how to remove comments from a bash script - bash

I'm trying to make a script that is getting a script file as a param. It should remove comments from the file and pipe it to another script. (with no temp file if possible)
at the beginning I was thinkig of doing this
cut -d"#" -f1 $1 | ./script_name
but it also clears a part of lines which aren't comments, because there are a few commands which uses # in them (counting string chars for example).
is there a way of doing it without a temp file?

You can use inline sed with better regex:
sed -i.bak '/^[[:blank:]]*#/d "$1"
^[[:blank:]]*# will match # only if is preceded by optional spaces at each line
-i.bak option will inline edit the input file with .bak as the extension of the backup file in case something goes wrong.

Here's one very bash-specific way of stripping comments from a script file. It also strips the she-bang line, if there was one (after all, it's a comment), and does some reformatting:
tmp_="() {
$(<script_name)
}" bash -c 'declare -f tmp_' | tail -n+2
This converts the script into a function, and uses the bash built-in declare to pretty-print the resulting function (the tail removes the function name, but not the surrounding braces; a more complicated post-process could remove them, too, if that were judged necessary).
The pretty-printing is done in a child bash process both to avoid polluting the execution environment with the temporary function and because the subprocess will effectively recognize the string value of the variable as a function.
Update:
Sadly, post shellshock the above no longer works. However, for patched bashes, the following probably does:
env "BASH_FUNC_tmp_%%=() {
$(<script_name)
}" bash -c 'declare -f tmp_' | tail -n+2
Also, note that this method does not strip comments which are internal to command or process substitution.

Yes, in general this type of problem can be solved without a temporary file.
This however will also depend on the complexity of the parsing required to determine when the comment delimiter character doesn't in fact introduce a comment.

Using python3 and install pygments
from pygments.lexers.shell import BashLexer
from pygments.token import Token, is_token_subtype
def delete_comments(fname):
src = open(fname, "r").read()
dst = open(fname, "w")
for token in BashLexer().get_tokens(src):
if not (is_token_subtype(token[0], Token.Comment)):
dst.write(token[1])
if token[0] == Token.Comment.Hashbang:
dst.write(token[1])
if __name__ == "__main__":
delete_comments("/path/to/your/shellScript.sh")

Related

How to remove duplicate with bash script command xargs when the string has some quotes ""?

I am a newbie in bash script.
Here is my environment:
Mac OS X Catalina
/bin/bash
I found here a mix of several commands to remove the duplicate string in a string.
I needed for my program which updates the .zhrc profile file.
Here is my code:
#!/bin/bash
a='export PATH="/Library/Frameworks/Python.framework/Versions/3.8/bin:/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin:/opt/local/bin:"'
myvariable=$(echo "$a" | tr ':' '\n' | sort | uniq | xargs)
echo "myvariable : $myvariable"
Here is the output:
xargs: unterminated quote
myvariable :
After some test, I know that the source of the issue is due to some quotes "" inside my variable '$a'.
Why am I so sure?
Because when I execute this code for example:
#!/bin/bash
a="/Library/Java/JavaVirtualMachines/jdk1.8.0_271.jdk/Contents/Home:/Library/Java/JavaVirtualMachines/jdk1.8.0_271.jdk/Contents/Home"
myvariable=$(echo "$a" | tr ':' '\n' | sort | uniq | xargs)
echo "myvariable : $myvariable"
where $a doesn't contain any quotes, I get the correct output:
myvariable : /Library/Java/JavaVirtualMachines/jdk1.8.0_271.jdk/Contents/Home
I tried to search for a solution for "xargs: unterminated quote" but each answer found on the web is for a particular case which doesn't correspond to my problem.
As I am a newbie and this line command is using several complex commands, I was wondering if anyone know the magic trick to make it work.
Basically, you want to remove duplicates from a colon-separated list.
I don't know if this is considered cheating, but I would do this in another language and invoke it from bash. First I would write a script for this purpose in zsh: It accepts as parameter a string with colon separtors and outputs a colon-separated list with duplicates removed:
#!/bin/zsh
original=${1?Parameter missing} # Original string
# Auxiliary array, which is set up to act like a Set, i.e. without
# duplicates
typeset -aU nodups_array
# Split the original strings on the colons and store the pieces
# into the array, thereby removing duplicates. The core idea for
# this is stolen from:
# https://stackoverflow.com/questions/2930238/split-string-with-zsh-as-in-python
nodups_array=("${(#s/:/)original}")
# Join the array back with colons and write the resulting string
# to stdout.
echo ${(j':')nodups_array}
If we call this script nodups_string, you can invoke it in your bash-setting as:
#!/bin/bash
a_path="/Library/Frameworks/Python.framework/Versions/3.8/bin:/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin:/opt/local/bin:"
nodups_a_path=$(nodups_string "$a_path")
my_variable="export PATH=$nodups_a_path"
echo "myvariable : $myvariable"
The overall effect would be literally what you asked for. However, there is still an open problem I should point out: If one of the PATH components happens to contain a space, the resulting export statement can not validly be executed. This problem is also inherent into your original problem; you just didn't mention it. You could do something like
my_variable=export\ PATH='"'$nodups_a_path"'"'
to avoid this. Of course, I wonder why you take such an effort to generat a syntactically valid export command, instead of simply building the PATH by directly where it is needed.
Side note: If you would use zsh as your shell instead of bash, and only want to keep your PATH free of duplicates, a simple
typeset -iU path
would suffice, and zsh takes care of the rest.
With awk:
awk -v RS=[:\"] 'NR > 1 { pth[$0]="" } END { for (i in pth) { if (i !~ /[[:space:]]+/ && i != "" ) { printf "%s:",i } } }' <<< "$a"
Set the record separator to : and double quotes. Then when the number record is greater than one, set up an array called pth with the path as the index. At the end, loop through the array, re printing the paths separated with :

