How do I check if a Sql server string is null or empty

I want to check for data but ignore it if it's null or empty. Currently the query is as follows...

Select              
Coalesce(listing.OfferText, company.OfferText, '') As Offer_Text,         
from tbl_directorylisting listing  
 Inner Join tbl_companymaster company            
  On listing.company_id= company.company_id

But I want to get company.OfferText if listing.Offertext is an empty string, as well as if it's null.

What's the best performing solution?

From stackoverflow

• SELECT
     CASE WHEN LEN(listing.OfferText) > 0 THEN listing.OfferText 
          ELSE COALESCE(Company.OfferText, '') END 
     AS Offer_Text,
  
  ...
  

  In this example, if listing.OfferText is NULL, the LEN() function should also return NULL, but that's still not > 0.

• Select              
  CASE
      WHEN listing.OfferText is null or listing.OfferText = '' THEN company.OfferText
      ELSE COALESCE(Company.OfferText, '')
  END As Offer_Text,         
  from tbl_directorylisting listing  
   Inner Join tbl_companymaster company            
    On listing.company_id= company.company_id
  

• You can use ISNULL and check the answer against the known output:

  SELECT case when ISNULL(col1, '') = '' then '' else col1 END AS COL1 FROM TEST

• Here's a solution, but I don't know if it's the best....

  Select              
  Coalesce(Case When Len(listing.Offer_Text) = 0 Then Null Else listing.Offer_Text End, company.Offer_Text, '') As Offer_Text,         
  from tbl_directorylisting listing  
   Inner Join tbl_companymaster company            
    On listing.company_id= company.company_id
  

• Here is another solution:

  SELECT Isnull(Nullif(listing.offertext, ''), company.offertext) AS offer_text, 
  FROM   tbl_directorylisting listing 
         INNER JOIN tbl_companymaster company 
           ON listing.company_id = company.company_id
  

• I think this:

  SELECT 
    ISNULL(NULLIF(listing.Offer_Text, ''), company.Offer_Text) AS Offer_Text
  FROM ...
  

  is the most elegant solution.