在 T-SQL 中使用任意数量的参数

A basic proof-of-concept... Actual code would be less, but since I don't know your table/field names, this is the full code, so anyone can verify it works, tweak it, etc.

--Search Parameters

DECLARE @SearchString VARCHAR(MAX)
SET @SearchString='Oakland;City;Planning' --Using your example search
DECLARE @Delim CHAR(1)
SET @Delim=';' --Using your deliminator from the example

--I didn't know your table name, so I'm making it... along with a few extra rows...

DECLARE @Projects TABLE (ProjectID INT, ProjectName VARCHAR(200))
INSERT INTO @Projects (ProjectID, ProjectName) SELECT 1, 'Oakland City Planning'
INSERT INTO @Projects (ProjectID, ProjectName) SELECT 2, 'Oakland City Construction'
INSERT INTO @Projects (ProjectID, ProjectName) SELECT 3, 'Skunk Works'
INSERT INTO @Projects (ProjectID, ProjectName) SELECT 4, 'Oakland Town Hall'
INSERT INTO @Projects (ProjectID, ProjectName) SELECT 5, 'Oakland Mall'
INSERT INTO @Projects (ProjectID, ProjectName) SELECT 6, 'StackOverflow Answer Planning'

--*** MAIN PROGRAM CODE STARTS HERE ***

DECLARE @Keywords TABLE (Keyword VARCHAR(MAX))

DECLARE @index int 
SET @index = -1 

--Each keyword gets inserted into the table
--Single keywords are handled, but I did not add code to remove duplicates
--since that affects performance only, not the result.

WHILE (LEN(@SearchString) > 0) 
  BEGIN  
    SET @index = CHARINDEX(@Delim , @SearchString)  
    IF (@index = 0) AND (LEN(@SearchString) > 0)  
      BEGIN   
        INSERT INTO @Keywords VALUES (@SearchString)
          BREAK  
      END  
    IF (@index > 1)  
      BEGIN   
        INSERT INTO @Keywords VALUES (LEFT(@SearchString, @index - 1))   
        SET @SearchString = RIGHT(@SearchString, (LEN(@SearchString) - @index))  
      END  
    ELSE 
      SET @SearchString = RIGHT(@SearchString, (LEN(@SearchString) - @index)) 
END


--This way, only a project with all of our keywords will be shown...

SELECT * 
FROM @Projects
WHERE ProjectID NOT IN (SELECT ProjectID FROM @Projects Projects INNER JOIN @Keywords Keywords ON CHARINDEX(Keywords.Keyword,Projects.ProjectName)=0)

我决定将几个不同的答案混合在一起:-P