Let's take a little example:
$ cat source.sh
#!/bin/bash
echo "I'm file source-1"
. source-2.sh
And:
$ cat source-2.sh
#!/bin/bash
echo "I'm file source-2"
Now run:
$ ./source.sh
I'm file source-1
I'm file source-2
If I'll change the call of the second file in first:
$ cat source.sh
#!/bin/bash
echo "I'm file source-1"
source source-2.sh
It will have the same effect as using dot.
What is difference between these methods?
The only difference is in portability.
. is the POSIX-standard command for executing commands from a file; source is a more-readable synonym provided by Bash and some other shells. Bash itself, however, makes no distinction between the two.
There is no difference.
From the manual:
source
source filename
A synonym for . (see Bourne Shell Builtins).
Related
Most of the time, an alias works well, but some times, the command is executed by other programs, and they find it in the PATH, in this situation an alias not works as well as a real file.
e.g.
I have the following alias:
alias ghc='stack exec -- ghc'
And I want to translate it into an executable file, so that the programs which depending on it will find it correctly. And the file will works just like the alias does, including how it process it's arguments.
So, is there any tool or scripts can help doing this?
Here is my solution, I created a file named ghc as following:
#!/bin/sh
stack exec -- ghc "$#"
The reason why there is double quote around $# is explained here: Propagate all arguments in a bash shell script
So, is there any tool or scripts can help doing this?
A lazy question for a simple problem... Here's a function:
alias2script() {
if type "$1" | grep -q '^'"$1"' is aliased to ' ; then
alias |
{ sed -n "s#.* ${1}='\(.*\)'\$##\!/bin/sh\n\1 \"\${\#}\"#p" \
> "$1".sh
chmod +x "$1".sh
echo "Alias '$1' hereby scriptified. To run type: './$1.sh'" ;}
fi; }
Let's try it on the common bash alias ll:
alias2script ll
Output:
Alias 'll' hereby scriptified. To run type: './ll.sh'
What's inside ll.sh:
cat ll.sh
Output:
#!/bin/sh
ls -alF "${#}"
https://www.youtube.com/watch?v=bu3_RzzEiVo
My intention is to experiment with shell scripts in a file. (FreeBSD 10.2)
I create a file named script.sh
cat > script.sh
set dir = `pwd`
echo The date today is `date`
echo The current directory is $dir
[Ctrl-d]
After giving it execution authority, I run the command
sh script.sh
I get
Why the directory is not displayed?
Then I make a change.
cat > script.sh
set dir = `pwd`
echo The date today is `date`
echo The current directory is `pwd`
[Ctrl-d]
This time , it works fine. The directory is shown successfully.
I would like to know why ? Could anyone tell me ?
The answer from TessellatingHeckler was on the right track.
From the man page for sh(1):
set [-/+abCEefIimnpTuVvx] [-/+o longname] [-c string] [-- arg ...]
The set command performs three different functions:
With no arguments, it lists the values of all shell variables.
If options are given, either in short form or using the long
``-/+o longname'' form, it sets or clears the specified options
as described in the section called Argument List Processing.
If you want a command for setting environment variables, that command would be setvar, which you'd use as follows:
setvar dir `pwd`
This is, however, uncommon usage. The more common synonym for this would be:
dir=`pwd`
or
dir=$(pwd)
Note that there are no spaces around the equals sign.
Note also that if you choose to use the setvar command, it's a good idea to put your value inside quotes. The following produces an error:
$ mkdir foo\ bar
$ cd foo\ bar
$ setvar dir `pwd`
Instead, you would need:
$ setvar dir "`pwd`"
Or more clearly:
$ dir="$(pwd)"
Note that you may also need to export your variables. The export command is used to mark a variable that should be passed along to sub shells that the running shell spawns. An example should make this more clear:
$ foo="bar"
$ sh -c 'echo $foo'
$ export foo
$ sh -c 'echo $foo'
bar
One other thing I'll add is that it's common and unnecessary to use date as you're doing in your script, since that command is able to produce its own formatted output. Try this:
$ date '+The date today is %+'
For date options, you can man date and man strftime.
Last tip: when using echo, put things in quotes. You'll produce less confusing and more reasonable output. Note:
$ foo="`printf 'a\nb\n'`"
$ echo $foo
a b
$ echo "$foo"
a
b
Hope this helps!
We now to find the directory of a shell script using dirname and $0, but this doesn't work when the script is inluded in another script.
Suppose two files first.sh and second.sh:
/tmp/first.sh :
#!/bin/sh
. "/tmp/test/second.sh"
/tmp/test/second.sh :
#!/bin/sh
echo $0
by running first.sh the second script also prints first.sh. How the code in second.sh can find the directory of itself? (Searching for a solution that works on bash/csh/zsh)
There are no solution that will work equally good in all flavours of shells.