Linux script text substitutions

I want to make a few configuration files (for homeassistant) that are very similar to each other. I am aiming to use a template file as the base and put in a few substitution strings at the top of the file and use a bash script to read the substitutions and run sed with the applicable strings.
i.e.
# substitutions
# room = living_room
# switch = hallway_motion
# delay = 3
automations:
foo......
.........
entity_id: $switch
When I run the script it will look for any line beginning with a # that has a word (key) and then an = and another word (maybe string) (value) and replace anywhere that key with a $ in front is in the rest of the file.
Like what is done by esphome. https://esphome.io/guides/configuration-types.html#substitutions
I am getting stuck at finding the "keys" in the file. How can I script this so it can find all the "keys" recursively?
Or is there something that does this, or something similar, out there already?
You can do this with sed in two stages. The first stage will generate a second stage sed script to fill in your template. I'd make a small adjustment to your syntax and recommend that you require curly braces around your variable name. In other words, write your variable expansions like this:
# foo = bar
myentry: ${foo}
This makes it easier to avoid pitfalls when you have one variable name that's a prefix of another (e.g., foo and foobar).
#!/bin/bash
in="$1"
stage2=$(mktemp)
trap 'rm -f "$stage2"' EXIT
sed -n -e 's,^# \([[:alnum:]_]\+\) = \([^#]\+\),s#\${\1}#\2#g,p' "$in" > "$stage2"
sed -f "$stage2" "$in"
Provide a filename as the first argument, and it will print the filled out template on stdout.
This example code is pretty strict about white space on variable definition lines, but that can obviously be adjusted to your liking.

