I can't see to get it working :
echo $VERSIONNUMBER
i get : v0.9.3-beta
VERSIONNUMBERNAME=${VERSIONNUMBER:1}
echo $VERSIONNUMBERNAME
I get : 0.9.3-beta
VERSION=${VERSIONNUMBERNAME/./_}
echo $VERSION
I get : 0_9.3-beta
I want to have : 0_9_3-beta
I've been googling my brains out I can't make heads or tails of it.
Ideally I'd like to remove the v and replace the periods with underscores in one line.
Let's create your variables:
$ VERSIONNUMBER=v0.9.3-beta
$ VERSIONNUMBERNAME=${VERSIONNUMBER:1}
This form only replaces the first occurrence of .:
$ echo "${VERSIONNUMBERNAME/./_}"
0_9.3-beta
To replace all occurrences of ., use:
$ echo "${VERSIONNUMBERNAME//./_}"
0_9_3-beta
Because this approach avoids the creation of pipelines and subshells and the use of external executables, this approach is efficient. This approach is also unicode-safe.
Documentation
From man bash:
${parameter/pattern/string}
Pattern substitution. The pattern is expanded to produce a pattern
just as in pathname expansion. Parameter is expanded and the longest
match of pattern against its value is replaced with
string. If pattern begins with /, all matches of pattern are replaced
with string. Normally only the first match is replaced. If
pattern begins with #, it must match at the beginning of the expanded
value of parameter. If pattern begins with %, it must match at the
end of the expanded value of parameter. If string
is null, matches of pattern are deleted and the / following pattern
may be omitted. If the nocasematch shell option is enabled, the
match is performed without regard to the case of alphabetic
characters. If parameter is # or *, the substitution operation is
applied to each positional parameter in turn, and the
expansion is the resultant list. If parameter is an array variable
subscripted with # or *, the substitution operation is
applied to each member of the array in turn, and the expansion is the
resultant list.
(Emphasis added.)
You can combine pattern substitution with tr:
VERSION=$( echo ${VERSIONNUMBER:1} | tr '.' '_' )
Related
I am attempting to write a bash script (poorly) and need assistance in stripping characters from a variable.
variable is defined as $managementipmask= 111.111.111.111/24
I need to strip the /24 from the end of the variable.
Thanks in advance.
Use parameter expansion to remove everything from the first /:
$ k="111.111.111.111/24"
$ echo "${k%%/*}"
111.111.111.111
See this resource on parameter expansion for additional details:
http://mywiki.wooledge.org/BashGuide/Parameters#Parameter_Expansion
${parameter%pattern}
The 'pattern' is matched against the end of 'parameter'. The result is
the expanded value of 'parameter' with the shortest match deleted.
${parameter%%pattern}
As above, but the longest match is deleted.
So you can delete from the last / using a single %:
$ k="111.111.111.111/24/23"
$ echo "${k%/*}"
111.111.111.111/24
Another way:
k="111.111.111.111/24"
echo "${k/%\/24/}"
It replaces last /24 with empty string.
From Bash Manual:
${parameter/pattern/string}
The pattern is expanded to produce a pattern just as in filename expansion. Parameter is expanded and the longest match of pattern
against its value is replaced with string. If pattern begins with ‘/’,
all matches of pattern are replaced with string. Normally only the
first match is replaced. If pattern begins with ‘#’, it must match at
the beginning of the expanded value of parameter. If pattern begins
with ‘%’, it must match at the end of the expanded value of parameter.
If string is null, matches of pattern are deleted and the / following
pattern may be omitted. If parameter is ‘#’ or ‘’, the substitution
operation is applied to each positional parameter in turn, and the
expansion is the resultant list. If parameter is an array variable
subscripted with ‘#’ or ‘’, the substitution operation is applied to
each member of the array in turn, and the expansion is the resultant
list.
I am new to shell scripting:
I have following:
old=/dev/sda
new=/dev/sda5
Given these variables, I need to extract 5 from "new" string.
How should I go about it? sed? awk?
