If I have to extract a part of the string, what should be the command to do that in Makefile.
Consider that I need a number which is a part of the pwd.
$(shell pwd) -- /xxx/www/yyy22/zzz
How would I extract the 22 from the pwd ?
Your question is not exact: what if there are more than one set of numbers in the string? Do you want them all? Just the first one? Just the last one?
In any event, you can't do this with built-in make functions. You'll have to use the shell; for example to return all the numbers in the string:
numbers := $(shell pwd | sed 's/[^0-9]//g')
Related
Some make variables have an effect on the generation of the targets and thus one may want to rebuild if their value changes, for instance because they were specified explicitly on the command line. With GNU Make, it is relatively easy to do. For instance one can do this:
CHECK_CFLAGS:=.last-cflags.$(shell echo $(CFLAGS) | md5sum | awk '{ print $$1 }')
foo.o: foo.c $(CHECK_CFLAGS)
.last-cflags.%:
-rm .last-cflags.*
touch $#
(Obviously if you are using target specific value, things become more complex). Is there a way to achieve the same desired effect with standard (POSIX) Make? If not, with the BSD variant of Make?
If you can assume GNU make 4.0 or above, you can use the != assignment operator which was implemented in GNU make for portability with BSD make.
From the GNU make manual:
The shell assignment operator '!=' can be used to execute a shell
script and set a variable to its output. This operator first evaluates
the right-hand side, then passes that result to the shell for execution.
If the result of the execution ends in a newline, that one newline is
removed; all other newlines are replaced by spaces. The resulting
string is then placed into the named recursively-expanded variable.
Note there is one subtle difference between != and := $(shell ...): in the former the resulting variable is a recursive variable which means it will be re-expanded on use. So you need to be careful if your script might emit characters that are special to make (the $ basically).
I am using
curl -0 https://cdn.(website).com/[range].(extension) -o "#1.(extension)"
to download files in a range of filename values.
How can I specify the range in a base higher than base 10 (I'm aiming for base 36 to encompass 0-9, a-z)?
ex. [000-zzz]
I don't know of an easy way to print numbers in a nonstandard base in bash, but you could use bash's brace expansion to generate the range textually. You can nest brace expansions, so {{0..9},{a..z}} will give you the digits followed by the lowercase letters. Then stack three of those to get all three-item sequences.
Note that this is a different feature from curl's bracket and brace expansion, so you can't use it in the output filename (the #1 in your example). But you can make a shell for loop, and use the shell variable to create the output filename. Something like this:
for itemnum in {{0..9},{a..z}}{{0..9},{a..z}}{{0..9},{a..z}}; do
curl -0 https://cdn.(website).com/${itemnum}.(extension) -o "${itemnum}.(extension)"
done
I want to search the below text file for words that ends in _letter, and get the whole portion upto "::". There is no space between any letter
blahblah:/blahblah::abc_letter:/blahblah/blahblah
blahblah:/blahblah::cd_123_letter:/blahblah/blahblah
blahblah:::/blahblah::24_cde_letter:/blahblah/blahblah
blahblah::/blahblah::45a6_letter:/blahblah/blahblah
blahblah:/blahblah::fgh_letter:/blahblah/blahblah
blahblah:/blahblah::789_letter:/blahblah/blahblah
I tried
egrep -o '*_letter'
and
egrep -o "*_letter"
But it only returns the word _letter
then I want to feed the input to the parametre of a shell script for loop. So the script will look like following
for i in [grep command]
mkdir $i
end
It will create the following directories
abc_letter/
cd_123_letter/
24_cde_letter/
45a6_letter/
fgh_letter/
789_letter/
ps: The result between :: and _letter doesn't contain any special character, only alphanumeric character
also my system doesn't have perl
Assuming no spaces or new-lines:
for i in $(sed 's/^.*:\([^/]*_letter\):.*$/\1/g' infile); do
mkdir $i
done
To extract after : to _letter strings from a file.txt and use them in your for loop, you can use the following egrep and revise your: script.sh, like this:
#!/bin/bash
for i in $(egrep -o "[^:]+_letter" file.txt); do
mkdir -p $i
done
Then you run ./script.sh, and later you check with ls, you see:
$ ls -1
24_cde_letter
45a6_letter
789_letter
abc_letter
cd_123_letter
fgh_letter
file.txt
script.sh
Explanation
Your original egrep -o '*_letter' probably just confused bash filename expansion with regular expression,
In bash, *something uses star globbing character to match * = anything here + something.
However in regular expression star * means the preceding character zero or more times. Since * is at the beginning of what you wrote, there is nothing before it, so it does not match anything there.
The only thing egrep can match is _letter, and since we are using the -o option it only displays the match, on an individual line, and thus why you originally only saw a line of _letter matches
Our new changes:
egrep pattern starts with [^ ... ], a negation, matches the opposite of what characters you put within. We put : within.
The + says to match the preceding one or more times.
So combined, it says look for anything-but-:, and do this one or more times.
Thus of course it matches anything after :, and keeps matching, until the next part of the pattern
The next part of the pattern is just _letter
egrep -o so only matched text will be shown, one per line
So in this way, from lines such as:
blahblah:/blahblah::abc_letter:/blahblah/blahblah
It successfully extracts:
abc_letter
Then, changes to your bash script:
Bash command substitution $() to have the results of the egrep command sent to the for-loop
for i value...; do ... done syntax
mkdir -p just a convenience in case you are re-testing, it will not error if directory was already made.
