I was following a tutorial on openvpn and it needed to execute a command . ./vars. It displays a message. On reading the file I found that it executes a echo command in file and disregards everything else in file. On adding other echo statement, it also gets executed. So i would like some basic explanation on this. Is this something related to bash only?
. and source are synonymous: it just runs the file line-by-line in the current shell. ./vars is just a path to some file named vars in the current directory. So all that command does is run the file vars line-by-line.
As for the rest of your question, I don't really understand what you're asking. Can you clarify?
The . at the start has an explanation here, and the ./filename is a relative reference to the file.
. sources or imports a bash script to the current script you are working on ( if you are creating a script ) or to the current tty ( terminal or command line ) if you are working with the commandline interface. The script it is sourcing must be an executable ( it should have the -x flag set )
./filename means find filename in the present directory am in. if you execute only ./filename in tty ( terminal or command line ) or inside a script, it finds filename and check if filename is executable, then it runs filename . You should take not that ./ means the present working directory. Using two dots ( ../filename ) instead of one with filename tells the bash parser go to the previous directory before the present one am in
using . ./filename you are telling the bash parser to import or source (.) filename (./filename ) in the directory you are currently on
Related
I would like to know what does the command source do. I have tried:
whatis
$ whatis source
source: nothing appropriate.
man
$ man source
No manual entry for source
source (-h, --help, etc...)
$ source
source: not enough arguments
But it seems no documentation about it.
I commonly use it to save any changed on my dotfiles, but what does it exactly do? Why there is not documentation about it?
source is a bash shell built-in command that executes the content of the file passed as an argument, in the current shell. It has a synonym in . (period).
Syntax
. filename [arguments]
source filename [arguments]
From the source manual
source filename [arguments]
Read and execute commands from filename in the current shell environment and
return the exit status of the last command executed from filename. If
filename does not contain a slash, file names in PATH are used to find the
directory containing filename. The file searched for in PATH need not be
executable. When bash is not in posix mode, the current directory is
searched if no file is found in PATH. If the sourcepath option to the short
builtin command is turned off, the PATH is not searched. If any arguments
are supplied, they become the positional parameters when filename is
executed. Otherwise the positional parameters are unchanged. The return
status is the status of the last command exited within the script (0 if no
commands are executed), and false if filename is not found or cannot be
read.
Be careful! ./ and source are not quite the same.
./script runs the script as an executable file, launching a new shell to run it
source script reads and executes commands from filename in the current shell environment
Note: ./script is not . script, but . script == source script
Is there any difference between source in bash after all?
This is not a problem affecting me in any way, but just for curiosity...
I have added export PATH=$PATH:/usr/local/go/bin:$GOPATH/bin
to my ~/.profile to include a new directory into my bash search.
Then, I ran $ source ~/.profile to reload may shell and I checked my path using $ echo $PATH
The question is:
- why every time I ran source ~/.profile, it appends the same information again,
- how can I clear it?
What I have tried:
- Tried running it multiple times and it keeps adding the same
- Tried to figure out what does the source command does but could not find where it is which source
First question:
why every time I ran source ~/.profile, it appends the same
information again
Simply, source <FILE> does not reload your shell. It only
executes all commands saved in <FILE> as if they were typed directly
by you in the terminal.
Second question:
how can I clear it?
To reload shell open a new terminal
window/tab. Doing just bash or exec bash won't work because a new
process will inherit its parent environment.
Third question:
Tried to figure out what does the source command does but could not
find where it is which source
As I explained once here https://unix.stackexchange.com/a/202326/72304:
All commands that can be run in Bash without typing an explicit path
to it such as ./command can be divided into two parts: Bash shell
builtins and external commands. Bash shell builtins come installed
with Bash and are part of it while external commands are not part of
Bash. This is important because Bash shell builtins are documented
inside man bash and their documentation can be also invoked with help
command while external commands are usually documented in their own
manpages or take some king of -h, --help flag. To check whether a
command is a Bash shell builtin or an external command:
$ type local
local is a shell builtin
It will display how command would be interpreted if used as a command
name (from help type). Here we can see that local is a shell builtin.
Let's see another example:
$ type vim
vim is /usr/bin/vim
In your case:
$ type source
source is a shell builtin
Now we know it's not an external command but a shell bultin (this is why which does not find it) so we need to use help to see what it does:
$ help source
source: source filename [arguments]
Execute commands from a file in the current shell.
Read and execute commands from FILENAME in the current shell. The
entries in $PATH are used to find the directory containing FILENAME.
If any ARGUMENTS are supplied, they become the positional parameters
when FILENAME is executed.
Exit Status:
Returns the status of the last command executed in FILENAME; fails if
FILENAME cannot be read.
source executes the content of the file passed as argument in the current shell.
It appends the same information again because export is appending a string to PATH, without checking anything (it is not checking if the substring that you want to append is already in the variable).
To avoid appending to PATH every time, you should save the values of your PATH without referring to itself, e.g.:
export PATH=/usr/local/bin:/usr/bin:/bin:/usr/local/games:/usr/local/go/bin:$GOPATH/bin
Edit:
To check if the directory is already in PATH:
if [[ ":$PATH:" != *":/usr/local/go/bin:$GOPATH/bin:"* ]]; then
export PATH=$PATH:/usr/local/go/bin:$GOPATH/bin
fi
I found the below snippet at the .sh file of my project to define some path :
PGMPATH=`pwd|sed -e "s#/survey1##" `
What does the above line means ?