Tried using:
partitionno=$(echo $new | sed 's/$old//g')
To get the correct result with the least change to your command, try:
partitionno=$(echo "$new" | sed "s|$old||g")
There are two key points here:
Shell variables are not expanded inside single quotes. So '$old' remains as the original four characters: $, o, l, and d. For the shell variables to be expanded, use double quotes.
sed "s/$old//g" still won't work because there are too many slashes. A substitute command uses three slashes. After the shell expands $old, there are five slashes. The solution is to use a different delimiter for the substitute command. I chose | above because | is not likely to be in a file name.
Using Shell Parameter Expansion:
$ old=/dev/sda
$ new=/dev/sda5
$ echo "${new#$old}"
5
${parameter#word}
${parameter##word}
The word is expanded to produce a pattern just as in filename expansion (see Filename Expansion). If the pattern matches the beginning of the expanded value of parameter, then the result of the expansion is the expanded value of parameter with the shortest matching pattern (the ‘#’ case) or the longest matching pattern (the ‘##’ case) deleted. If parameter is ‘#’ or ‘’, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with ‘#’ or ‘’, the pattern removal operation is applied to each member of the array in turn, and the expansion is the resultant list.
I am wrinting a sed invocation using a shell variable. The variable contains a path name with file name:
sed "file name is '$variable'"
...
variable=/path/path/file.txt
The problem is that I don't need the /path/path/ part. I need just file.txt part in output.
Also my path is dynamic so I am guessing that I need to search (somehow) in a string for a first slash from the ending. How do I do that?
You can use basename to do that:
basename /tmp/a.jpg
a.jpg
You can use the shell's variable substitution feature to remove parts matching a glob pattern:
$ variable=/path/path/file.txt
$ echo ${variable##*/} # Remove longest left part matching "*/"
file.txt
From the bash manual:
${parameter#word}
${parameter##word}
The word is expanded to produce a pattern just as in pathname expansion. If
the pattern matches the beginning of the value of parameter, then the result
of the expansion is the expanded value of parameter with the shortest match-
ing pattern (the ‘‘#’’ case) or the longest matching pattern (the ‘‘##’’
case) deleted. If parameter is # or *, the pattern removal operation is
applied to each positional parameter in turn, and the expansion is the
resultant list. If parameter is an array variable subscripted with # or *,
the pattern removal operation is applied to each member of the array in
turn, and the expansion is the resultant list.
${parameter%word}
${parameter%%word}
The word is expanded to produce a pattern just as in pathname expansion. If
the pattern matches a trailing portion of the expanded value of parameter,
then the result of the expansion is the expanded value of parameter with the
shortest matching pattern (the ‘‘%’’ case) or the longest matching pattern
(the ‘‘%%’’ case) deleted. If parameter is # or *, the pattern removal
operation is applied to each positional parameter in turn, and the expansion
is the resultant list. If parameter is an array variable subscripted with #
or *, the pattern removal operation is applied to each member of the array
in turn, and the expansion is the resultant list.
Can somebody explain how echo "${PWD/#$HOME/~}" would print ~ in case the PWD evaluates to $HOME. Never read about such replacement using echo. What is going on here?
It is not echo it is your shell makes Parameter Expansion using ${parameter/pattern/string} syntax:
The pattern is expanded to produce a pattern just as in filename
expansion. Parameter is expanded and the longest match of pattern
against its value is replaced with string. If pattern begins with ‘/’,
all matches of pattern are replaced with string. Normally only the
first match is replaced. If pattern begins with ‘#’, it must match at
the beginning of the expanded value of parameter. If pattern begins
with ‘%’, it must match at the end of the expanded value of parameter.
If string is null, matches of pattern are deleted and the / following
pattern may be omitted. If parameter is ‘#’ or ‘*’, the substitution
operation is applied to each positional parameter in turn, and the
expansion is the resultant list. If parameter is an array variable
subscripted with ‘#’ or ‘*’, the substitution operation is applied to
each member of the array in turn, and the expansion is the resultant
list.
It doesn't look like POSIX supports it.
