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Closed 9 years ago.
I am trying to solve below problem using regular expression. My input string is something like this
"MTSGNN0002( 1), MTSGNN0028( 645), MTSGNN0050( 10)"
and I want output like this
"MTSGNN0002,MTSGNN0028,MTSGNN0050"
It should delete all charecter which comes between brackets. Kindly help me out in solving this.
Sometimes it's simpler to find what you want than delete what you don't:
s = "MTSGNN0002( 1), MTSGNN0028( 645), MTSGNN0050( 10)"
s.scan(/MT\w+/).join(',') # Change MT to whatever suits your data.
Look at this regexp:
([a-zA-Z0-9]*)\(.*\)
http://rubular.com/r/maZNs0mDkv
From there on it's easy ;-)
Here you go..
\([^\)]*\)|\s
As per your question.. The above replaces the brackets and the text inside and remove the spaces.
Example and Source Demo:
Here i have got one more solution
([(\b]\s.[0-9]*\b\))
http://rubular.com/r/9NyoU3RKUT
replace mathes of this regex \([ \d]+\) with empty string
Related
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Closed 9 years ago.
I have a string like this /4/LM2301_800.mp4.
I would like to use a regex to strip the /4/ and the _800 from the file name.
I am currently trying to use the strip command. Can anyone point me in the right direction?
Your file could be deeply nested inside a directory structure:
path = "/4/LM2301_800.mp4"
path.sub(/^(\/.)*\/(.*)_\d+\.mp4/, $2)
=> "LM2301" # you already know these are mp4 files, so you could add the suffix
Or:
path = "/4/LM2301_800.mp4"
path.sub(/^(\/.)*\/(.*)_\d+\.(.*)/, $2+'.'+$3)
=> "LM2301.mp4"
" /4/LM2301_800.mp4".scan(/.*\/(.*?)_\d*(.*)/).join
=> "LM2301.mp4"
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 9 years ago.
I want to store all the filenames inside a folder into an array. What's the best way to do this?
You can use this:
files = Dir.foreach(dir).select { |x| File.file?("#{dir}/#{x}") }
This returns the filenames, i.e. without folder.
If you need the complete path, use something like this:
files = Dir.foreach(dir) \
.map { |x| File.expand_path("#{dir}/#{x}") } \
.select { |x| File.file?(x) }
You can use:
files = Dir.entries(directory)
that returns an array containing all the filenames in the given directory.
Take a look in the Ruby Doc for more information.
you can also use files=Dir.glob(*).
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Closed 10 years ago.
I am making a application(desktop app), and need to fetch ids from DB and pass these ids to another desktop application.
How to achieve this?
Your question is your answer. You can use command line argument passing in windows application. As to how to do this is written below.
You can Create a process and call the application to start and pass those arguments:
Process pro=new Process();
pro.StartInfo.FileName = #"ApplicationName.exe";
pro.StartInfo.Arguments = arg1 + " " + arg2;
pro.Start();
pro.WaitForExit();
in the second application get these values by following the below code:
var arguments = Environment.GetCommandLineArgs();
if (arguments.Length > 1)
{
productIdFirst = Convert.ToInt32(arguments[1]);
productIdSecond = Convert.ToInt32(arguments[2]);
}
If I understand you correctly then you're talking about a website and you could pass it a query string. If that is the case then you should check out http://msdn.microsoft.com/en-us/library/system.web.httprequest.querystring.aspx :)
I agree to Martim, that it is no clear if your tags "Windows" not perhaps mean "Website". But if you really mean "Windows" like you've tagged, you can use COM as Middleware. A german article already has great explanation with code samples. Languages can be translated.
http://www.activevb.de/tutorials/tut_middleware/middleware.html
Download Source Code
http://www.activevb.de/tutorials/tut_middleware/downloads/Middleware.zip
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 10 years ago.
I am new to CI.
Now i am try to develop the comment system with reply option.
So far i have developed the insert comment and reply comment.
But i don't know how to fetch reply comments under the main comment.
Please help me.
Thanks to all
Use a recursive function! For that you will need to do something like this:
Get the first comment.
Call the recursive function. The ID of the comment must be passed as an argument.
The function must get the comments & call itself in order to get all the comments.
A pseudo-code of this schema will be:
myComment = getComment();
recursiveComments(myComment);
function recursiveComments(currentComment){
print(currentComment);
replies = getReplies(currentComment['idComment']);
foreach(replies as reply){
recursiveComments(reply);
}
}
In the pseudo-code, I'm assuming that you get a row_array with the getComment() function & that this comment is the "head" of all the comments, like the initial post.
After that, I call the recursive function which gets all the replies & sub-replies for each comment. Note the I fetch a single reply in each call to the function.
Hope this gives you an idea! (=
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Closed 11 years ago.
I have list of website, I want all website with url http://www.tutorspree.com/tutor/#{AnyNumber) where after 'tutor' there could be any number. How do I write Regex for this?
I would define it with:
Regexp.new("^http://www\.tutorspree\.com/tutor/\\d+$")
This avoids having to escape the forward slashes. It would be used on a list of URLs like this:
tutor_re = Regexp.new("^http://www\.tutorspree\.com/tutor/\\d+$")
list = [ "http://nomatch.com/", "http://www.tutorspree.com/tutor/1", "http://www.tutorspree.com/tutor/2" ]
matches = list.select { |url| tutor_re.match url }
Something like:
http:\/\/www\.tutorspree\.com\/tutor\/\#\d+
ought to work.