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我在 Oracle 10gR2 中有一个现有表，我在其中添加了一个XMLTYPE列。现在，我需要用一个 XML 填充该字段，该 XML 由每行字段中的混合值组成，但我无法正确获取语法。下面是为每一行生成所需 XML 的正确工作的 SELECT 语句：

SELECT XMLELEMENT("TBLOrigs",
  XMLELEMENT("TBLOrig", XMLATTRIBUTES(Field1 AS "Field1"),
  XMLFOREST(Field2 AS "Field2", Field3 AS "Field3", Field4 AS "Field4", Field5 AS "Field5"),
    XMLELEMENT("Dets",
    XMLFOREST(Field6, Field7 AS "Field7")))) AS XmlCol
  FROM TBLOrigin Orig1
WHERE Field1 <= 10

我正在尝试的更新是这样的，但它不起作用：

UPDATE TBLOrigin
SET AUXFIELD = (
  SELECT XMLELEMENT("TBLOrigs",
  XMLELEMENT("TBLOrig", XMLATTRIBUTES(Field1 AS "Field1"),
  XMLFOREST(Field2 AS "Field2", Field3 AS "Field3", Field4 AS "Field4", Field5 AS "Field5"),
    XMLELEMENT("Dets",
    XMLFOREST(Field6, Field7 AS "Field7")))) AS XmlCol
  FROM TBLOrigin Orig1
  WHERE Orig1.Field1 = Orig2.Field1
  AND Orig1.Field2 = Orig2.Field2
  AND Orig1.Field3 = Orig2.Field3
  AND Orig1.Field4 = Orig2.Field4
  AND Orig1.Field5 = Orig2.Field5)
FROM TBLOrigin Orig2
WHERE Field1 <= 10

我正在尝试使用从其字段生成的 XML 内容更新每一行。我不断收到以下错误：

Erro na Linha de Comando:13 Coluna:47
Relatório de Erro:
Erro de SQL: ORA-00933: SQL command not properly ended
00933. 00000 -  "SQL command not properly ended"
*Cause:    
*Action:

作为参考，下面是 Transact Sql 中的工作更新

UPDATE dbo.TBLOrigin
SET AuxInfo = (SELECT 
  Field1 as "@Field1",
  Field2 as "Field2", 
  Field3 as "Field3", 
  Field4 as "Field4", 
  Field5 as "Field5", 
  Field6 as "Dets/Field6", 
  Field7 as "Dets/Field7"
FROM dbo.TBLOrigin AS Orig1
WHERE Orig1.Field1 = Orig2.Field1
AND Orig1.Field2 = Orig2.Field2
AND Orig1.Field3 = Orig2.Field3
AND Orig1.Field4 = Orig2.Field4
AND Orig1.Field5 = Orig2.Field5
  FOR XML PATH('TBLOrig'), TYPE, ROOT('TBLOrigs'))
FROM dbo.TBLOrigin Orig2

非常感谢任何帮助

4

1 回答 1

1

Oracle 不执行 UPDATE...FROM

以下工作吗？

UPDATE TBLOrigin
SET AUXFIELD = 
       XMLELEMENT("TBLOrigs",
         XMLELEMENT("TBLOrig", XMLATTRIBUTES(Field1 AS "Field1"),
         XMLFOREST(Field2 AS "Field2", Field3 AS "Field3", 
                   Field4 AS "Field4", Field5 AS "Field5"),
         XMLELEMENT("Dets",
            XMLFOREST(Field6, Field7 AS "Field7")))) AS XmlCol
WHERE Field1 <= 10

如果没有，您可以发布表定义。

于 2011-03-03T02:32:38.543 回答