insert text allocated in a variable before the first empty line

I have some text files $f resembling the following
function
%blah
%blah
%blah
code here
I want to append the following text before the first empty line:
%
%This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike
%3.0 Unported License. See notes at the end of this file for more information.
I tried the following:
top=$(cat ./PATH/text.txt)
top="${top//$'\n'/\\n}"
sed -i.bak 's#^$#'"$top"'\\n#' $f
where the second line (I think) preserves the new line in the text and the third line (I think) substitutes the first empty line with the text plus a new empty line.
Two problems:
1- My code appends the following text:
%n%This work is licensed under the Creative Commons
Attribution-NonCommercial-ShareAlike n%3.0 Unported License. See notes
at the end of this file for more information.\n
2- It appends it at end of the file.
Can someone please help me understand the problems with my code?
If you are using GNU sed, following would work.
Use ^$ to find the empty line and then use sed to replace/put the text that you want.
# Define your replacement text in a variable
a="%\n%This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike\n%3.0 Unported License. See notes at the end of this file for more information."
Note, $a should include those \n that will be directly interpreted by sed as newlines.
$ sed "0,/^$/s//$a/" inputfile.txt
In the above syntax, 0 represents the first occurrence.
Output:
function
%blah
%blah
%
%This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike
%3.0 Unported License. See notes at the end of this file for more information.
%blah
code here
test
You've included bash and sed tags in your question. Since I can't seem to come up with a way of doing this in sed, here's a bash-only solution. It's likely to perform the worst of all working solutions you might find.
The following works with your sample input:
$ while read -r x; do [[ -z "$x" ]] && cat boilerplate; printf '%s\n' "$x"; done < src
This will however insert the boilerplate before EVERY blank line, which is probably not what you're after. Instead, we should probably make this more than a one-liner:
#!/usr/bin/env bash
y=true
while read -r x; do
if [[ -z "$x" ]] && $y; then
cat boilerplate
y=false
fi
printf '%s\n' "$x"
done < src
Note that unlike the code in your question, this doesn't store your boilerplate in a variable, it just cats it "at the right time".
Note that this sends the combined output to stdout. If your goal is to modify the original file, you'll need to wrap this in something that moves around temporary files. (Note that sed's -i option also doesn't really edit files in place, it only hides the moving-around-temp-files from you.)
The following alternatives are probably a better idea.
A similar solution to the bash one might be achieved with better performance using awk:
awk 'NR==FNR{b=b $0 ORS;next} /^$/&&!y{printf "%s",b;y++} 1' boilerplate src
This awk solution obviously reads your boilerplate into a variable, though it's not a shell variable.
Notwithstanding non-standard platform-specific extensions, awk does not have any facility for editing files "in place" either. A portable solution using awk would still need to push temp files around.
And of course, the following old standard of ed is great to keep in your back pocket:
printf 'H\n/^$/\n-\n.r boilerplate\nw\nq\n' | ed src
In bash, of course, you could always use heretext, which might be clearer:
$ ed src <<< $'H\n/^$/\n-\n.r boilerplate\nw\nq\n'
The ed command is non-stream version of sed. Or rather, sed is the stream version of ed, which has been around since before the dinosaurs and is still going strong.
The commands we're using are separated by newlines and fed to ed's standard input. You can discard stdout if you feel the urge. The commands shown here are:
H - instruct ed to print more useful errors, if it gets any.
/^$/ - search for the first occurrence of a newline.
- - GO BACK ONE LINE. Awesome, right?
.r boilerplate - Read your boilerplate at the current line,
w - and write the file.
q - Quit.
Note that this does not keep a .bak file. You'll need to do that yourself if you really want one.
And if, as you suggested in comments, the filename you're reading is to be constructed from a variable, note that variable expansion does not happen inside format quoting ($' .. '). You can either switch quoting mechanisms mid-script:
ed "$file" <<< $'H\n/^$/\n-\n.r ./TATTOO_'"$currn"$'/top.txt\nw\nq\n'
Or you could put ed script in a variable constructed by printf
printf -v scr 'H\n/^$/\n-\n.r ./TATTOO_%s/top.txt\nw\nq\n' "$currn"
ed "$file" <<< "$scr"`
Adding the text to a variable so you can interpolate the variable is wasteful and an unnecessary complication. sed can easily read the contents of a file by itself.
sed -i.bak '1r./PATH/text.txt' "$f"
Unfortunately, this part of sed is poorly standardized, so you may have to experiment a little bit. Some dialects require a newline (perhaps, or perhaps not, preceded by a backslash) before the filename.
sed -i.bak '1r\
./PATH/text.txt' "$f"
(Notice also the double quotes around the file name. You generally always want double quotes around variables which contain file names. More here.)
Adapting the recipe from here we can extend this to apply to the first empty line instead of the first line.
sed -i.bak -e '/^$/!b' -e 'r./PATH/text.txt' -e :a -e '$!{' -e n -e ba -e } "$f"
This adds the boilerplate after the first empty line but perhaps that's acceptable. Refactoring it to replace it or add an empty line after should not be too challenging anyway. (Maybe use sed -n and instead explicitly print everything except the empty line.)
In brief terms, this skips to the end (simply prints) up until we find the first empty line. Then, we read and print the file, and go into a loop which prints the remainder of the file without returning to the beginning of the script.
sed that I think works. Uses files for the extra bit to be inserted.
b='##\n## comment piece\n##'
sed --posix -ne '
1,/^$/ {
/^$/ {
x;
/^true$/ !{
x
s/^$/true/
i\
'"$b"'
};
x;
s/^.*$//
}
}
p
' file1
with the examples using ranges of 1,/^$/, an empty first line would result in the disclaimer being printed twice. To avoid this, I've set it up to put a flag in the hold space ( x; s/^$/true/ ) that I can swap to the pattern space to check whether its the first blank. Once theres a match for blank line, i\ inserts the comment ($b) in front of the pattern space.
Thanks to ghoti for the initial plan.

How do I use `sed` to alter a variable in a bash script?

