if - else condition with common code

[John Jai]

Dear ranchers,

I am having an if - else scenario which looks like below.

if(flag.equals("A")){
			//TODO - Do task for match "A"			
		}else if(flag.equals("B")){
			//TODO - Do task for match "B"
		}else if(flag.equals("C")){
			//TODO - Do task for match "A" and match "B"
		}

If the flag is equals to "C" then i have to place the same code of condition "A" and "B" inside "C". The code for condition A and B is not that big to be made into a separate function. Is there a way I can make "C" does the work of "A" + "B" without duplicate code?

Thanks

[Mike Simmons]

John Jai wrote:The code for condition A and B is not that big to be made into a separate function.

I suspect this is not true. You can put the code for A and B into separate functions. It's not a big deal, really. Even without closures. If it seems like a big deal to you, consider this: how much easier would it have been to just do it in the first place, compared to composing a post here to ask about it, wait and read the responses, and reply appropriately? Is creating TWO WHOLE FUNCTIONS really so difficult?

[John Jai]

Ok.. i just wanted to know if there is another option...

My code looks like this now

if("A".equals(flag)){
			try{
				logger.info("Starting to chunk the Parent Document with Classify_As_Batch classification property -> " + document);
				ChunkDocumentSAX chunkSAX = new ChunkDocumentSAX(document, dtdFilePath, workDirectory, chunkAttribute, chunkAttributeValue); 
				chunkSAX.chunkDocument();
				childDocumentsList = chunkSAX.getChildDocumentsList();
				if(childDocumentsList != null && childDocumentsList.size() > 0){
					for(String childDocument : childDocumentsList){
						logger.info("Adding Chunk Child Document for Classification -> " + childDocument);
						documentsList.add(childDocument);
					}
				}else {
					logger.error("No Chunk Documents created & hence ignorign the parent document from classification -> " + document);
				}
			}catch (Exception e){
				e.printStackTrace();
				logger.error("Exception occurred while trying to chunk the parent document -> " + document);
				throw e;
			}				
		}else if("B".equals(flag)){			
			documentsList.add(document);	
			
			logger.info("Adding Document with Classify_As_Document classification property for Classification -> " + document);
		}else if("C".equals(flag)){
			try{
				logger.info("Starting to chunk the Parent Document with Classify_As_Batch classification property -> " + document);
				ChunkDocumentSAX chunkSAX = new ChunkDocumentSAX(document, dtdFilePath, workDirectory, chunkAttribute, chunkAttributeValue); 
				chunkSAX.chunkDocument();
				childDocumentsList = chunkSAX.getChildDocumentsList();
				if(childDocumentsList != null && childDocumentsList.size() > 0){
					for(String childDocument : childDocumentsList){
						logger.info("Adding Chunk Child Document for Classification -> " + childDocument);
						documentsList.add(childDocument);
					}
				}else {
					logger.error("No Chunk Documents created & hence ignorign the parent document from classification -> " + document);
				}
			}catch (Exception e){
				e.printStackTrace();
				logger.error("Exception occurred while trying to chunk the parent document -> " + document);
				throw e;
			}				
			documentsList.add(document);
		}

[Rob Spoor]

if (flag.equals("A") || flag.equals("C"))
{   
    //TODO - Do task for match "A"             
}
if (flag.equals("B") || flag.equals("C")) // no else if
{
    //TODO - Do task for match "B"   
}

As long as "A" and "B" are not the same this should work just fine.

[Aniruddh Joshi]

Rob Spoor wrote:

if (flag.equals("A") || flag.equals("C"))
{   
    //TODO - Do task for match "A"             
}
if (flag.equals("B") || flag.equals("C")) // no else if
{
    //TODO - Do task for match "B"   
}

As long as "A" and "B" are not the same this should work just fine.

Kudos Rob !
logic before code

[Carey Brown]

John Jai wrote:

if(flag.equals("A")){
			//TODO - Do task for match "A"			
		}else if(flag.equals("B")){
			//TODO - Do task for match "B"
		}else if(flag.equals("C")){
			//TODO - Do task for match "A" and match "B"
		}

int BIT_A = 1;
int BIT_B = 2;
int bits = 0;
if( flag.equals("A") ) bits |= BIT_A;
if( flag.equals("B") ) bits |= BIT_B;
if( flag.equals("C") ) bits |= (BIT_A|BIT_B);

if( (bits & BIT_A) != 0 )
    //TODO task for "A"
if( (bits & BIT_B) != 0 )
    //TODO task for "B"

Bit fiddling is not for everyone but this approach can work for much more complicated rules.

[Rob Spoor]

That's basically what I already did except with bits.