In bash you can use BASH_SOURCE:
$(dirname "$BASH_SOURCE")
Example:
$ cat /tmp/1.sh
. /tmp/sub/2.sh
$ cat /tmp/sub/2.sh
echo $BASH_SOURCE
$ bash /tmp/1.sh
/tmp/sub/2.sh
As you can see, the script prints the name of 2.sh,
although you start /tmp/1.sh, that includes 2.sh with the source command.
I must note, that this solution will work only in bash. In Bourne-shell (/bin/sh) it is impossible.
In csh/tcsh/zsh you can use $_ instead of BASH_SOURCE.
In bash i get the executing script's parent folder name by this line
SCRIPT_PARENT=`readlink -f ${BASH_SOURCE%/*}/..`
Is there any way to achieve this in zsh in a way that works both in zsh and bash?
Assume i have got a file /some/folder/rootfolder/subfolder/script with the contents:
echo `magic-i-am-looking-for`
I want it to behave this way:
$ cd /some/other/folder
$ . /some/folder/rootfolder/subfolder/script
/some/folder/rootfolder
$ . ../../folder/rootfolder/subfolder/script
/some/folder/rootfolder
$ cd /some/folder/rootfolder
$ . subfolder/script
/some/folder/rootfolder
$ cd subfolder
$ . script
/some/folder/rootfolder
This should work in bash and zsh. My first implements this behavior, but does due to $BASH_SOURCE not work in zsh.
So basically its:
Is there a way to emulate $BASH_SOURCE in zsh, that also works in bash?
I now realized that $0 in zsh behaves like $BASH_SOURCE in bash. So using $BASH_SOURCE when available and falling back to $0 solves my problem:
${BASH_SOURCE:-$0}
There is a little zsh edge case left, when sourcing from $PATH like:
zsh> cat ../script
echo \$0: $0
echo \$BASH_SOURCE: $BASH_SOURCE
echo '${BASH_SOURCE:-$0}:' ${BASH_SOURCE:-$0}
zsh> . script
$0: script
$BASH_SOURCE:
${BASH_SOURCE:-$0}: script
bash> . script
$0: bash
$BASH_SOURCE: /home/me/script
${BASH_SOURCE:-$0}: /home/me/script
I could do a which script but this would not play nice with other cases
While it would be easy to do this in zsh, it is just as easy to use pure bash which is able to be evaluated in zsh. If you cannot use any command that may or may not be on your path, then you can only use variable alteration to achieve what you want:
SCRIPT_SOURCE=${0%/*}
This is likely to be a relative path. If you really want the full path then you will have to resort to an external command (you could implement it yourself, but it would be a lot of work to avoid using a very available command):
SCRIPT_SOURCE=$(/bin/readlink -f ${0%/*})
This doesn't depend on your $PATH, it just depends on /bin/readlink being present. Which it almost certainly is.
Now, you wanted this to be a sourced file. This is fine, as you can just export any variable you set, however if you execute the above then $0 will be the location of the sourced file and not the location of the calling script.
This just means you need to set a variable to hold the $0 value which the sourced script knows about. For example:
The script you will source:
echo ${LOCATION%/*}
The script that sources that script:
LOCATION=$0
<source script here>
But given that the ${0%/*} expansion is so compact, you could just use that in place of the script.
Because you were able to run the command from your $PATH I'll do something like that:
SCRIPT_PARENT=$(readlink -f "$(which $0)/..")
Is that your desired output?
$0 expands to the name of the shell script.
$ cat ./sample-script
#!/bin/bash
echo $0
$ chmod 700 ./sample-script
$ ./sample-script
./sample-script
If the shell script is invoked via a symbolic link, $0 expands to its name:
$ ln -s ./sample-script symlinked-script
$ ./symlinked-script
./symlinked-script
How could I get the name of an alias? Here `$0' expands again to the filename:
$ alias aliased-script=./sample-script
$ aliased-script
./sample-script
Aliases are pretty dumb, according to the man page
...Aliases are expanded when a command is read, not when it is executed...
so since bash is basically just replacing a string with another string and then executing it, there's no way for the command to know what was expanded in the alias.
I imagine you already know this, but for the record the answer is: you need cooperation by the code implementing the alias.
alternate_name () {
MY_ALIAS_WAS=alternate_name real_name "$#"
}
or, if you really want to use the superseded alias syntax:
alias alternate_name="MY_ALIAS_WAS=alternate_name real_name"
...and then...
$ cat ~/bin/real_name
#!/bin/sh
echo $0, I was $MY_ALIAS_WAS, "$#"
bash does not make this available. This is why symlinks are used to invoke multiplex commands, and not aliases.