So altogether it helps to extract the pattern you wanted and generate directories with those names.
How can I truncate the ../ or .. characters from string in bash
So, If I have strings
str1=../lib
str2=/home/user/../dir1/../dir2/../dir3
then how I can get string without any .. characters in a string like after truncated result should be
str1=lib
str2=/home/user/dir1/dir2/dir3
Please note that I am not interesting in absolute path of string.
You don't really need to fork a sub-shell to call sed. Use bash parameter expansion:
echo ${var//..\/}
str1=../lib
str2=/home/user/../dir1/../dir2/../dir3
echo ${str1//..\/} # Outputs lib
echo ${str2//..\/} # Outputs /home/user/dir1/dir2/dir3
You could use:
pax> str3=$(echo $str2 | sed 's?\.\./??g') ; echo $str3
/home/user/dir1/dir2/dir3
Just be aware (as you seem to be) that's a different path to the one you started with.
If you're going to be doing this infrequently, forking an external process to do it is fine. If you want to use it many times per second, such as in a tight loop, the internal bash commands will be quicker:
pax> str3=${str2//..\/} ; echo $str3
/home/user/dir1/dir2/dir3
This uses bash pattern substitution as described in the man page (modified slightly to adapt to the question at hand):
${parameter/pattern/string}
The 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.
If string is null, matches of pattern are deleted and the / following pattern may be omitted.
You can use sed to achieve it
sed 's/\.\.\///g'
For example
echo $str2 | sed 's/\.\.\///g'
OP => /home/user/dir1/dir2/dir3
package_version := $(version)x0d$(date)
what is the x0d part between version and date vars? is it just string?
What $(dotin_files:.in=) does below
code
dotin_files := $(shell find . -type f -name \*.in)
dotin_files := $(dotin_files:.in=)
what this means $(dotin_files:=.in)
code
$(dotin_files): $(dotin_files:=.in)
$(substitute) $#.in > $#
can target contain multiple files?
what is the meaning of declaring target variable as PHONY?
code
.PHONY: $(dotin_files)
In the regex replacement code below
code
substitute := perl -p -e 's/#([^#]+)#/defined $$ENV{$$1} ? $$ENV{$$1} : $$&/ge'
what are $$ENV{$$1} and $$&? I guess it's Perl scope...
thanks for your time
Variable Expansion
$() is variable expansion in make, this should just be string substitution - if your makefile is
version=1
date=1.1.10
package_version:=$(version)x0d$(date)
then the variable package_version will expand to 1x0d1.1.10.
Substitution
The syntax $(var:a=b) is a substitution reference and will expand to var with a suffix a substituted with b.
For example, in
foobar:= foo bar
faabar:=$(foobar:oo=aa)
$(faabar) will expand to the string faa bar.
Multiple Targets
Multiple targets in a make rule is equivalent to having n rules with a single target, eg
foo bar:foo.c bar.c
$(CC) -o $# $^
is equivalent to
foo:foo.c bar.c
$(CC) -o $# $^
bar:foo.c bar.c
$(CC) -o $# $^
remember that any variables here are expanded.
Phony Targets
The .PHONY target declares that a rule doesn't produce an actual file, so it will always be built. As always, variables are expanded first. In your case this will expand to something like
.PHONY: foo bar
Escaping
A dollar sign is an escape character in makefiles, the $$ in your perl example is a literal $, eg substitute will be the string
perl -p -e 's/#([^#]+)#/defined $ENV{$1} ? $ENV{$1} : $&/ge'
The dollar signs here are processed by perl, and probably give environment variables (I don't know perl).
x0d part between version and date vars, is it just string?
Yes.
What $(dotin_files:.in=) does below
Removes the .in from the filenames found with the shell find.
what this means $(dotin_files:=.in)
I think you meant $(dotin_files:.in=). As already answered, within the variable dotin_files it replaces any occurrence of ".in" with an empty string(the part between the "=" and ")".
can target contain multiple files?
Yes
what is the meaning of declaring target variable as PHONY?
make will ignore targets time-stamp and consider them as new
thus rebuilding them each time.
In the regex replacement code below what are $$ENV{$$1} and $$&?
To avoid expansion of $ENV, the $ is doubled, think of '%' in C format strings, thus the string
perl -p -e 's/#([^#]+)#/defined $$ENV{$$1} ? $$ENV{$$1} : $$&/ge'
when called as a shell command will become:
perl -p -e 's/#([^#]+)#/defined $ENV{$1} ? $ENV{$1} : $&/ge'
$ENV is the perl Environment hash, $1 I think it's a backreference in the s/// regexp group.
Michael, you've been asking a lot of basic Makefile questions, and the ones you're asking now are ones you should be able to answer for yourself by experiment.
can target contain multiple files?
Try it:
dotin_files := foo.in bar.in
$(dotin_files):
#echo $#
Now try make foo.in and make bar.in. What happens?
What $(dotin_files:.in=) does
It's a substitution reference. Try it yourself and see what happens, like this:
dotin_files := foo.in bar.in
dotin_files := $(dotin_files:.in=)
all:
#echo $(dotin_files)
What did it do?
substitute := perl -p -e 's/#([^#]+)#/defined $$ENV{$$1} ? $$ENV{$$1} : $$&/ge'
what are $$ENV{$$1} and $$&? I guess it's Perl scope...
Let's take a look:
all:
#echo $(substitute)
If you want to know more about Perl, or regexs, or find, or make, or whatever, feel free to ask here, but please take a little time to try to figure it out first.