This question already has answers here:
What is the meaning of the ${0##...} syntax with variable, braces and hash character in bash?
(4 answers)
Closed 2 years ago.
I am learning the bash script materials on http://www.tldp.org/LDP/abs/html/index.html
and stuck in the Example 7-7:
http://tldp.org/LDP/abs/html/comparison-ops.html#EX14
There is an ${filename##*.} != "gz", this probably means
that the $filename does not end with .gz, but I do not
know the meaning of ## here. Could anyone help me?
Thanks!
Used in a variable expansion, ${string##sub} removes the longest matching substring sub from string (# removes the shortest matching substring by contrast).
In your case, yes - this will return the string after the first . from the filename, giving the file extension.
If you search for ## in this documentation, you'll find an explanation (along with other similar commands).
In the context of filenames, is trying to find the extension in the variable filename
filename="*.log"
echo ${filename##*.}
log
We are attaining the part of the string filename after "*."
## is a used for to remove a substring from a variable. For more info check this page.
For eg. if filename=/home/user.name/folder.1/test.gz, then ${filename##*.} will give you output as gz.
Related
This question already has answers here:
What is the meaning of the ${0##...} syntax with variable, braces and hash character in bash?
(4 answers)
Closed 11 months ago.
I found the following code in a shell script, but I am unsure of what the test condition is evaluating for
if test "${SOME_VAR#*$from asdf/qwer}" != "$SOME_VAR"; then
echo "##zxcv[message text='some text.' status='NORMAL']";
fi
The combination #*$ does not mean anything special. There are three special symbols and each of them has its own meaning in this context.
${SOME_VAR#abc} is a parameter expansion. Its value is the value of $SOME_VAR with the shortest prefix that matches abc removed.
In your example, abc is *${from} asdf/qwer. That means anything * followed by the value of variable $from (which is dynamic and replaced when the expression is evaluated), followed by a space and followed by asdf/qwer.
All in all, if the value of $SOME_VAR starts with a string that ends in ${from} asdf/qwer then everything before and including asdf/qwer is removed and the resulting value is passed as the first argument to test.
Type man bash in your terminal to read the documentation of bash or read it online.
This question already has answers here:
What does the curly-brace syntax ${var%.*} mean?
(3 answers)
Closed 2 years ago.
I'm learning a bit about running a bash script in a linux terminal, specifically in the context of converting audio video files.
I came across this command here on SO that does exactly what I want. However, I'd like to understand it better:
for i in *.avi; do ffmpeg -i "$i" "${i%.*}.mp4"; done
Now, this is obviously a for-loop and I get the first * wildcard. I get the do block. But what I don't quite understand is ${i%.*}. Specifically, what does the %.* bit do in the output location? Why not use ${i}.mp4 instead?
It's called parameter expansion and it removes everything starting from the last dot (ie. extension). Try the following:
$ i="foo.bar.baz"
$ echo ${i%.*}
foo.bar
Author of the original code ("${i%.*}.mp4") apparently wanted to replace original extension with .mp4 so the original extension is removed and .mp4 is appended.
Parameter expansion
${parameter%word}
${parameter%%word}
The word is expanded to produce a pattern and matched according to the rules described below (see Pattern Matching). If the pattern matches a trailing portion of the expanded value of parameter, then the result of the expansion is the value of parameter with the shortest matching pattern (the ‘%’ case) or the longest matching pattern (the ‘%%’ case) deleted.
This question already has answers here:
Command not found error in Bash variable assignment
(5 answers)
Closed 5 years ago.
I am trying to find te longest word in a given file. Before I check the lengtgh of each word I need to remove all of the following tokens {,.:} that may be attached (once or more) to the word. so for example, for this text:
:,cat dog, encyclopedia; remove:.,
i need the result:
cat dog encyclopedia remove
I am trying this, but I get a "command not found":
longest=0
for word in $(<$1)
do
#new_word = $(echo "${word//[.,:]/}")
new_word = "${word//[.,:]/}"
len=${#new_word}
if (( len > longest ))
then
longest=$len
longword=$new_word
fi
done
echo The longest word is $longword and its length is $longest.
thank you.
Your use of parameter expansion replacement pattern is correct.
The problem is that there must not be any whitespace around = while declaring variables in bash (any shell in general).
So, the following should work:
new_word="${word//[.,:]/}"
As an aside, use a while read ... construct to loop over the lines in a file, using for is pretty fragile.
This question already has an answer here:
Bash: manipulating with strings (percent sign)
(1 answer)
Closed 6 years ago.
I've got the following variable set in bash:
ver=$(/usr/lib/virtualbox/VBoxManage -v | tail -1)
then I have the following variable which I do not quite understand:
pkg_ver="${ver%%r*}"
Could anyone elaborate on what this does, and how pkg_ver is related to the original ver value?
It is a bash parameter expansion syntax to extract text from end of string upto first occurrence of r
name="Ivory"
printf "%s\n" "${name%%r*}"
Ivo
${PARAMETER%%PATTERN}
This form is to remove the described pattern trying to match it from the end of the string. The operator "%" will try to remove the shortest text matching the pattern, while "%%" tries to do it with the longest text matching.
You will get everything from variable ver until first "r" character and it will be stored inside pkg_ver.
export ver=aaarrr
echo "${ver%%r*}"
aaa
This question already has answers here:
explain the linux regex for getting filename
(2 answers)
Closed 8 years ago.
I am tring to understand the bash script.
I am seeing ##* / expression with bash variable.
i.e ${foo##*/}
Can someone please tell me why we use that expression?
It's called "Parameter expansion". The variable $foo is searched for a substring matching the pattern */ (i.e. anything up to a slash) from the beginning (#), and what remains in the variable is returned. Doubling the #-sign makes the matching greedy, i.e. it tries to find the longest possible match.