如何在SQL Server中將平臺轉換爲層次結構XML？

我有一個平坦的表是這樣的：如何在SQL Server中將平臺轉換爲層次結構XML？

DIV_ID DIV_NAME MARKET_ID MARKET_NAME TRANSIT_ID TRANS_NAME 

D1  "D1 NAME" M1  "M1 NAME" T1  "T1 NAME" 
D1  "D1 NAME" M1  "M1 NAME" T2  "T2 NAME"

想將它與層次結構轉換爲XML，一個不重複的父母：

<DIVISIONS> 
<DIVISION DIV_ID="D1" DIV_NAME="D1 NAME"> 
    <MARKET MARKET_ID="M1" MARKET_NAME="M1 NAME"> 
    <TRANSIT TRANSIT_ID="T1" TRANSIT_NAME="T1 NAME"> 
    <TRANSIT TRANSIT_ID="T1" TRANSIT_NAME="T1 NAME"> 
    </MARKET> 
    <MARKET MARKET_ID="M2" MARKET_NAME="M2 NAME"> 
    ... 
    </MARKET> 
    </DIVISION> 
<DIVISION DIV_ID="" DIV_NAME=""> 
... 
</DIVISION> 
</DIVISIONS>

看來SQL可以通過直接生成它「FOR XML「與一些參數。我嘗試了一些，但仍然無法得到我想要的。

任何幫助表示讚賞。謝謝！

編輯：我發現了一個解決方案，但SQL Server輸出'& lt;'和'& gt;'而不是嵌套元素中的「<」，「>」。這裏是我的代碼：

SELECT DISTINCT DIV_ID AS '@DIV_ID', DIV_NAME AS '@DIV_NAME', 
    (SELECT DISTINCT MARKET_ID AS '@MARKET_ID', MARKET_NAME AS '@MARKET_NAME', 
    (SELECT TRANSIT_ID AS '@TRANSIT_ID', TRANSIT_NAME AS '@TRANSIT_NAME' 
    FROM HX_DIVISIONS T3 
    WHERE T3.MARKET_ID = T2.MARKET_ID 
    FOR XML PATH ('TRANSIT') 
    ) TRANSITS 
    FROM HX_DIVISIONS T2 
    WHERE T2.DIV_ID = T1.DIV_ID 
    FOR XML PATH ('MARKET')) MARKETS 
FROM HX_DIVISIONS T1 
FOR XML PATH('DIVISION'), ROOT('DIVISIONS')

我試圖用REPLACE（XXX '& LT;'， '<'）的另一篇文章中建議，但它不工作。

輸出是這樣的：

<DIVISIONS> 
    <DIVISION DIV_ID="D1" DIV_NAME="D1 NAME"> 
    <MARKETS>&lt;MARKET MARKET_ID="M1" MARKET_NAME="M1 NAME"&gt;&amp;lt;TRANSITS&gt;&lt;TRANSIT TRANSIT_ID="T1" TRANSIT_NAME="T1 NAME"/&gt; 
.... 
</DIVISIONS>

來源

2016-01-20 abacusit

編輯你的問題，顯示你有什麼到目前爲止已經試過。 –

您的解決方案非常接近...您需要在FOR XML PATH後添加「，TYPE」才能將結果作爲XML獲取，而不是作爲嵌套的文本。但是，這樣做，DISTINCT將無法工作......在我的回答中，我首先使用CTE來獲取DISTINCT DIV_ID值並從那裏嵌套。而你最內在的SELECT（對於transits必須包含DIV_ID在... WHERE ... – Shnugo

嗨，我看到你剛剛在... ...這仍然是開放的嗎？你需要任何進一步的幫助嗎？ – Shnugo

試試這樣說：

DECLARE @tbl TABLE(DIV_ID VARCHAR(10), DIV_NAME VARCHAR(10), TRANSIT_ID VARCHAR(10), TRANSIT_NAME VARCHAR(10), MARKET_ID VARCHAR(10), MARKET_NAME VARCHAR(10)); 
INSERT INTO @tbl VALUES 
('D1','D1 NAME','T1','T1 NAME','M1','M1 NAME') 
,('D1','D1 NAME','T1','T1 NAME','M2','M2 NAME'); 

WITH DistinctDIVIDs AS 
(
    SELECT DISTINCT DIV_ID,DIV_NAME 
    FROM @tbl 
) 
SELECT DIVs.DIV_ID AS [@DIV_ID] 
     ,DIVs.DIV_NAME AS [@DIV_NAME] 
     ,(
      SELECT markets.MARKET_ID AS [@MARKET_ID] 
       ,markets.MARKET_NAME AS [@MARKET_NAME] 
       ,(
        SELECT transits.TRANSIT_ID AS [@TRANSIT_ID] 
          ,transits.TRANSIT_NAME AS [@TRANSIT_NAME] 
        FROM @tbl AS transits 
        WHERE transits.DIV_ID=DIVs.DIV_ID AND transits.MARKET_ID=markets.MARKET_ID 
        FOR XML PATH('TRANSIT'),TYPE 
       ) 
      FROM @tbl AS markets 
      WHERE markets.DIV_ID=DIVs.DIV_ID 
      FOR XML PATH('MARKET'),TYPE 
     ) 
FROM DistinctDIVIDs AS DIVs 
FOR XML PATH('DIVISION'),ROOT('DIVISIONS')

結果

<DIVISIONS> 
    <DIVISION DIV_ID="D1" DIV_NAME="D1 NAME"> 
    <MARKET MARKET_ID="M1" MARKET_NAME="M1 NAME"> 
     <TRANSIT TRANSIT_ID="T1" TRANSIT_NAME="T1 NAME" /> 
    </MARKET> 
    <MARKET MARKET_ID="M2" MARKET_NAME="M2 NAME"> 
     <TRANSIT TRANSIT_ID="T1" TRANSIT_NAME="T1 NAME" /> 
    </MARKET> 
    </DIVISION> 
</DIVISIONS>

來源

2016-01-21 11:19:41 Shnugo

如何在SQL Server中將平臺轉換爲層次結構XML？

回答

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