I've seen this line in many shell scripts but I don't understand the effect it has. Could someone explain please?
tempfile=`tempfile 2>/dev/null` || tempfile=/tmp/test$$
It creates a temporary file and puts the path to it in the $tempfile variable.
`tempfile 2>/dev/null`
runs the tempfile command (man tempfile) and discards any error messages. If it succeeds, it returns the name of the newly created temporary file. If it fails, it returns non-zero, in which case the next part of the command runs.
For a command this || that, that only runs if this fails, i.e. returns non-zero.
$$ is a variable in bash that expands to the process ID of the shell. (Compare the results of ps and echo $$.) So tempfile=/tmp/test$$ will expand to something like tempfile=/tmp/test2278.
Presumably, later in the script, something writes to $tempfile.
The shell has a separate namespace for command and variables (making it a Lisp-2, LOL) which is exploited in your script line. tempfile is a command which is run to compute the value of the tempfile variable which is unrelated to it in any way. tempfile produces a pathname suitable for use as the name of a temporary file. 2> /dev/null redirects any error message from tempfile into /dev/null (2 is the standard error file descriptor). The command1 || command2 logic means, "execute command2 if command1 fails". If we can't get a temporary name from tempfile, then we use /tmp/test$$, where $$ is a special built-in shell parameter which expands to the shell's own process ID.
tempfile creates a temporay file with a file name similar to /tmp/tmp.XXXXXX
2>/dev/null redirects the command output to the /dev/null device, which just throws it away. This redirection just ignore any errors on creating a temporary file.
|| chains two commands together. If the first fails, the second is executed. If the first succeeds nothing else happens.
$$ is the pid of the current shell, which means that if the tempfile command fails the tempfile variable will still contain a string in the form /tmp/test6052 if the process' pid is 6052.
The first part of the line, up to the ||, runs the program tempfile and captures standard output in the variable tempfile, throwing errors away. There's an exit status, too: either zero for success or non-zero for failure (either failure to execute the tempfile command or failure reported by the tempfile command when it is run).
The || means "if the LHS (left-hand side) failed then do the RHS (right-hand side)".
So, if the tempfile command had a problem, the RHS will be used, assigning a simpler temporary file name to tempfile (the variable).
Overall, it is equivalent to:
if tempfile=`tempfile 2>/dev/null`
then : OK
else tempfile=/tmp/test$$
fi
Only it is on one line, not four.
The idea is, I'm sure, to get something in $tempfile whether or not the tempfile command exists on the machine.
Did you look at man tempfile?
That line is trying to use tempfile(1) to generate a temporary filename, storing it in $tempfile. If that fails (the "||", "or" part), it falls back to an explicit filename of /tmp/test$$, where $$ is the PID of the executing script.
Related
This question already has answers here:
Read user input inside a loop
(6 answers)
Closed 5 years ago.
First post here! I really need help on this one, I looked the issue on google, but can't manage to find an useful answer for me. So here's the problem.
I'm having fun coding some like of a framework in bash. Everyone can create their own module and add it to the framework. BUT. To know what arguments the script require, I created an "args.conf" file that must be in every module, that kinda looks like this:
LHOST;true;The IP the remote payload will connect to.
LPORT;true;The port the remote payload will connect to.
The first column is the argument name, the second defines if it's required or not, the third is the description. Anyway, long story short, the framework is supposed to read the args.conf file line by line to ask the user a value for every argument. Here's the piece of code:
info "Reading module $name argument list..."
while read line; do
echo $line > line.tmp
arg=`cut -d ";" -f 1 line.tmp`
requ=`cut -d ";" -f 2 line.tmp`
if [ $requ = "true" ]; then
echo "[This argument is required]"
else
echo "[This argument isn't required, leave a blank space if you don't wan't to use it]"
fi
read -p " $arg=" answer
echo $answer >> arglist.tmp
done < modules/$name/args.conf
tr '\n' ' ' < arglist.tmp > argline.tmp
argline=`cat argline.tmp`
info "Launching module $name..."
cd modules/$name
$interpreter $file $argline
cd ../..
rm arglist.tmp
rm argline.tmp
rm line.tmp
succes "Module $name execution completed."
