Save your file, usually with the .sh extension (but not required) and you can run it like:
sh foo.sh
Or you could change the permissions to make it executable:
chmod u+x foo.sh
Then run it like:
./foo.sh
Lots of resources available on this site and the web for more info, if needed.
echo 'hello' && echo 'world'
Just separate your commands with &&
We can run multiple commands in shell by using ; as separator between multiple commands
For example,
ant clean;ant
If we use && as separator then next command will be running if last command is successful.
you can also use a semicolon ';' and run multiple commands, like :
$ls ; who
Yep, just put all your commands in one file and then
bash filename
This will run the commands in sequence. If you want them all to run in parallel (i.e. don't wait for commands to finish) then add an & to the end of each line in the file
If you want to use multiple commands at command line, you can use pipes to perform the operations.
grep "Hello" <file-name> | wc -l
It will give number of times "Hello" exist in that file.
Sure. It's called a "shell script". In bash, put all the commands in a file with the suffix "sh". Then run this:
chmod +x myfile.sh
then type
. ./myFile
or
source ./myfile
or just
./myfile
To have the commands actually run at the same time you can use the job ability of zsh
$ zsh -c "[command1] [command1 arguments] & ; [command2] [command2 arguments]"
Or if you are running zsh as your current shell:
$ ping google.com & ; ping 127.0.0.1
The ; is a token that lets you put another command on the same line that is run directly after the first command.
The & is a token placed after a command to run it in the background.
Related
I am trying to create a simple script to zip a list of files each into its own zip file. The files are big, so I a trying to send the to background using ampersand. It works as I can see the temporary files filling up and after some time the files are created, but issuing the 'jobs' command does not list the jobs. What am I doing wrong?
#!/bin/ksh
for file in $*;do
bash -c "zip -q $file.zip $file" &
done
NATIVE CSH SOLUTION
As I said earlier, shell scripts execute in a subshell and the parent shell will not be able to list the jobs of a subshell. In order to use jobs, the jobs need to be running in the same shell.
This can be achieved by source-ing the file. Since your default shell is csh the file should contain these lines according to the csh syntax
# not a script. no need for shebang
# sourcing this file **in csh** will
# start quiet zip jobs in the background
# for all files in the working dir (*)
foreach file in (*)
zip -q "$file.zip" "$file" &
end
Keeping this file in an easily accessible location and running source /path/to/file will give you what you need.
This is the only way to do it in csh for the following reasons:
cannot be a shell script. jobs will not be possible
csh does not support shell functions
setting alias not easy due csh's foreach syntax
But also consider a few of these alternatives
A. The organisation allows for changing the login shell
Change the shell to one that allows shell functions (e.g. to bash)
chsh -s `which bash` $USER
Logout and login or simply execute bash (or your shell of choice) to start a new shell
Check you are in the right shell echo $0
Add a function to your user-level login script (~/.bashrc for bash)
# executing this command appends a bash function named `zipAll` to ~/.bashrc
# modify according to your choice of shell
cat << 'EOF' >> ~/.bashrc
zipAll() {
for file in *; do
zip -q "$file.zip" "$file" &
done
}
EOF
The function zipAll should be available from the next login onwards.
B. The organisation does not allow changing login shell
Simply execute bash (or your shell of choice) to start a new shell
Follow steps A3 to A4
Temporarily switch to a new shell with bash (or your shell of choicd) when you need this function
C. B; but you want to use bash (or other shell)
I do not know if this is a good solution. Hopefully someone will point out the ill-effects of it. Hopefully your organisation simply allows you to change the login shell
Seeing as your default shell is csh, add a line to ~/.cshrc to start bash (or your choice of shell)
echo 'bash --login' >> ~/.cshrc
Follow steps A2 to A4
Copy necessary lines from existing ~/.cshrc to ~/.bashrc (or the file corresponding to your shell)
Confusion regarding zip usage was oversight on my part. Apologies.
NB: The syntax zip -q $file $file.zip does not work with my version. But I retain it assuming that it works on OP's system
PS: The command that works with my version of zip is zip -q $file.zip file
I am trying to run a single command using bash in a sh script. There is no way to use bash for the script, I have to use sh. However, I need to run a bash-only command in sh.
Basically, I want something like the following:
bash --command_in "echo foobar"
Is this possible? I don't want to make a second script file just to run that one command in bash (like bash my_script.bash).
Derp, it's the -c flag. This wasn't easy to Google, and the --help is prety brief.
I have the following file directory
/home/aero/airplane/case1
/home/aero/airplane/case2
..
/home/aero/airplane/casex
I have a shell script that performs some simple commands. I don't want to "cd" into each directory and run the shell script (runscript.sh). Is there a way to run it in a simpler fashion, where I type something along the lines of:
runscript.sh /case*
Thanks for your help!
You can have a second script that run runscript.sh with all the cases:
#In the new shell run every case
runscript.sh /home/aero/airplane/case1
runscript.sh /home/aero/airplane/case2
..
runscript.sh /home/aero/airplane/casex
You can do something like the following:
Create a file named runscript.sh inside /home/aero/airplane
#!/bin/bash
MAINPATH=/home/aero/airplane
CHILD_DIR=$(ls | grep -v runscript.sh)
for i in $MAINPATH
do
cd $MAINPATH/$i
pwd
cd ..
done
Run:
sh runscript.sh
See that the script enters all the directories running it a single time.
Now, you can change the "pwd" line to do whatever you want inside all the directories.
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 !.
I am trying to run a command in a script, something like this one:
ssh user#host:/bin/echo > /home/path/file.log
Now when I run this command on a command line it works fine, however when put in a script (shell or ruby ) it cribs saying:
/bin/sh: /home/path/*.log: No such file or directory
Am I missing something?
Thanks!
Update:
It's weird that same thing is not being executed now even on the shell when I use putty. I have verified that the path and file exists on remote machine which is being ssh'ed into.
You need to loop over the files. If it works from the command line then your interactive shell is not a standard shell.
for f in /home/path/*.log; do
:>"$f"
done
Note also the use of a null command; in many shells, you don't need a command at all. Your echo puts an unattractive empty line at the beginning of each file.
If you are attempting to run this remotely, you will need to quote it:
ssh user#remote 'for f in /home/path/*.log; do :>"$f"; done'
Its working fine when I put quotes:
ssh user#host:"/bin/echo > /home/path/file.log"