Still celebrating National IT Professionals Day with 3 months of free Premium Membership. Use Code ITDAY17

x
?
Solved

help with regular expressions

Posted on 2006-07-20
11
Medium Priority
?
256 Views
Last Modified: 2007-12-19
Hi,

I require some help with the regular exp in java.

I have a String say
String test="regular +expr + java+ example";

I will get the String as  req parameter.
If I find space before and after " + " I will have to remove that and replace that to test="regular +expr+java+example";(without any spaces)
if there is space after "+ " then the String should become test="regular +expr+java+example";(without any spaces)
If I have space only before " +"it should stay as it is test="regular +expr+java+example";

The only condition when the "+" will remain unaltered is when it is preceded by space(s) and immediately followed by text with no space in between.i.e(regular +expr).

Please help me.

Thanks.

0
Comment
Question by:ZOOMAY
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 5
  • 2
  • 2
  • +2
11 Comments
 
LVL 8

Expert Comment

by:colr__
ID: 17145855
if (!test.contains(" +"))
   test.removeAll(" ");
0
 
LVL 12

Expert Comment

by:enachemc
ID: 17145868
search for \+\s+
replace with +

You might be required to escape the "\" character in Java.
0
 
LVL 8

Expert Comment

by:colr__
ID: 17145873
In fact it might be this:

if (!test.contains(" +") && !test.contains("+ "))
   test.removeAll(" ");

Is this what you are looking for?
0
Build and deliver software with DevOps

A digital transformation requires faster time to market, shorter software development lifecycles, and the ability to adapt rapidly to changing customer demands. DevOps provides the solution.

 
LVL 30

Expert Comment

by:Mayank S
ID: 17145890
And since String is immutable:

String newTest = test.removeAll ( " " ) ;
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 17145945
test.replaceAll("\\+\\s*", "\\+");
0
 
LVL 86

Accepted Solution

by:
CEHJ earned 120 total points
ID: 17145948
Of course that would be

test = test.replaceAll("\\+\\s*", "\\+"));
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 17145960
(without the final bracket ;-))
0
 
LVL 12

Assisted Solution

by:enachemc
enachemc earned 80 total points
ID: 17146021
CEHJ,
it should be
test.replaceAll("\\+\\s+", "\\+");

because if you use * instead of +, there might be a lot of useless replaces (like replace "+" with "+")
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 17146073
>>there might be a lot of useless replaces

I doubt it, as that would be fairly incompetent of 'them' to try to replace a token with an identical one, but you're right - it'd be better to err on the safe side
0
 

Author Comment

by:ZOOMAY
ID: 17213681
Thanks a lot for the help.
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 17213690
:-)
0

Featured Post

Concerto's Cloud Advisory Services

Want to avoid the missteps to gaining all the benefits of the cloud? Learn more about the different assessment options from our Cloud Advisory team.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

In this post we will learn how to connect and configure Android Device (Smartphone etc.) with Android Studio. After that we will run a simple Hello World Program.
In this post we will learn how to make Android Gesture Tutorial and give different functionality whenever a user Touch or Scroll android screen.
Viewers learn about the scanner class in this video and are introduced to receiving user input for their programs. Additionally, objects, conditional statements, and loops are used to help reinforce the concepts. Introduce Scanner class: Importing…
This tutorial will introduce the viewer to VisualVM for the Java platform application. This video explains an example program and covers the Overview, Monitor, and Heap Dump tabs.
Suggested Courses

721 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question