As you can see, it's supposed to ask the user a value for every argument... But:
1) The read command seems to not be executing. It just skips it, and the argument has no value
2) Despite the fact that the args.conf file contains 3 lines, the loops seems to be executing just a single time. All I see on the screen is "[This argument is required]" just one time, and the module justs launch (and crashes because it has not the required arguments...).
Really don't know what to do, here... I hope someone here have an answer ^^'.
Thanks in advance!
(and sorry for eventual mistakes, I'm french)
Alpha.
As #that other guy pointed out in a comment, the problem is that all of the read commands in the loop are reading from the args.conf file, not the user. The way I'd handle this is by redirecting the conf file over a different file descriptor than stdin (fd #0); I like to use fd #3 for this:
while read -u3 line; do
...
done 3< modules/$name/args.conf
(Note: if your shell's read command doesn't understand the -u option, use read line <&3 instead.)
There are a number of other things in this script I'd recommend against:
Variable references without double-quotes around them, e.g. echo $line instead of echo "$line", and < modules/$name/args.conf instead of < "modules/$name/args.conf". Unquoted variable references get split into words (if they contain whitespace) and any wildcards that happen to match filenames will get replaced by a list of matching files. This can cause really weird and intermittent bugs. Unfortunately, your use of $argline depends on word splitting to separate multiple arguments; if you're using bash (not a generic POSIX shell) you can use arrays instead; I'll get to that.
You're using relative file paths everywhere, and cding in the script. This tends to be fragile and confusing, since file paths are different at different places in the script, and any relative paths passed in by the user will become invalid the first time the script cds somewhere else. Worse, you aren't checking for errors when you cd, so if any cd fails for any reason, then entire rest of the script will run in the wrong place and fail bizarrely. You'd be far better off figuring out where your system's root directory is (as an absolute path), then referencing everything from it (e.g. < "$module_root/modules/$name/args.conf").
Actually, you're not checking for errors anywhere. It's generally a good idea, when writing any sort of program, to try to think of what can go wrong and how your program should respond (and also to expect that things you didn't think of will also go wrong). Some people like to use set -e to make their scripts exit if any simple command fails, but this doesn't always do what you'd expect. I prefer to explicitly test the exit status of the commands in my script, with something like:
command1 || {
echo 'command1 failed!' >&2
exit 1
}
if command2; then
echo 'command2 succeeded!' >&2
else
echo 'command2 failed!' >&2
exit 1
fi
You're creating temp files in the current directory, which risks random conflicts (with other runs of the script at the same time, any files that happen to have names you're using, etc). It's better to create a temp directory at the beginning, then store everything in it (again, by absolute path):
module_tmp="$(mktemp -dt module-system)" || {
echo "Error creating temp directory" >&2
exit 1
}
...
echo "$answer" >> "$module_tmp/arglist.tmp"
(BTW, note that I'm using $() instead of backticks. They're easier to read, and don't have some subtle syntactic oddities that backticks have. I recommend switching.)
Speaking of which, you're overusing temp files; a lot of what you're doing with can be done just fine with shell variables and built-in shell features. For example, rather than reading line from the config file, then storing them in a temp file and using cut to split them into fields, you can simply echo to cut:
arg="$(echo "$line" | cut -d ";" -f 1)"
...or better yet, use read's built-in ability to split fields based on whatever IFS is set to:
while IFS=";" read -u3 arg requ description; do
(Note that since the assignment to IFS is a prefix to the read command, it only affects that one command; changing IFS globally can have weird effects, and should be avoided whenever possible.)
Similarly, storing the argument list in a file, converting newlines to spaces into another file, then reading that file... you can skip any or all of these steps. If you're using bash, store the arg list in an array:
arglist=()
while ...
arglist+=("$answer") # or ("#arg=$answer")? Not sure of your syntax.
done ...
"$module_root/modules/$name/$interpreter" "$file" "${arglist[#]}"
(That messy syntax, with the double-quotes, curly braces, square brackets, and at-sign, is the generally correct way to expand an array in bash).
