disk="/dev/sda"
local dev_node=${disk##*/}
dev_node is assigned with "sda".
Also,
partition="/dev/sda3"
echo ${partition%%[0-9]*}
It returns /dev/sda and remove 3.
I did not understand the functionality of ##*/ and %%[0-9]*in the above commands. I tried searching but could not get enough information.
Please explain and provide any links to tutorial related to this.
This is a very good manual / tutorial. What concerns your question:
${string##substring} Deletes longest match of $substring from front of
$string.
and
${string%%substring} Deletes longest match of $substring from back of
$string.
applied to your example: removing the longest substring matching */ from /dev/sda results in sda
This procedure is commonly described as parameter expansion.
In your case ## and %% are operators that extract part of the string.
## deletes longest match of defined substring starting at the start of given string.
%% does the same, except it starts from back of the string.
Good guide is here.
Related
I have tried below one:
A=https://xyz.site
echo -e ${A//:*}
Result: https
Please describe me that, how this ${A//:*} term results https or http and what's the concept behind it, share a article or pdf if possible.
For Worldwide web [www]
Its preety simple to extract this one:
A=www.google.com
echo -e ${A::3}
Result: www
${parameter:offset:length} — This is referred to as Substring Expansion. In your example ${A::3} means ${A:0:3} and returns the first 3 characters of the variable A.
${parameter/pattern/string} — This notation replaces the first match of pattern with a string. If pattern begins with /, all matches of pattern are replaced with string. In your example ${A//:*} means ${A//:*/} and it replaces all patterns :* with an empty string.
It's a weird problem
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120"
And two strings below:
s1="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\[Content_Types].xml"
s2="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\_rels\.rels"
When I use the command below:
s1.strip(to_be_stripped)
s2.strip(to_be_stripped)
I get these outputs:
'[Content_Types].x'
'_rels\\.'
If I use lstrip(), they will be:
'[Content_Types].xml'
'_rels\\.rels'
Which is the right outputs.
However, if we replace all Project Known with zeus_pipeline:
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120"
And:
s2="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120\\_rels\.rels"
s2.lstrip(to_be_stripped)will be '.rels'
If I use / instead of \\, nothing goes wrong. I am wondering why this problem happens.
strip isn't meant to remove full strings exactly. Rather, you give it a string, and every character in that string is removed from the start and of the string to be stripped.
In your case, the variable to_be_stripped contains the characters m and l, so those are stripped from the end of s1. However, it doesn't contain the character x, so the stripping stops there and no characters beyond that are removed.
Check out this question. The accepted answer is probably more extensive than you need - I like another user's suggestion of using replace instead of strip. This would look like:
s1.replace(to_be_stripped, "")
for my $item (#array) {
if (index($item, '$n') != -1) {
print "HELLO\n";
}
}
Problem is: Perl critic gives below policy violation.
String may require interpolation at line 168, near '$item, '$n''. (Severity: 1)
Please advise how do I fix this?
In this case the analyzer either found a bug or is plain wrong in flagging your code.
Are you looking for a literal "$n" in $item, or for what $n variable evaluates to?
If you want to find the literal $n characters then there is nothing wrong with your code
If you expect $item to contain the value stored in $n variable then allow it to be evaluated,
if (index($item, $n) != -1)
If this is indeed the case but $n may also contain yet other escaped sequences or encodings which you need as literal characters (so to suppress their evaluation) then you may need to do a bit more, depending of what exactly may be in that variable.
In case you do need to find characters $ followed by n (what would explain a deliberate act of putting single quotes around a variable) you need to handle the warning.
For the particular policy that is violated see Perl::Critic::Policy::ValuesAndExpressions
This policy warns you if you use single-quotes or q// with a string that has unescaped metacharacters that may need interpolation.
To satisfy the policy you'd need to use double quotes and escape the $, for example qq(\$n). In my opinion this would change the fine original code segment into something strange to look at.
If you end up wanting to simply silence the warning see documentation, in Bending The Rules
A comment. The tool perlcritic is useful but you have to use it right. It's a static code analyzer and it doesn't know what your program is doing, so to say; it can catch bad practices but can't tell you how to write programs. Many of its "policies" are unsuitable for particular code.
The book that it is based on says all this very nicely in its introduction. Use sensibly.
When I look at the question where this comes from it appears that you are looking for index at which substrings were matched, so you need the content of $n variable, not literal "$n". Then perlcritic identified a bug in the code, good return for using it!
I have a string, such as time=1234, and I want to extract just the number after the = sign. However, this number could be in the range of 0 and 100000 (eg. - time=1, time=23, time=99999, etc.).
I've tried things like $(string:5:8}, but this will only work for examples of a certain length.
How do I get the substring of everything after the = sign? I would prefer to do it without outside commands like cut or awk, because I will be running this script on devices that may or may not have that functionality. I know there are examples out there using outside functions, but I am trying to find a solution without the use of such.
s=time=1234
time_int=${s##*=}
echo "The content after the = in $s is $time_int"
This is a parameter expansion matching everything matching *= from the front of the variable -- thus, everything up to and including the last =.
If intending this to be non-greedy (that is, to remove only content up to the first = rather than the last =), use ${s#*=} -- a single # rather than two.
References:
The bash-hackers page on parameter expansion
BashFAQ #100 ("How do I do string manipulations in bash?")
BashFAQ #73 ("How can I use parameter expansion? How can I get substrings? [...])
BashSheet quick-reference, paramater expansion section
if time= part is constant you can remove prefix by using ${str#time=}
Let's say you have str='time=123123' if you execute echo ${str#time=} you would get 123123
I have a string like hello /world today/
I need to replace /world today/ with /MY NEW STRING/
Reading the manual I have found
newString = string.match("hello /world today/","%b//")
which I can use with gsub to replace, but I wondered is there also an elegant way to return just the text between the /, I know I could just trim it, but I wondered if there was a pattern.
Try something like one of the following:
slashed_text = string.match("hello /world today/", "/([^/]*)/")
slashed_text = string.match("hello /world today/", "/(.-)/")
slashed_text = string.match("hello /world today/", "/(.*)/")
This works because string.match returns any captures from the pattern, or the entire matched text if there are no captures. The key then is to make sure that the pattern has the right amount of greediness, remembering that Lua patterns are not a complete regular expression language.
The first two should match the same texts. In the first, I've expressly required that the pattern match as many non-slashes as possible. The second (thanks lhf) matches the shortest span of any characters at all followed by a slash. The third is greedier, it matches the longest span of characters that can still be followed by a slash.
The %b// in the original question doesn't have any advantages over /.-/ since the the two delimiters are the same character.
Edit: Added a pattern suggested by lhf, and more explanations.