如何排序在SQL Server

Question

通過其數字varchar數據類型列我在SQL Server中包含這樣的數據的列：

1.1.1.QuestionText 
1.1.1.Question 
1.1.1(a).Questions 
1.1.2.Questionswithtext 
1.1.2(b).Text 
10.1.1.Answers 
2.1.1.Questions 
2.2.2.QuestionText 

如何按升序排列顯示這個？

Answer 1

你發生有一個四部分命名，所以你可以使用parsename()。結果如下：

order by (case when isnumeric(arsename(col, 4)) = 1 then cast(parsename(col, 4) as int) 
      end), 
     (case when isnumeric(parsename(col, 3)) = 1 then cast(parsename(col, 3) as int) 
      end), 
     (case when isnumeric(parsename(col, 2)) = 1 then cast(parsename(col, 2) as int) 
      end), 
     (case when isnumeric(parsename(col, 1)) = 1 then cast(parsename(col, 1) as int) 
      end), 
     col 

這並不完美，因爲您似乎有奇怪的放置圓括號。但它可能足以滿足你的目的。

Answer 2

試試這個：

 CREATE TABLE #TMP 
     (
     name nvarchar(30) 
     ) 

     INSERT INTO #TMP (name) VALUES ('1.1.1.QuestionText') 
     INSERT INTO #TMP (name) VALUES ('1.1.1.Question') 
     INSERT INTO #TMP (name) VALUES ('1.1.1(a).Questions') 
     INSERT INTO #TMP (name) VALUES ('1.1.2.Questionswithtext') 
     INSERT INTO #TMP (name) VALUES ('1.1.2(b).Text') 
     INSERT INTO #TMP (name) VALUES ('10.1.1.Answers') 
     INSERT INTO #TMP (name) VALUES ('2.1.1.Questions') 
     INSERT INTO #TMP (name) VALUES ('2.2.2.QuestionText') 

     select 
     SUBSTRING(replace(name, '.',''),1, 
       PATINDEX('%[^0-9]%', 
        replace(name, '.','') 
       ) -1 
      ) as replaced, name from #TMP order by 
     CAST( 
      SUBSTRING(replace(name, '.',''),1, 
       PATINDEX('%[^0-9]%', 
        replace(name, '.','') 
       ) -1 
      ) 
     as int) 

它的工作原理總是當它從數字開始。它僅由此第一個數字排序。如果您想通過數字第一和文本的其餘命令後，你可以（在順序我把在選擇同樣喜歡展現訂購零件）更改爲上次查詢：

 select 
     SUBSTRING(replace(name, '.',''),1, 
       PATINDEX('%[^0-9]%', 
        replace(name, '.','') 
       ) -1 
      ) as replacedDigit, 
      SUBSTRING(replace(name, '.',''), 
       PATINDEX('%[^0-9]%', 
        replace(name, '.','') 
       ), 100) as replacedText, 

      name from #TMP order by 
     CAST( 
      SUBSTRING(replace(name, '.',''),1, 
       PATINDEX('%[^0-9]%', 
        replace(name, '.','') 
       ) -1 
      ) 
     as int), SUBSTRING(replace(name, '.',''), 
       PATINDEX('%[^0-9]%', 
        replace(name, '.','') 
       ), 100) 

Answer 3

試試這個

SELECT * 
FROM t 
ORDER BY 
     CASE PATINDEX('%[^0-9]%', PARSENAME(name, 4)) 
      WHEN 0 THEN RIGHT('0000' + PARSENAME(name, 4), 4) 
      WHEN 1 THEN PARSENAME(name, 1) 
      ELSE RIGHT('0000' + SUBSTRING(PARSENAME(name, 2), 1, PATINDEX('%[^0-9]%', PARSENAME(name, 2)) - 1), 4) + SUBSTRING(PARSENAME(name, 2), PATINDEX('%[^0-9]%', PARSENAME(name, 2)), LEN(name)) 
     END 
    , CASE PATINDEX('%[^0-9]%', PARSENAME(name, 3)) 
      WHEN 0 THEN RIGHT('0000' + PARSENAME(name, 3), 4) 
      WHEN 1 THEN PARSENAME(name, 1) 
      ELSE RIGHT('0000' + SUBSTRING(PARSENAME(name, 2), 1, PATINDEX('%[^0-9]%', PARSENAME(name, 2)) - 1), 4) + SUBSTRING(PARSENAME(name, 2), PATINDEX('%[^0-9]%', PARSENAME(name, 2)), LEN(name)) 
     END 
    , CASE PATINDEX('%[^0-9]%', PARSENAME(name, 2)) 
      WHEN 0 THEN RIGHT('0000' + PARSENAME(name, 2), 4) 
      WHEN 1 THEN PARSENAME(name, 1) 
      ELSE RIGHT('0000' + SUBSTRING(PARSENAME(name, 2), 1, PATINDEX('%[^0-9]%', PARSENAME(name, 2)) - 1), 4) + SUBSTRING(PARSENAME(name, 2), PATINDEX('%[^0-9]%', PARSENAME(name, 2)), LEN(name)) 
     END 
    , CASE PATINDEX('%[^0-9]%', PARSENAME(name, 1)) 
      WHEN 0 THEN RIGHT('0000' + PARSENAME(name, 1), 4) 
      WHEN 1 THEN PARSENAME(name, 1) 
      ELSE RIGHT('0000' + SUBSTRING(PARSENAME(name, 2), 1, PATINDEX('%[^0-9]%', PARSENAME(name, 2)) - 1), 4) + SUBSTRING(PARSENAME(name, 2), PATINDEX('%[^0-9]%', PARSENAME(name, 2)), LEN(name)) 
     END 

對於這一點：

1.1.1.Question 
1.1.1.QuestionText 
1.1.1(a).Questions 
1.1.2.Questionswithtext 
1.1.2(b).Text 
2.1.1.Questions 
2.2.2.QuestionText 
10.1.1.Answers 

Answer 4

這應該工作：

SELECT name 
FROM (SELECT name, SUBSTRING(name, 1, 
           LEN(name) - PATINDEX('%[0-9]%', REVERSE(name)) + 1) n 
      FROM  #tmp 
     ) v 
ORDER BY CAST(PARSENAME(n, 4) AS INT), 
     CAST(PARSENAME(n, 3) AS INT), 
     CAST(PARSENAME(n, 2) AS INT), 
     CAST(PARSENAME(n, 1) AS INT) 

輸出：

name 
1.1.1.QuestionText 
1.1.1.Question 
1.1.1(a).Questions 
1.1.2.Questionswithtext 
1.1.2(b).Text 
2.1.1.Questions 
2.2.2.QuestionText 
10.1.1.Answers 

如果要通過後版本號的文本命令，那麼這將工作：

SELECT name 
FROM (SELECT name, SUBSTRING(name, 1, 
           LEN(name) - PATINDEX('%[0-9]%', REVERSE(name)) + 1) nv, 
          SUBSTRING(name, LEN(name) - PATINDEX('%[0-9]%', REVERSE(name)) + 2, 
           LEN(name)) nt        
      FROM  #tmp 
     ) v 
ORDER BY CAST(PARSENAME(nv, 4) AS INT), 
     CAST(PARSENAME(nv, 3) AS INT), 
     CAST(PARSENAME(nv, 2) AS INT), 
     CAST(PARSENAME(nv, 1) AS INT), 
     nt 

輸出：

name 
1.1.1(a).Questions 
1.1.1.Question 
1.1.1.QuestionText 
1.1.2(b).Text 
1.1.2.Questionswithtext 
2.1.1.Questions 
2.2.2.QuestionText 
10.1.1.Answers