I saw the line data=$(cat) in a bash script (just declaring an empty variable) and am mystified as to what that could possibly do.
I read the man pages, but it doesn't have an example or explanation of this. Does this capture stdin or something? Any documentation on this?
EDIT: Specifically how the heck does doing data=$(cat) allow for it to run this hook script?
#!/bin/bash
# Runs all executable pre-commit-* hooks and exits after,
# if any of them was not successful.
#
# Based on
# http://osdir.com/ml/git/2009-01/msg00308.html
data=$(cat)
exitcodes=()
hookname=`basename $0`
# Run each hook, passing through STDIN and storing the exit code.
# We don't want to bail at the first failure, as the user might
# then bypass the hooks without knowing about additional issues.
for hook in $GIT_DIR/hooks/$hookname-*; do
test -x "$hook" || continue
echo "$data" | "$hook"
exitcodes+=($?)
done
https://github.com/henrik/dotfiles/blob/master/git_template/hooks/pre-commit
cat will catenate its input to its output.
In the context of the variable capture you posted, the effect is to assign the statement's (or containing script's) standard input to the variable.
The command substitution $(command) will return the command's output; the assignment will assign the substituted string to the variable; and in the absence of a file name argument, cat will read and print standard input.
The Git hook script you found this in captures the commit data from standard input so that it can be repeatedly piped to each hook script separately. You only get one copy of standard input, so if you need it multiple times, you need to capture it somehow. (I would use a temporary file, and quote all file name variables properly; but keeping the data in a variable is certainly okay, especially if you only expect fairly small amounts of input.)
Doing:
t#t:~# temp=$(cat)
hello how
are you?
t#t:~# echo $temp
hello how are you?
(A single Controld on the line by itself following "are you?" terminates the input.)
As manual says
cat - concatenate files and print on the standard output
Also
cat Copy standard input to standard output.
here, cat will concatenate your STDIN into a single string and assign it to variable temp.
Say your bash script script.sh is:
#!/bin/bash
data=$(cat)
Then, the following commands will store the string STR in the variable data:
echo STR | bash script.sh
bash script.sh < <(echo STR)
bash script.sh <<< STR
Related
I have a requirement, where i need to pass multiple arguments to the script to trigger parallel process for each argument. Now i need to capture each process output in the separate log file.
for arg in test_{01..05} ; do bash test.sh "$arg" & done
Above piece of code can only give parallel processing for the input arguments. I tried with exec > >(tee "/path/of/log/$arg_filedate +%Y%m%d%H.log") 2>&1 and it was able to create single log file name with just date with empty output. Can someone suggest whats going wrong here or if there is any best way other than using parallel package
Try:
data_part=$(date +%Y%m%d%H)
for arg in test_{01..05} ; do bash test.sh "$arg" > "/path/to/log/${arg}_${data_part}.log" & done
If i use "$arg_date +%Y%m%d%H.log" it is creating a file with just date without arg
Yes, because $arg_ is parsed as a variable name
arg_=blabla
echo "$arg_" # will print blabla
echo "${arg_}" # equal to the above
To separate _ from arg use braces "${arg}_" would expand variable arg and add string _.
Can someone fix this for me.
It should copy a version log file to backup after moving to a repo directory
Then it automatically appends line given as input to the log file with some formatting.
That's it.
Assume existence of log file and test directory.
#!/bin/bash
cd ~/Git/test
cp versionlog.MD .versionlog.MD.old
LOGDATE="$(date --utc +%m-%d-%Y)"
read -p "MSG > " VHMSG |
VHENTRY="- **${LOGDATE}** | ${VHMSG}"
cat ${VHENTRY} >> versionlog.MD
shell output
virufac#box:~/Git/test$ ~/.logvh.sh
MSG > testing script
EOF
EOL]
EOL
e
E
CTRL + C to get out of stuck in reading lines of input
virufac#box:~/Git/test$ cat versionlog.MD
directly outputs the markdown
# Version Log
## version 0.0.1 established 01-22-2020
*Working Towards Working Mission 1 Demo in 0.1 *
- **01-22-2020** | discovered faker.Faker and deprecated old namelessgen
EOF
EOL]
EOL
e
E
I finally got it to save the damned input lines to the file instead of just echoing the command I wanted to enter on the screen and not executing it. But... why isn't it adding the lines built from the VHENTRY variable... and why doesn't it stop reading after one line sometimes and this time not. You could see I was trying to do something to tell it to stop reading the input.
