I don't understand the meaning of:
$ ./your_program <dino>wilma
I'm learning perl, and I do not understand how to do this. I am using PUTTY.
The $ ./your_program indicates that you should run a program your_program on your shell. It assumes you have Linux. The $ indicates your command prompt.
So if you have a Windows machine and a server or another computer with Linux that you connect to with PuTTY, you need to write your program on that machine.
Then you need to make it executable.
$ chmod u+x your_program
Now you can run it. Running a program that is executable in Linux is done by typing the name of the program into the shell. You just did that with chmod, and maybe with vim or emacs when you created the file. But because your program is not accessible from everywhere, you need to put the ./your_program so the shell knows that you want to run it inside of the current directory. That's what the . is for.
$ ./your_program wilma
The wilma is a command line argument. It will be passed to your program.
You could also run it with the perl interpreter without making it executable.
$ perl your_program wilma
You can name all your Perl programs with .pl at the end so it's easier for you to distinguish what type of file they are.
$ denotes the unix command prompt.
./ is the current path - by convention unix systems don't look for executable programs in the current working directory (the places it looks is defined by the PATH environment variable).
your_program is the name of the file you just created/saved.
The above will only work if your file is set "executable" - chmod u+x your_program. You can alternatively use perl your_program and achieve basically the same result.
<dino means 'open the file "dino" and feed it into this program on the standard input. (STDIN).
>wilma means open the file "wilma"; truncate it, and write the output of this program to this file.
STDIN is a unix concept that's 'standard input' - it can either be 'things you type' or the content of a file or command.
That might not make a lot of sense, but it's all about piping - you can:
cat file | grep someword | sed 's/oneword/anotherword/'
That opens a file ( with cat) filters all the lines containing someword and then does a pattern replacement on it.
cat will "send" file to grep on STDIN.
It seems to be a quotation from chapter 5.6 of Learn Perl, the whole quote is:
...In that way, the user can type a command like this one at the shell
prompt:
$ ./your_program <dino >wilma
That command tells the shell that the program's input should be read
from the file dino, and the output should go to the file wilma. As
long as the program blindly reads its input from STDIN, processes it
(in whatever way we need), and blindly writes its output to STDOUT,
this will work just fine.
http://perl.find-info.ru/perl/027/learnperl4-chp-5-sect-6.html
Perhaps a Chinese translation might be of use to the OP 文海梅:
http://www.biostatistic.net/thread-4903-1-1.html
Related
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
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 read other threads enter link description herethat discuss .bat to L/unix conversions, but none has been satisfactory. I have also tried a lot of hack type approach in writing my own scripts.
I have the following example.bat script that is representative of the kind of script I want to run on unix.
Code:
echo "Example.bat"
perl script1 param.in newParam.in
perl script2 newParam.in stuff.D2D stuff.D2C
program.exe stuff.D2C
perl script3 stuff.DIS results.out
My problem is I don't know how to handle the perl and program.exe in the unix bash shell. I have tried putting them in a system(), but that did not work. Can someone please help me?
Thank you!
Provided that you have an executable file named program.exe somewhere in your $PATH (which you well might — Unix executables don't have to end in .exe, but nothing says they can't), the code you've pasted is a valid shell script. If you save it in a file named, say, example.bat, you can run it by typing
sh example.bat
into the shell prompt.
Of course, Unix shell scripts are usually given the suffix .sh — or no suffix at all — rather than .bat. Also, if you want your script to be executable directly, by typing just
example.sh
rather than sh example.sh, you need to do three things:
Start the script with a "shebang" line: a line that begins with #! and the full path to the shell interpreter you want to use to run it (e.g. /bin/sh for the basic Bourne shell), like this:
#!/bin/sh
echo "This is a shell script."
# ... more commands here ...
Mark your script as executable using the chmod command, e.g.
chmod a+rx example.sh
Put your script somewhere along your $PATH. On Unix, the default path will not normally contain the current directory ., so you can't execute programs from the current directory just by typing their name. You can, however, run them by specifying an explicit path, e.g.
./example.sh # runs example.sh from the current directory
To find out what your $PATH is, just type echo $PATH into the shell.
I am wondering that how can I Use the file command and determine if a file is a script or not.for example in usr bin I want to know which file is script or not. actually i don't want write any script just i need a command for determine that.
You can certainly trust file to find any script in the directory you specify:
file /usr/bin/* | grep script
Or, if you prefer to do it yourself and you are using bash you can do:
for f in /usr/bin/*; do r=$(head -1 $f | grep '^#! */') && echo "$f: $r"; done
which uses the shebang to determine the interpreter and thus the script entity.
This should work (assuming that you're using BASH):
for f in `ls`; do file $f|grep "executable"; done
Update- I just validated that this works for C shell scripts, BASH, Perl, and Ruby. It also ignores file permissions (meaning that even if a file doesn't have the executable bit set, it still works). This seems to be do to the file command looking for a command interpreter (bash, perl, etc…)
file can't guarantee to tell you anything about a text file, if it doesn't know how to interpret it.
You may need to do a combination of things. jschorr's answer should probably work for the stuff in /bin, but another way to test a file might be to check whether a text file is executable.
stat -c "%A" myfilename | grep x
If that returns anything, then your file has execute permissions on it. So if file gets you a description that tells you it's plain text (like "ASCII text"), and there are execute permissions on the file, then it's a pretty good bet that it's a script file.
Not perfect, but I don't think anything will be.
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.