I'm trying to use enscript to print PDFs from Mutt, and hitting character encoding issues. One way around them seems to be to just use sed to replace the problem characters: sed -ir 's/[“”]/"/g' {input}
My test input file is this:
“very dirty”
we’re
I'm hoping to get "very dirty" and we're but instead I'm still getting
â\200\234very dirtyâ\200\235
weâ\200\231re
I found a nice little post on printing to PDFs from Mutt that I used as a starting point. I have a bash script that I point to from my .muttrc with set print_command="$HOME/.mutt/print.sh" -- the script currently reads about like this:
#!/bin/bash
input="$1" pdir="$HOME/Desktop" open_pdf=evince
# Straighten out curly quotes
sed -ir 's/[“”]/"/g' $input
sed -ir "s/[’]/'/g" $input
tmpfile="`mktemp $pdir/mutt_XXXXXXXX.pdf`"
enscript --font=Courier8 $input -2r --word-wrap --fancy-header=mutt -p - 2>/dev/null | ps2pdf - $tmpfile
$open_pdf $tmpfile >/dev/null 2>&1 &
sleep 1
rm $tmpfile
It does a fine job of creating a PDF (and works fine if you give it a file as an argument) but I can't figure out how to fix the curly quotes.
I've tried a bunch of variations on the sed line:
input=sed -r 's/[“”]/"/g' $input
$input=sed -ir "s/[’]/'/g" $input
Per the suggestion at Can I use sed to manipulate a variable in bash? I also tried input=$(sed -r 's/[“”]/"/g' <<< $input) and I get an error: "Syntax error: redirection unexpected"
But none manages to actually change $input -- what is the correct syntax to change $input with sed?
Note: I accepted an answer that resolved the question I asked, but as you can see from the comments there are a couple of other issues here. enscript is taking in a whole file as a variable, not just the text of the file. So trying to tweak the text inside the file is going to take a few extra steps. I'm still learning.
On Editing Variables In General
BashFAQ #21 is a comprehensive reference on performing search-and-replace operations in bash, including within variables, and is thus recommended reading. On this particular case:
Use the shell's native string manipulation instead; this is far higher performance than forking off a subshell, launching an external process inside it, and reading that external process's output. BashFAQ #100 covers this topic in detail, and is well worth reading.
Depending on your version of bash and configured locale, it might be possible to use a bracket expression (ie. [“”], as your original code did). However, the most portable thing is to treat “ and ” separately, which will work even without multi-byte character support available.
input='“hello ’cruel’ world”'
input=${input//'“'/'"'}
input=${input//'”'/'"'}
input=${input//'’'/"'"}
printf '%s\n' "$input"
...correctly outputs:
"hello 'cruel' world"
On Using sed
To provide a literal answer -- you almost had a working sed-based approach in your question.
input=$(sed -r 's/[“”]/"/g' <<<"$input")
...adds the missing syntactic double quotes around the parameter expansion of $input, ensuring that it's treated as a single token regardless of how it might be string-split or glob-expanded.
But All That May Not Help...
The below is mentioned because your test script is manipulating content passed on the command line; if that's not the case in production, you can probably disregard the below.
If your script is invoked as ./yourscript “hello * ’cruel’ * world”, then information about exactly what the user entered is lost before the script is started, and nothing you can do here will fix that.
This is because $1, in that scenario, will only contain “hello; ’cruel’ and world” are in their own argv locations, and the *s will have been replaced with lists of files in the current directory (each such file substituted as a separate argument) before the script was even started. Because the shell responsible for parsing the user's command line (which is not the same shell running your script!) did not recognize the quotes as valid at the time when it ran this parsing, by the time the script is running, there's nothing you can do to recover the original data.
Abstract: The way to use sed to change a variable is explored, but what you really need is a way to use and edit a file. It is covered ahead.
Sed
The (two) sed line(s) could be solved with this (note that -i is not used, it is not a file but a value):
input='“very dirty”
we’re'
sed 's/[“”]/\"/g;s/’/'\''/g' <<<"$input"
But it should be faster (for small strings) to use the internals of the shell:
input='“very dirty”
we’re'
input=${input//[“”]/\"}
input=${input//[’]/\'}
printf '%s\n' "$input"
$1
But there is an underlying problem with your script, you are trying to clean an input received from the command line. You are using $1 as the source of the string. Once somebody writes:
./script “very dirty”
we’re
That input is lost. It is broken into shell's tokens and "$1" will be “very only.
But I do not believe that is what you really have.
file
However, you are also saying that the input comes from a file. If that is the case, then read it in with:
input="$(<infile)" # not $1
sed 's/[“”]/\"/g;s/’/'\''/g' <<<"$input"
Or, if you don't mind to edit (change) the file, do this instead:
sed -i 's/[“”]/\"/g;s/’/'\''/g' infile
input="$(<infile)"
Or, if you are clear and certain that what is being given to the script is a filename, like:
./script infile
You can use:
infile="$1"
sed -i 's/[“”]/\"/g;s/’/'\''/g' "$infile"
input="$(<"$infile")"
Other comments:
Then:
Quote your variables.
Do not use the very old `…` syntax, use $(…) instead.
Do not use variables in UPPER case, those are reserved for environment variables.
And (unless you actually meant sh) use a shebang (first line) that targets bash.
The command enscript most definitively requires a file, not a variable.
Maybe you should use evince to open the PS file, there is no need of the step to make a pdf, unless you know you really need it.
I believe that is better use a file to store the output of enscript and ps2pdf.
Do not hide the errors printed by the commands until everything is working as desired, then, just call the script as:
./script infile 2>/dev/null
Or as required to make it less verbose.
Final script.
If you call the script with the name of the file that enscript is going to use, something like:
./script infile
Then, the whole script will look like this (runs both in bash or sh):
#!/usr/bin/env bash
Usage(){ echo "$0; This script require a source file"; exit 1; }
[ $# -lt 1 ] && Usage
[ ! -e $1 ] && Usage
infile="$1"
pdir="$HOME/Desktop"
open_pdf=evince
# Straighten out curly quotes
sed -i 's/[“”]/\"/g;s/’/'\''/g' "$infile"
tmpfile="$(mktemp "$pdir"/mutt_XXXXXXXX.pdf)"
outfile="${tmpfile%.*}.ps"
enscript --font=Courier10 "$infile" -2r \
--word-wrap --fancy-header=mutt -p "$outfile"
ps2pdf "$outfile" "$tmpfile"
"$open_pdf" "$tmpfile" >/dev/null 2>&1 &
sleep 5
rm "$tmpfile" "$outfile"

