We often have the situation, that string values start with a whitespace. How can I check and return all strings which start with a whitespace?
I found a similar question: Velocity, what's the most efficient way to check if a string is empty and not null
But I don't know how to exactly do it, as I have never done anything with Velocity before.
Don't forget that you can call Java methods on Velocity references. So nothing stops you from doing:
#foreach($str in $myStrings)
#if($str.startsWith(' '))
## do something with $str
#end
#end
Related
I just want to capture the part of the string in nbnbaasd<sd which appears before any a.
I want it to return nbnb as the match.
/.+(?!a)/.match("nbnbaasd<sd") # returns the whole string
Just use a negated character set:
/[^a]+/.match("nbnbaasd<sd")
It's far more efficient than the look-ahead method.
See it here in action: http://regexr.com?32288
It returns the whole string because indeed, "nbnbaasd<sd" is not followed by an "a".
Try this.
/.+?(?=a)/.match("nbnbaasd<sd")
(You do not actually need to use a lookahead to achieve this, but perhaps you've simplified your problem and in your real problem you do need a zero-width assertion for some reason. So this is a solution as close as possible to the one you've attempted.)
I'm too ambitious or is there a way do this
to add a string if not present ?
and
remove a the same string if present?
Do all of this using Regex and avoid the if else statement
Here an example
I have string
"admin,artist,location_manager,event_manager"
so can the substring location_manager be added or removed with regards to above conditions
basically I'm looking to avoid the if else statement and do all of this plainly in regex
"admin,artist,location_manager,event_manager".test(/some_regex/)
The some_regex will remove location_manager from the string if present else it will add it
Am I over over ambitions
You will need to use some sort of logic.
str += ',location_manager' unless str.gsub!(/location_manager,/,'')
I'm assuming that if it's not present you append it to the end of the string
Regex will not actually add or remove anything in any language that I am aware of. It is simply used to match. You must use some other language construct (a regex based replacement function for example) to achieve this functionality. It would probably help to mention your specific language so as to get help from those users.
Here's one kinda off-the-wall solution. It doesn't use regexes, but it also doesn't use any if/else statements either. It's more academic than production-worthy.
Assumptions: Your string is a comma-separated list of titles, and that these are a unique set (no duplicates), and that order doesn't matter:
titles = Set.new(str.split(','))
#=> #<Set: {"admin", "artist", "location_manager", "event_manager"}>
titles_to_toggle = ["location_manager"]
#=> ["location_manager"]
titles ^= titles_to_toggle
#=> #<Set: {"admin", "artist", "event_manager"}>
titles ^= titles_to_toggle
#=> #<Set: {"location_manager", "admin", "artist", "event_manager"}>
titles.to_a.join(",")
#=> "location_manager,admin,artist,event_manager"
All this assumes that you're using a string as a kind of set. If so, you should probably just use a set. If not, and you actually need string-manipulation functions to operate on it, there's probably no way around except for using if-else, or a variant, such as the ternary operator, or unless, or Bergi's answer
Also worth noting regarding regex as a solution: Make sure you consider the edge cases. If 'location_manager' is in the middle of the string, will you remove the extraneous comma? Will you handle removing commas correctly if it's at the beginning or the end of the string? Will you correctly add commas when it's added? For these reasons treating a set as a set or array instead of a string makes more sense.
No. Regex can only match/test whether "a string" is present (or not). Then, the function you've used can do something based on that result, for example replace can remove a match.
Yet, you want to do two actions (each can be done with regex), remove if present and add if not. You can't execute them sequentially, because they overlap - you need to execute either the one or the other. This is where if-else structures (or ternary operators) come into play, and they are required if there is no library/native function that contains them to do exactly this job. I doubt there is one in Ruby.
If you want to avoid the if-else-statement (for one-liners or expressions), you can use the ternary operator. Or, you can use a labda expression returning the correct value:
# kind of pseudo code
string.replace(/location,?|$/, function($0) return $0 ? "" : ",location" )
This matches the string "location" (with optional comma) or the string end, and replaces that with nothing if a match was found or the string ",location" otherwise. I'm sure you can adapt this to Ruby.
to remove something matching a pattern is really easy:
(admin,?|artist,?|location_manager,?|event_manager,?)
then choose the string to replace the match -in your case an empty string- and pass everything to the replace method.