In your case, it replaces the value of $HOME envvar (not the string '$HOME' literally) with ~ in the output if PWD envvar starts with it.
Sorry for the lame bash question, but I can't seem to be able to work it out.
I have the following simple case:
I have variable like artifact-1.2.3.zip
I would like to get a sub-string between the hyphen and the last index of the dot (both exclusive).
My bash skill are not too strong. I have the following:
a="artifact-1.2.3.zip"; b="-"; echo ${a:$(( $(expr index "$a" "$b" + 1) - $(expr length "$b") ))}
Producing:
1.2.3.zip
How do I remove the .zip part as well?
The bash man page section titled "Variable Substitution" describes using ${var#pattern}, ${var##pattern}, ${var%pattern}, and ${var%%pattern}.
Assuming that you have a variable called filename, e.g.,
filename="artifact-1.2.3.zip"
then, the following are pattern-based extractions:
% echo "${filename%-*}"
artifact
% echo "${filename##*-}"
1.2.3.zip
Why did I use ## instead of #?
If the filename could possibly contain dashes within, such as:
filename="multiple-part-name-1.2.3.zip"
then compare the two following substitutions:
% echo "${filename#*-}"
part-name-1.2.3.zip
% echo "${filename##*-}"
1.2.3.zip
Once having extracted the version and extension, to isolate the version, use:
% verext="${filename##*-}"
% ver="${verext%.*}"
% ext="${verext##*.}"
% echo $ver
1.2.3
% echo $ext
zip
$ a="artifact-1.2.3.zip"; a="${a#*-}"; echo "${a%.*}"
‘#pattern’ removes pattern so long as it matches the beginning of $a.
The syntax of pattern is similar to that used in filename matching.
In our case,
* is any sequence of characters.
- means a literal dash.
Thus #*- matches everything up to, and including, the first dash.
Thus ${a#*-} expands to whatever $a would expand to,
except that artifact- is removed from the expansion,
leaving us with 1.2.3.zip.
Similarly, ‘%pattern’ removes pattern so long as it matches the end of the expansion.
In our case,
. a literal dot.
* any sequence of characters.
Thus %.* is everything including the last dot up to the end of the string.
Thus if $a expands to 1.2.3.zip,
then ${a%.*} expands to 1.2.3.
Job done.
The man page content for this is as follows (at least on my machine, YMMV):
${parameter#word}
${parameter##word}
The word is expanded to produce a pattern just as in pathname
expansion. If the pattern matches the beginning of the value of
parameter, then the result of the expansion is the expanded
value of parameter with the shortest matching pattern (the ``#''
case) or the longest matching pattern (the ``##'' case) deleted.
If parameter is # or *, the pattern removal operation is applied
to each positional parameter in turn, and the expansion is the
resultant list. If parameter is an array variable subscripted
with # or *, the pattern removal operation is applied to each
member of the array in turn, and the expansion is the resultant
list.
${parameter%word}
${parameter%%word}
The word is expanded to produce a pattern just as in pathname
expansion. If the pattern matches a trailing portion of the
expanded value of parameter, then the result of the expansion is
the expanded value of parameter with the shortest matching pat-
tern (the ``%'' case) or the longest matching pattern (the
``%%'' case) deleted. If parameter is # or *, the pattern
removal operation is applied to each positional parameter in
turn, and the expansion is the resultant list. If parameter is
an array variable subscripted with # or *, the pattern removal
operation is applied to each member of the array in turn, and
the expansion is the resultant list.
HTH!
EDIT
Kudos to #x4d for the detailed answer.
Still think people should RTFM though.
If they don't understand the manual,
then post another question.
Using Bash RegEx feature:
>str="artifact-1.2.3.zip"
[[ "$str" =~ -(.*)\.[^.]*$ ]] && echo ${BASH_REMATCH[1]}
I think you can do this:
string=${a="artifact-1.2.3.zip"; b="-"; echo ${a:$(( $(expr index "$a" "$b" + 1) - $(expr length "$b") ))}}
substring=${string:0:4}
The last step removes the last 4 characters from the string. There's some more info on here.