After some realizing a thing I had done in the script was by accident... I tried to fix it and saw that the | at the end of the read command was seemingly the only reason the script did any of what it did save to the file in the first place.
I would have done this in python3 if I had know this script wouldn't be the simplest thing I had ever done. Now I just have to know how you do it after all the time spent on it so that I can remember never to think a shell script will save time again.
Use printf to write a string to a file. cat tries to read from a file named in the argument list. And when the argument is - it means to read from standard input until EOF. So your script is hanging because it's waiting for you to type all the input.
Don't put quotes around the path when it starts with ~, as the quotes make it a literal instead of expanding to the home directory.
Get rid of | at the end of the read line. read doesn't write anything to stdout, so there's nothing to pipe to the following command.
There isn't really any need for the VHENTRY variable, you can do that formatting in the printf argument.
#!/bin/bash
cd ~/Git/test
cp versionlog.MD .versionlog.MD.old
LOGDATE="$(date --utc +%m-%d-%Y)"
read -p "MSG > " VHMSG
printf -- '- **%s** | %s\n' "${LOGDATE}" "$VHMSG" >> versionlog.MD
I have a BASH script that has a long set of arguments and two ways of calling it:
my_script --option1 value --option2 value ... etc
or
my_script val1 val2 val3 ..... valn
This script in turn compiles and runs a large FORTRAN code suite that eventually produces a netcdf file as output. I already have all the metadata in the netcdf output global attributes, but it would be really nice to also include the full run command one used to create that experiment. Thus another user who receives the netcdf file could simply reenter the run command to rerun the experiment, without having to piece together all the options.
So that is a long way of saying, in my BASH script, how do I get the last command entered from the parent shell and put it in a variable? i.e. the script is asking "how was I called?"
I could try to piece it together from the option list, but the very long option list and two interface methods would make this long and arduous, and I am sure there is a simple way.
I found this helpful page:
BASH: echoing the last command run
but this only seems to work to get the last command executed within the script itself. The asker also refers to use of history, but the answers seem to imply that the history will only contain the command after the programme has completed.
Many thanks if any of you have any idea.
You can try the following:
myInvocation="$(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 $*?
In bash from the CLI I can do:
$ ERR_TYPE=$"OVERLOAD"
$ echo $ERR_TYPE
OVERLOAD
$ read ${ERR_TYPE}_ERROR
1234
$ echo $OVERLOAD_ERROR
1234
This works great to set my variable name dynamically; in a script it doesn't work. I tried:
#!/bin/env bash
ERR_TYPE=("${ERR_TYPE[#]}" "OVERLOAD" "PANIC" "FATAL")
for i in "${ERR_TYPE[#]}"
do
sh -c $(echo ${i}_ERROR=$"1234")
done
echo $OVERLOAD_ERROR # output is blank
# I also tried these:
# ${i}_ERROR=$(echo ${i}_ERROR=$"1234") # command not found
# read ${i}_ERROR=$(echo ${i}_ERROR=$"1234") # it never terminates
How would I set a variable as I do from CLI, but in a script? thanks
When you use dynamic variables names instead of associative arrays, you really need to question your approach.
err_type=("OVERLOAD" "PANIC" "FATAL")
declare -A error
for type in "${err_type[#]}"; do
error[$type]=1234
done
Nevertheless, in bash you'd use declare:
declare "${i}_error=1234"
Your approach fails because you spawn a new shell, passing the command OVERLOAD_ERROR=1234, and then the shell exits. Your current shell is not affected at all.
Get out of the habit of using ALLCAPSVARNAMES. One day you'll write PATH=... and then wonder why your script is broken.
If the variable will hold a number, you can use let.
#!/bin/bash
ERR_TYPE=("OVERLOAD" "PANIC" "FATAL")
j=0
for i in "${ERR_TYPE[#]}"
do
let ${i}_ERROR=1000+j++
done
echo $OVERLOAD_ERROR
echo $PANIC_ERROR
echo $FATAL_ERROR
This outputs:
1000
1001
1002
I'd use eval.
I think this would be considered bad practice though (it had some thing to do with the fact that eval is "evil" because it allows bad input or something):
eval "${i}_ERROR=1234"
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.