The other operation you suggested was more difficult to achieve with regex only. Maybe someone knows a better answer
I have a string "\nbed.bed_id,\nbed.bed_label,\nbed.room_id,\nbed.pool_bed, nbed.record_state\n"and I need to split it by white space and comma.
I tried split(/,?\s+/) which works but also leaves a "" at the beginning.
Using split($;) doesn't. What I'm looking for is say split(/,?$;/) is there a way to retain the default functionality and just add to it?
(p.s I know I can do this split[1..-1], there are so many ways to do things in ruby).
update:
My issue was with $; I wasn't sure really what it was and thought it had special meaning, because as a variable irb>$; #=> nil. Now it may be that I missed it, or that the documentation has been updated but, ruby-doc.org says "If pattern is omitted, the value of $; is used. If $; is nil (which is the default), str is split on whitespace as if ` ‘ were specified."
As well, $; is from Perl or awk, known as the SUBSEP, and a further explanation as to why the beginning is stripped away with $; is here Why is split(' ') trying to be (too) smart?
You can't avoid split() from returning some empty elements at the start or end in this case?
Try rejecting empty strings from the array:
string.split(/,?\s+/).reject &:empty?
With using split u can do it
str = "\nbed.bed_id,\nbed.bed_label,\nbed.room_id,\nbed.pool_bed, nbed.record_state\n"
st = str.split(/,?\s+/)
st.shift
st
following a RoR security tutorial (here), i wrote something along the lines of
##private_re = //
def secure?
action_name =~ ##private_re
end
the idea is that in the base case, this shouldn't match anything, and return nil. problem is that it doesn't. i've worked around for the time being by using a nonsensical string, but i'd like to know the answer.
The empty regular expression successfully matches every string.
Examples of regular expressions that will always fail to match:
/(?=a)b/
/\Zx\A/
/[^\s\S]/
It is meant to not change the behavior of the controller in any way, as // will match every string.
The idea is that ##private is meant to be set in the controller to match things you DO want to be private. Thus, that code is meant to do nothing, but when combined with
##private = /.../ in the controller, gives you a nice privacy mechanism.
I've been trying to construct a ruby regex which matches trailing spaces - but not indentation placeholders - so I can gsub them out.
I had this /\b[\t ]+$/ and it was working a treat until I realised it only works when the line ends are [a-zA-Z]. :-( So I evolved it into this /(?!^[\t ]+)[\t ]+$/ and it seems like it's getting better, but it still doesn't work properly. I've spent hours trying to get this to work to no avail. Please help.
Here's some text test so it's easy to throw into Rubular, but the indent lines are getting stripped so it'll need a few spaces and/or tabs. Once lines 3 & 4 have spaces back in, it shouldn't match on lines 3-5, 7, 9.
some test test
some test test
some other test (text)
some other test (text)
likely here{ dfdf }
likely here{ dfdf }
and this ;
and this ;
Alternatively, is there an simpler / more elegant way to do this?
If you're using 1.9, you can use look-behind:
/(?<=\S)[\t ]+$/
but unfortunately, it's not supported in older versions of ruby, so you'll have to handle the captured character:
str.gsub(/(\S)[\t ]+$/) { $1 }
Your first expression is close, and you just need to change the \b to a negated character class. This should work better:
/([^\t ])[\t ]+$
In plain words, this matches all tabs and spaces on lines that follow a character that is not a tab or a space.
Wouldn't this help?
/([^\t ])([\t ]+)$/
You need to do something with the matched last non-space character, though.
edit: oh, you meant non blank lines. Then you would need something like /([^\s])\s+/ and sub them with the first part
I'm not entirely sure what you are asking for, but wouldn't something like this work if you just want to capture the trailing whitespaces?
([\s]+)$
or if you only wanted to capture tabs
([ \t]+)$
Since regexes are greedy, they'll capture as much as they can. You don't really need to give them context beforehand if you know what you want to capture.
I still am not sure what you mean by trailing indentation placeholders, so I'm sorry if I'm misunderstanding.
perhaps this...
[\t|\s]+?$
or
[ ]+$