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.
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 is file globbing?
(1 answer)
Stop shell wildcard character expansion?
(4 answers)
command line * linux [duplicate]
(1 answer)
Closed 4 years ago.
I stumbled upon this while working through the exercises in K&R2. Why does echo * prints the names of all files in the current directory? More generally, when I write a C program that takes command-line arguments, and when I give it * as an argument, it puts the names of all files in its parent directory in to the argument vector. Why does this happen? What is so special about *?
I could not find anything about this in the internet.
This is called globbing. Here's a detailed description. Other wildcards include ? for one character, [abc] for one of a set of characters, and [a-z] for one of a range of characters. This is built into various shells, including Bash.
In response to your comment "I think echo is written in C" — this doesn't matter a bit. Once source code is compiled into an executable containing machine code, it doesn't matter what language it was written in.
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 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 ${filename##*.} will give you output as gz.
This question already has an answer here:
Bash: manipulating with strings (percent sign)
(1 answer)
Closed 6 years ago.
I've got the following variable set in bash:
ver=$(/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