Reference of PGMPATH is used as below :
LIBS="${LIBS}:${PGMPATH}/edmz-par-api_1.4.jar"
LIBS="${LIBS}:${PGMPATH}/commons-logging.jar"
If it is telling the path where the jar file is located , please explain how it works .
So first you should know that this is two commands - pwd and sed -e "s#/survey1##" - and these two commands are being run together in a pipeline. That is, the output of the first command is being sent to the second command as input.
That is, in general, what | means in unix shell scripts.
So then, what do each of these commands do? pwd stands for "print working directory" and prints the current directory (where you ran the script from, unless the script itself had any cd commands in it).
sed is a command that's really a whole separate programming language that people do many simple text-processing commands with. The simple sed program you have here - s#/survey1## - strips the string /survey1 out of its input, and prints the result.
So the end result is that the variable PGMPATH becomes the current directory with /survey1 stripped out of it.
I am attempting to work with an existing library of code but have encountered an issue. In short, I execute a shell script (let's call this one A) whose first act is to call another script (B). Script B is in my current directory (a requirement of the program I'm using). The software's manual makes reference to bash, however comments in A suggest it was developed in ksh. I've been operating in bash so far.
Inside A, the line to execute B is simply:
. B
It uses the "dot space" syntax to call the program. It doesn't do anything unusual like sudo.
When I call A without dot space syntax, i.e.:
./A
it always errors saying it cannot find the file B. I added pwd, ls, whoami, echo $SHELL, and echo $PATH lines to A to debug and confirmed that B is in fact right there, the script is running with the same $SHELL as I am at the command prompt, the script is the same user as I am, and the script has the same search path $PATH as I do. I also verified if I do:
. B
at the command line, it works just fine. But, if I change the syntax inside A to:
./B
instead, then A executes successfully.
Similarly, if I execute A with dot space syntax, then both . B and ./B work.
Summarizing:
./A only works if A contains ./B syntax.
. A works for A with either ./B or . B syntax.
I understand that using dot space (i.e. . A) syntax executes without forking to a subshell, but I don't see how this could result in the behavior I'm observing given that the file is clearly right there. Is there something I'm missing about the nuances of syntax or parent/child process workspaces? Magic?
UPDATE1: Added info indicating that the script may have been developed in ksh, while I'm using bash.
UPDATE2: Added checking to verify $PATH is the same.
UPDATE3: The script says it was written for ksh, but it is running in bash. In response to Kenster's answer, I found that running bash -posix then . B fails at the command line. That indicates that the difference in environments between the command line and the script is that the latter is running bash in a POSIX-compliant mode, whereas the command line is not. Looking a little closer, I see this in the bash man page:
When invoked as sh, bash enters posix mode after the startup files are read.
The shebang for A is indeed #!/bin/sh.
In summary, when I run A without dot space syntax, it's forking to its own subshell, which is in POSIX-compliant mode because the shebang is #!/bin/sh (instead of, e.g., #!/bin/bash. This is the critical difference between the command line and script runtime environments that leads to A being unable to find B.
Let's start with how the command path works and when it's used. When you run a command like:
ls /tmp
The ls here doesn't contain a / character, so the shell searches the directories in your command path (the value of the PATH environment variable) for a file named ls. If it finds one, it executes that file. In the case of ls, it's usually in /bin or /usr/bin, and both of those directories are typically in your path.
When you issue a command with a / in the command word:
/bin/ls /tmp
The shell doesn't search the command path. It looks specifically for the file /bin/ls and executes that.
Running ./A is an example of running a command with a / in its name. The shell doesn't search the command path; it looks specifically for the file named ./A and executes that. "." is shorthand for your current working directory, so ./A refers to a file that ought to be in your current working directory. If the file exists, it's run like any other command. For example:
cd /bin
./ls
would work to run /bin/ls.
Running . A is an example of sourcing a file. The file being sourced must be a text file containing shell commands. It is executed by the current shell, without starting a new process. The file to be sourced is found in the same way that commands are found. If the name of the file contains a /, then the shell reads the specific file that you named. If the name of the file doesn't contain a /, then the shell looks for it in the command path.
. A # Looks for A using the command path, so might source /bin/A for example
. ./A # Specifically sources ./A
So, your script tries to execute . B and fails claiming that B doesn't exist, even though there's a file named B right there in your current directory. As discussed above, the shell would have searched your command path for B because B didn't contain any / characters. When searching for a command, the shell doesn't automatically search the current directory. It only searches the current directory if that directory is part of the command path.
In short, . B is probably failing because you don't have "." (current directory) in your command path, and the script which is trying to source B is assuming that "." is part of your path. In my opinion, this is a bug in the script. Lots of people run without "." in their path, and the script shouldn't depend on that.
Edit:
You say the script uses ksh, while you are using bash. Ksh follows the POSIX standard--actually, KSH was the basis for the POSIX standard--and always searches the command path as I described. Bash has a flag called "POSIX mode" which controls how strictly it follows the POSIX standard. When not in POSIX mode--which is how people generally use it--bash will check the current directory for the file to be sourced if it doesn't find the file in the command path.
If you were to run bash -posix and run . B within that bash instance, you should find that it won't work.
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.