Unix: Remove date from filename using sed without modifying existing one

I have a legacy code which transmits the file only if it has date within the command. But client transmission doesnt want date to be appended to filename. legacy code cannot be modified since many other transmision depends on it. So my requirement is i want to have date parameter in the command but again the same has to be removed using a single command.
Condition in legacy code:
grep '\`date' $COMMAND
Note: COMMAND will contain the complete command defined below and not the filename (not CMD output).
So ideally my command should have `date added. I added a command like this below.
CMD=`echo prefix_filename.txt | sed 's/^prefix_//'`_`date +%m%d%Y`
The above command is used to remove prefix_ and send filename. Here i get output as filename.txt_09232016. Since legacy code logic only checks if command has `date in it, i added it. Is there a way to remove the date again in the same command so that my output will be filename.txt
Current output:
filename.txt_09232016
Expected output:
filename.txt
Get the file name before date part:
echo 'filename.txt_09232016' | grep -o '^.*\.txt'
Or remove date from the end of the file:
echo 'filename.txt_09232016' | sed 's/_[0-9]\+$//'
There are a number of things you can do to improve/simplify your code. The main thing is that bash have very nice built-in string manipulation. Another is that you should probably use $(...) instead of `...` notation:
CMD=`echo prefix_filename.txt | sed 's/^prefix_//'`_`date +%m%d%Y`
Can be replaced with
ORIG=prefix_filename.txt
CMD=${ORIG#prefix_}_$(date +%m%d%Y)
Continuing,
echo $CMD
NODATE=${CMD%_*}
echo $NODATE
This prints
filename.txt_09232016
filename.txt
The construct ${var#pattern} removes the shortest occurrence of pattern from the start of your variable: in this case, prefix_. Similarly, the construct ${var%pattern} removes the shortest occurrence of pattern from the end of your string: in this case _*.
In the first case, you could have used ${var##pattern} since prefix_ is a fixed string. However, in the second case you could not use ${var%%pattern}, since you want to make sure you only truncate starting at the last underscore, not the first one and the date is specified as a dynamic pattern.
Just as an FYI, the links point to www.tldp.org, which has the best Bash manual I have come across by far. It gets dense sometimes, but the explanations are generally worth it in the end.
Just do that:
echo filename.txt_09232016 | sed s/_[^_]*$//
Here, you are replacing (by nothing) ' _ ' and all subsequent characters, until the end of the string ($), since they are all different (^) of ' _ '.

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