If you can't count on bash extensions like arrays, you can at least do it the old messy way with a plain variable:
arglist=""
while ...
arglist="$arglist $answer" # or "$arglist $arg=$answer"? Not sure of your syntax.
done ...
"$module_root/modules/$name/$interpreter" "$file" $arglist
... but this runs the risk of arguments being word-split and/or expanded to lists of files.
I am trying to access a string returned by a shell script which was called from a parent shell script. Something like this:
ex.sh:
echo "Hemanth"
ex2.sh:
sh ex.sh
if [ $? == "Hemanth" ]; then
echo "Hurray!!"
else
echo "Sorry Bro!"
fi
Is there a way to do this? Any help would be appreciated.
Thank you.
Use a command substitution syntax on ex2.sh
valueFromOtherScript="$(sh ex.sh)"
printf "%s\n" "$valueFromOtherScript"
echo by default outputs a new-line character after the string passed, if you don't need it in the above variable use printf as
printf "Hemanth"
on first script. Also worth adding $? will return only the exit code of the last executed command. Its values are interpreted as 0 being a successful run and a non-zero on failure. It will NEVER have a string value as you tried to use.
A Bash script does not really "return" a string. What you want to do is capture the output of a script (or external program, or function, they all act the same in this respect).
Command substitution is a common way to capture output.
captured_output="$(sh ex.sh)"
This initializes variable captured_output with the string containing all that is output by ex.sh. Well, not exactly all. Any script (or command, or function) actually has two output channels, usually called "standard out" (file descriptor number 1) and "standard error" (file descriptor number 2). When executing from a terminal, both typically end up on the screen. But they can be handled separately if needed.
For instance, if you want to capture really all output (including error messages), you would add a "redirection" after your command that tells the shell you want standard error to go to the same place as standard out.
captured_output="$(sh ex.sh 2>&1)"
If you omit that redirection, and the script outputs something on standard error, then this will still show on screen, and will not be captured.
Another way to capture output is sending it to a file, and then read back that file to a variable, like this :
sh ex.sh > output_file.log
captured_output="$(<output_file.log)"
A script (or external program, or function) does have something called a return code, which is an integer. By convention, a value of 0 means "success", and any other value indicates abnormal execution (but not necessarily failure) : the meaning of that return code is not standardized, it is ultimately specific to each script, program or function.
This return code is available in the $? special shell variable immediately after the execution terminates.
sh ex.sh
return_code=$?
echo "Return code is $return_code"
I have a BASH script that has a long set of arguments and two ways of calling it:
my_script --option1 value --option2 value ... etc
or
my_script val1 val2 val3 ..... valn
This script in turn compiles and runs a large FORTRAN code suite that eventually produces a netcdf file as output. I already have all the metadata in the netcdf output global attributes, but it would be really nice to also include the full run command one used to create that experiment. Thus another user who receives the netcdf file could simply reenter the run command to rerun the experiment, without having to piece together all the options.
So that is a long way of saying, in my BASH script, how do I get the last command entered from the parent shell and put it in a variable? i.e. the script is asking "how was I called?"
I could try to piece it together from the option list, but the very long option list and two interface methods would make this long and arduous, and I am sure there is a simple way.
I found this helpful page:
BASH: echoing the last command run
but this only seems to work to get the last command executed within the script itself. The asker also refers to use of history, but the answers seem to imply that the history will only contain the command after the programme has completed.
Many thanks if any of you have any idea.
You can try the following:
myInvocation="$(printf %q "$BASH_SOURCE")$((($#)) && printf ' %q' "$#")"
$BASH_SOURCE refers to the running script (as invoked), and $# is the array of arguments; (($#)) && ensures that the following printf command is only executed if at least 1 argument was passed; printf %q is explained below.
