I have the following script created by some self-claimed bash expert:
SCRIPT_LOCATION="$(readlink -f $0)"
SCRIPT_DIRECTORY="$(dirname ${SCRIPT_LOCATION})"
export PYTHONPATH="${PYTHONPATH}:${SCRIPT_DIRECTORY}/util"
That runs nicely on my local Ubuntu 16.04. Now I wanted to use it on our RH 7.2 servers; and there I got an error message from readlink; about being called with bad parameters.
Then I figured: on Ubuntu, $0 gives "bash"; whereas on RH, it gives "-bash".
EDIT: script is invoked as . ourscript.sh
Questions:
Any idea why that is?
When I change my script to use a hardcoded readlink -f bash the whole things works. Are there "better" ways for fixing this?
Feel free to also explain what readlink -f bash is actually doing ;-)
As the script is sourced the readlink -f $0 is pointless as it will just show you the command used to run the shell you are currently using.
To explain the difference in command lets look at the bash man page:
A login shell is one whose first character of argument zero is a -, or one started with the --login option.
When bash is invoked as an interactive login shell, or as a non-interactive shell with the --login option, it first reads and executes commands from the file /etc/profile, if that file exists. After reading that file, it looks for ~/.bash_profile, ~/.bash_login, and ~/.profile, in that order, and reads and executes commands from the first one that exists and is readable. The --noprofile option may be used when the shell is started to inhibit this behavior.
So guessing ubuntu starts with the noprofile option.
As for readlink, we can again look at the man page
-f, --canonicalize
canonicalize by following every symlink in every component of the given name recursively; all but the last component must exist
Therefore it follows symlinks to the base.
Using readlink -f with any non qualified path will result in it just appending the last arg to your current working directory which will not actually show where the script is run.
Try putting any random string instead of bash after it and will see the script is unaffected.
e.g
readlink -f dafsfdsf
Returns
/home/me/testscript/dafsfdsf
Related
This is not a problem affecting me in any way, but just for curiosity...
I have added export PATH=$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'm using the 'at' command in order to create 3 directories, just a dumb bash script:
#!/bin/bash
for i in {1..3}
do
mkdir dir$i
done
Everything is ok if I execute that script directly on terminal, but when I use 'at' command as follows:
at -f g.sh 18:06
It only creates one directory named dir{1..3}, taking interval not as an interval but as a list with one element {1..3}. According to this I think my mistake is using bash script due to at executes commands using /bin/sh but I'm not sure.
Please tell me if I'm right and I would appreciate some alternative to my code since even it is useless I'm curious to know what's wrong with at and bash.
The #! line only affects what happens when you run a script as a program (e.g. using it as a command in the shell). When you use at, it's not being run as a program, it's simply used as the standard input to /bin/sh, so the shebang has no effect.
You could do:
echo './g.sh' | at 18:06
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 have a simple script cmakeclean to clean cmake temp files:
#!/bin/bash -f
rm CMakeCache.txt
rm *.cmake
which I call like
$ cmakeclean
And it does remove CMakeCache.txt, but it doesn't remove cmake_install.cmake:
rm: *.cmake: No such file or directory
When I run it like:
$ . cmakeclean
it does remove both.
What is the difference and can I make this script work like an usual linux command (without . in front)?
P.S.
I am sure the both times is same script is executed. To check this I added echo meme in the script and rerun it in both ways.
Remove the -f from your #!/bin/bash -f line.
-f prevents pathname expansion, which means that *.cmake will not match anything. When you run your script as a script, it interprets the shebang line, and in effect runs /bin/bash -f scriptname. When you run it as . scriptname, the shebang is just seen as a comment line and ignored, so the fact that you do not have -f set in your current environment allows it to work as expected.
. script is short for source script which means the current shell executes the commands in the script. If there's an exit in there, the current shell will exit (and e. g. the terminal window will close).
This is typically used to modify the environment of the current shell (set variables etc.).
script asks the shell to fork itself, then exec the given script in the child process, and then wait in the father for the termination of the child. If there's an exit in the script, this will be executed by the child shell and thus only terminate this. The father shell stays intact and unaltered by this call.
This is typically used to start other programs from the current shell.
Is this about ClearCase? What did you do in your poor life where you've been assigned to work in the deepest bowels of hell?
For years, I was a senior ClearCase Administer. I haven't touched it in over a decade. My life is way better now. The sky is bluer, bird songs are more melodious, and my dread over coming to work every day is now a bit less.
Getting back to your issue: It's hard to say exactly what's going on. ClearCase does some wacky things. In a dynamic view, the ClearCase repository on Unix systems is hidden in the shell's environment. Now you see it, now you don't.
When you run a shell script, it starts up a new environment. If a particular shell variable is not imported, it is invisible that shell script. When you merely run cmakeclean from the command line, you are spawning a new shell -- one that does not contain your ClearCase environment.
When you run a shell script with a dot prefix like . cmakeclean, you are running that shell script in the current shell which contains your ClearCase environment. Thus, it can see your ClearCase view.
If you're using a snapshot view, it is possible that you have a $HOME/.bashrc that's changing directories on you. When a new shell environment runs in BASH (the default shell in MacOS X and Linux), it first runs $HOME/.bashrc. If this sets a particular directory, then you end up in that directory and not in the directory where you ran your shell script. I use to see this when I too was involved in ClearCase hell. People setup their .kshrc script (it was the days before BASH and most people used Kornshell) to setup their views. Unfortunately, this made running any other shell script almost impossible to do.
I want to write a very simple script , which takes a process name , and return the tail of the last file name which contains the process name.
I wrote something like that :
#!/bin/sh
tail $(ls -t *"$1"*| head -1) -f
My question:
Do I need the first line?
Why isn't ls -t *"$1"*| head -1 | tail -f working?
Is there a better way to do it?
1: The first line is a so called she-bang, read the description here:
In computing, a shebang (also called a
hashbang, hashpling, pound bang, or
crunchbang) refers to the characters
"#!" when they are the first two
characters in an interpreter directive
as the first line of a text file. In a
Unix-like operating system, the
program loader takes the presence of
these two characters as an indication
that the file is a script, and tries
to execute that script using the
interpreter specified by the rest of
the first line in the file
2: tail can't take the filename from the stdin: It can either take the text on the stdin or a file as parameter. See the man page for this.
3: No better solution comes to my mind: Pay attention to filenames containing spaces: This does not work with your current solution, you need to add quotes around the $() block.
$1 contains the first argument, the process name is actually in $0. This however can contain the path, so you should use:
#!/bin/sh
tail $(ls -rt *"`basename $0`"*| head -1) -f
You also have to use ls -rt to get the oldest file first.
You can omit the shebang if you run the script from a shell, in that case the contents will be executed by your current shell instance. In many cases this will cause no problems, but it is still a bad practice.
Following on from #theomega's answer and #Idan's question in the comments, the she-bang is needed, among other things, because some UNIX / Linux systems have more than one command shell.
Each command shell has a different syntax, so the she-bang provides a way to specify which shell should be used to execute the script, even if you don't specify it in your run command by typing (for example)
./myscript.sh
instead of
/bin/sh ./myscript.sh
Note that the she-bang can also be used in scripts written in non-shell languages such as Perl; in the case you'd put
#!/usr/bin/perl
at the top of your script.