CREATE AS SELECT * 但从另一个表中获得一列

Question

我需要使用CREATE AS SELECT 语句“重新创建”超过 50 个表（在 Oracle 中）。但是，所有这些表都将使用另一表中的数据修改一列。有没有一种方法可以在不声明 SELECT 语句中的每一列的情况下实现这一点？

类似：

CREATE TABLE table_name_copy AS SELECT *, (SELECT col_name FROM other_table WHERE other_table.col_id = table_name.col_id) AS col_name FROM table_name`

基本上在所有表上，我都有一个需要用other_table 列中的数据替换的列。

Answer 1

这样生成 SQL 字符串：

SELECT 'CREATE TABLE table_name_copy AS SELECT '
    || LISTAGG (column_name, ', ') WITHIN GROUP (ORDER BY column_name)
    || ', (SELECT col_name FROM other_table 
           WHERE other_table.col_id = table_name.col_id) AS col_name'
    || ' FROM table_name'
FROM all_tab_cols
WHERE owner = 'OWNER'
AND table_name = 'TABLE_NAME'
AND column_name != 'COL_NAME'

如果你想运行上面的语句，你可以使用EXECUTE IMMEDIATE:

DECLARE
  v_sql VARCHAR2(10000);
BEGIN
  SELECT 'CREATE TABLE table_name_copy AS SELECT '
      || LISTAGG (column_name, ', ') WITHIN GROUP (ORDER BY column_name)
      || ', (SELECT col_name FROM other_table 
             WHERE other_table.col_id = table_name.col_id) AS col_name'
      || ' FROM table_name'
  INTO v_sql
  FROM all_tab_cols
  WHERE owner = 'OWNER'
  AND table_name = 'TABLE_NAME'
  AND column_name != 'COL_NAME';

  EXECUTE IMMEDIATE v_sql;
END;
/

Answer 2

如果col_id 列对于两个连接的表都是固定的，

您可以通过模式使用user_tab_columns 和user_tables 字典视图通过以下机制生成名为“table_name_copy”的新表：

declare
 v_ddl varchar2(4000); 
 v_cln varchar2(400); 
begin
 for c in ( select * 
              from user_tables t 
             where t.table_name in
             ( select c.table_name 
                 from user_tab_columns c
                where c.column_name = 'COL_ID'  ) 
             order by t.table_name )
 loop    
   v_ddl := 'create table '||c.table_name||'_copy as 
                      select ';
   for d in ( select listagg('t1.'||column_name, ',') within group ( order by column_name ) cln 
                from user_tab_columns 
               where table_name = c.table_name 
                 and column_name != 'COL_ID' )                   
   loop
    v_cln := v_cln||d.cln;  
   end loop;
       v_ddl := v_ddl||v_cln;                       
       v_ddl := v_ddl||', t2.col_id t2_id 
        from '||c.table_name||' t1 
        left outer join other_table t2 on ( t1.col_id = t2.col_id )';

       execute immediate v_ddl;                    
       v_ddl := null;
       v_cln := null;
 end loop;
end;

Answer 3

也许您可以使用简单的连接和星号来返回第一个表中的所有列，如下所示：

CREATE TABLE table_name_copy AS 
  SELECT * FROM (
    SELECT tab1.*, tab2.column_name 
    FROM table_name tab1 LEFT JOIN other_table tab2 ON tab1.col_id = tab2.col_id
  );

Answer 4

我会试试这个（但我没有要测试的 Oracle SQL，所以请不要怀疑）

CREATE TABLE table_name_copy AS 

SELECT * FROM (
               SELECT *, (SELECT col_name FROM other_table WHERE other_table.col_id = table_name.col_id) as col_name 

               FROM table_name`
               )

编辑： 然后运行

ALTER TABLE table_name_copy DROP COLUMN <old column>
删除不再需要的列