While this won't always be a verbatim copy of your command line, it'll be equivalent - the string you get is reusable as a shell command.
chepner points out in a comment that this approach will only capture what the original arguments were ultimately expanded to:
For instance, if the original command was my_script $USER "$(date +%s)", $myInvocation will not reflect these arguments as-is, but will rather contain what the shell expanded them to; e.g., my_script jdoe 1460644812
chepner also points that out that getting the actual raw command line as received by the parent process will be (next to) impossible. Do tell me if you know of a way.
However, if you're prepared to ask users to do extra work when invoking your script or you can get them to invoke your script through an alias you define - which is obviously tricky - there is a solution; see bottom.
Note that use of printf %q is crucial to preserving the boundaries between arguments - if your original arguments had embedded spaces, something like $0 $* would result in a different command.
printf %q also protects against other shell metacharacters (e.g., |) embedded in arguments.
printf %q quotes the given argument for reuse as a single argument in a shell command, applying the necessary quoting; e.g.:
$ printf %q 'a |b'
a\ \|b
a\ \|b is equivalent to single-quoted string 'a |b' from the shell's perspective, but this example shows how the resulting representation is not necessarily the same as the input representation.
Incidentally, ksh and zsh also support printf %q, and ksh actually outputs 'a |b' in this case.
If you're prepared to modify how your script is invoked, you can pass $BASH_COMMANDas an extra argument: $BASH_COMMAND contains the raw[1]
command line of the currently executing command.
For simplicity of processing inside the script, pass it as the first argument (note that the double quotes are required to preserve the value as a single argument):
my_script "$BASH_COMMAND" --option1 value --option2
Inside your script:
# The *first* argument is what "$BASH_COMMAND" expanded to,
# i.e., the entire (alias-expanded) command line.
myInvocation=$1 # Save the command line in a variable...
shift # ... and remove it from "$#".
# Now process "$#", as you normally would.
Unfortunately, there are only two options when it comes to ensuring that your script is invoked this way, and they're both suboptimal:
The end user has to invoke the script this way - which is obviously tricky and fragile (you could however, check in your script whether the first argument contains the script name and error out, if not).
Alternatively, provide an alias that wraps the passing of $BASH_COMMAND as follows:
alias my_script='/path/to/my_script "$BASH_COMMAND"'
The tricky part is that this alias must be defined in all end users' shell initialization files to ensure that it's available.
Also, inside your script, you'd have to do extra work to re-transform the alias-expanded version of the command line into its aliased form:
# The *first* argument is what "$BASH_COMMAND" expanded to,
# i.e., the entire (alias-expanded) command line.
# Here we also re-transform the alias-expanded command line to
# its original aliased form, by replacing everything up to and including
# "$BASH_COMMMAND" with the alias name.
myInvocation=$(sed 's/^.* "\$BASH_COMMAND"/my_script/' <<<"$1")
shift # Remove the first argument from "$#".
# Now process "$#", as you normally would.
Sadly, wrapping the invocation via a script or function is not an option, because the $BASH_COMMAND truly only ever reports the current command's command line, which in the case of a script or function wrapper would be the line inside that wrapper.
[1] The only thing that gets expanded are aliases, so if you invoked your script via an alias, you'll still see the underlying script in $BASH_COMMAND, but that's generally desirable, given that aliases are user-specific.
All other arguments and even input/output redirections, including process substitutiions <(...) are reflected as-is.
"$0" contains the script's name, "$#" contains the parameters.
Do you mean something like echo $0 $*?
My Aim -->
Files Listing from a command has to be read line by line and be used as part of another command.
Description -->
A command in linux returns
archive/Crow.java
archive/Kaka.java
mypmdhook.sh
which is stored in changed_files variable. I use the following while loop to read the files line by line and use it as part of a pmd command
while read each_file
do
echo "Inside Loop -- $each_file"
done<$changed_files
I am new to writing shell script but my assumption was that the lines would've been separated in the loop and printed in each iteration but instead I get the following error --
mypmdhook.sh: 7: mypmdhook.sh: cannot open archive/Crow.java
archive/Kaka.java
mypmdhook.sh: No such file
Can you tell me how I can just get the value as a string and not as a file what is opened. By the way, the file does exist which made me feel even more confused.(and later use it inside a command). I'd be happy with any kind of answer that helps me understand and resolve this issue.
