1

使用 oracle 10 我有一个到 sql server 的链接。我有一个查询（如下）运行良好，就像一个选择语句但是当我使用它来创建一个视图时（视图创建成功但是当我尝试在 sql developer 中查看视图中的数据时）我得到一个

数据库“链接名称”在此上下文中不支持 ROWID

问题是我将一个表中的多行组合成多列的子选择吗？

  SELECT
    upper(cus."cus_last_name")  AS lname,
    upper(cus."cus_first_name") AS fname,
    upper(cus."cus_initial")    AS minitial,
    upper(cus."maiden_name")   AS olname,
    pt."cus_dob"               AS dob,
    pt."cus_id"              AS id,
    (
      SELECT
        cus_inst_key."cus_key_value"
      FROM
        "cus_inst_key"@linkname cus_inst_key
      WHERE
        cus_inst_key."cus_id"             = cus."cus_id"
      AND cus_inst_key."cus_key_cd"       = 1
    ) AS CUSALTID,
    (
      SELECT
        pcus_inst_key."cus_key_value"
      FROM
        "cus_inst_key"@linkname cus_inst_key
      WHERE
        cus_inst_key."cus_id"             = cus."cus_id"
      AND cus_inst_key."cus_key_cd"       = 2
     ) AS CUSALTID2,
    (
      SELECT
        cus_inst_key."cus_key_value"
      FROM
        "cus_inst_key"@linkname cus_inst_key
      WHERE
        cus_inst_key."cus_id"             = cus."cus_id"
      AND cus_inst_key."cus_key_cd"       = 3
    ) AS CUSALTID3,
    (
      SELECT
        cus_inst_key."cus_key_value"
      FROM
        "cus_inst_key"@linkname cus_inst_key
      WHERE
        cus_inst_key."cus_id"             = cus."cus_id"
      AND cus_inst_key."cus_key_cd"       = 4
    ) AS CUSALTID4
  FROM
    "cus"@linkname cus;

谢谢你的帮助。

视图实际上是创建好的，由于某种原因，我只是不能在 sql developer 中使用数据选项卡。

4

0 回答 0