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


Java Replace String

Posted on 2010-09-12
Medium Priority
Last Modified: 2013-12-29
I want to do find and replace. For example we have a string with something like .
"  select * from table a, table b where =
a.input = ?
<input>*/   "

Now i want tofind all /*<input>  and <input>*/ and replace it with blank space or remove it.
Question by:shetty01
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
LVL 40

Expert Comment

by:Gurvinder Pal Singh
ID: 33657279

String str="select * from table a, table b where = /*<input> and a.input = ? <input>*/  ";
LVL 40

Expert Comment

by:Gurvinder Pal Singh
ID: 33657304
String str="select * from table a, table b where = /*<input> and a.input = ? <input>*/  ";

Expert Comment

ID: 33660313
Use The commons library.
In your case, commons-lang, and it's StringUtils class :

StringUtils.replace(originalString, stringToReplace, stringReplacement);

Replaces all occurences of stringToReplace in originalString by stringReplacement.

So if you want to remove <input> by nothing, th line goes like that :
String newSqlQuery = StringUtils.replace(sqlQuery, "<input>", "");

More generally, have a look at commons library, they are really usefull (commons-lang, commons-io, commons-dbcp, and so on...) Go to apache commons website to see how many they have.

It allows  you to use a well written, tested and proven to be efficient set of very usefull functions.

Who nowadays closes an IO stream without using IOUtils.closeSilentlt(ioStream);     :)

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.

LVL 16

Expert Comment

ID: 33660331
Does the /*<input> and <input>*/ are considered to be something like start and end tag? I think they are, so try this, because '*' must be avoided in this way :
LVL 86

Expert Comment

ID: 33660843
Try the following:
s = s.replaceAll("(?sm)(/\\*<input>.*?<input>\\*/)", "");	

Open in new window

LVL 86

Expert Comment

ID: 33660852
btw, the dup question you posted on this should be deleted

Author Comment

ID: 33664250
I found the solution. Somehow i posted it twice since first time my post did not work.  I got solution in my other post. thanks for help.
LVL 86

Accepted Solution

CEHJ earned 2000 total points
ID: 33664323
Please post then the code you're actually using. (The code you refer to won't work as is)

Featured Post

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.

Question has a verified solution.

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

This was posted to the Netbeans forum a Feb, 2010 and I also sent it to Verisign. Who didn't help much in my struggles to get my application signed. ------------------------- Start The idea here is to target your cell phones with the correct…
Introduction This article is the last of three articles that explain why and how the Experts Exchange QA Team does test automation for our web site. This article covers our test design approach and then goes through a simple test case example, how …
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 explains how to use the VisualVM tool for the Java platform application. This video goes into detail on the Threads, Sampler, and Profiler tabs.
Suggested Courses

704 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