Since you have data stored in a variable, use a "here string" instead of file redirection:
changed_files="archive/Crow.java
archive/Kaka.java
mypmdhook.sh"
while read each_file
do
echo "Inside Loop -- $each_file"
done <<< "$changed_files"
Inside Loop -- archive/Crow.java
Inside Loop -- archive/Kaka.java
Inside Loop -- mypmdhook.sh
Extremely important to quote "$changed_files" in order to preserve the newlines, so the while-read loop works as you expect. A rule of thumb: always quote variables, unless you knows exactly why you want to leave the quotes off.
What happens here is that the value of your variable $changed_files is substituted into your command, and you get something like
while read each_file
do
echo "Inside Loop -- $each_file"
done < archive/Crow.java
archive/Kaka.java
mypmdhook.sh
then the shell tries to open the file for redirecting the input and obviously fails.
The point is that redirections (e.g. <, >, >>) in most cases accept filenames, but what you really need is to give the contents of the variable to the stdin. The most obvious way to do that is
echo $changed_files | while read each_file; do echo "Inside Loop -- $each_file"; done
You can also use the for loop instead of while read:
for each_file in $changed_files; do echo "inside Loop -- $each_file"; done
I prefer using while read ... if there is a chance that some filename may contain spaces, but in most cases for ... in will work for you.
Rather than storing command's output in a variable use while loop like this:
mycommand | while read -r each_file; do echo "Inside Loop -- $each_file"; done
If you're using BASH you can use process substitution:
while read -r each_file; do echo "Inside Loop -- $each_file"; done < <(mycommand)
btw your attempt of done<$changed_files will assume that changed_files represents a file.
Today I tried a script in linux to get all files in one dir. It was pretty straightforward, but I found something interesting.
#!/bin/bash
InputDir=/home/XXX/
for file in $InputDir'*'
do
echo $file
done
The output is:
/home/XXX/fileA /home/XXX/fileB
But when I just input the dir directly, like:
#!/bin/bash
InputDir=/home/XXX/
for file in /home/XXX/*
do
echo $file
done
The output is:
/home/XXX/fileA
/home/XXX/fileB
It seems, in the first script, there was only one loop and all the file names were stored in the variable $file in the FIRST loop, separated by space. But in the second script, one file name was stored in $file just in one loop, and there were more than one loop. What is exactly the difference between these two scripts?
Thanks very much, maybe my question is a little bit naive..
The behavior is correct and "as expected".
for file in $InputDir'*' means assign "/home/XXX/*" to $file (note the quotes). Since you quoted the asterisk, it will not be executed at this time. When the shell sees echo $file, it first expands the variables and then it does glob expansion. So after the first step, it sees
echo /home/XXX/*
and after glob expansion, it sees:
echo /home/XXX/fileA /home/XXX/fileB
Only now, it will execute the command.
In the second case, the pattern /home/XXX/* is expanded before the for is executed and thus, each file in the directory is assigned to file and then the body of the loop is executed.
This will work:
for file in "$InputDir"*
but it's brittle; it will fail, for example, when you forget to add a / to the end of the variable $InputDir.
for file in "$InputDir"/*
is a little bit better (Unix will ignore double slashes in a path) but it can cause trouble when $InputDir is not set or empty: You'll suddenly list files in the / (root) folder. This can happen, for example, because of a typo:
inputDir=...
for file in "$InputDir"/*
Case matters on Unix :-)
To help you understand code like this, use set -x ("enable tracing") in a line before the code you want to debug.
The difference is the quoting of '*'. In the first case the loop only executes once, with $file equal to /home/XXX/* which then expands to all the files in the directory when passed to echo. In the second case it executes once per file, with $file equal to each file name in turn.
Bottom line - change:
for file in $InputDir'*'
to:
for file in $InputDir*
or, better, and to make it more readable - change:
InputDir=/home/XXX/
for file in $InputDir'*'
to:
InputDir=/home/XXX
for file in $InputDir/*