Question

我想使用UNION_ALL组合来自不同模式的表。这些表具有相同的模式，就像在这个玩具示例中一样：

class1.names
+----------+
| id | name|
+----------+
| 1  | jon |
| 2  | ann |
| 3  | rob |

class2.names
+----------+
| id | name|
+----------+
| 1  | rav |
| 2  | meg |
| 3  | mat |

我可以将类列表硬编码到数组中，或者更优选地，使用如下查询获取它们：

SELECT DISTINCT(TABLE_SCHEMA)
FROM INFORMATION_SCHEMA.TABLES 
WHERE TABLE_TYPE = 'BASE TABLE'

我想像这样组合表格：

SELECT *, 'class1' as class FROM class1.names
UNION_ALL
SELECT *, 'class2' as class FROM class2.names
UNION_ALL
etc.

但是模式级查询中会有比SELECT *, 'class1'...更多的内容，所以我想使用循环或其他系统方法来实现这一点。

我正在查看动态sql或使用GROUP_CONCAT与＆＃39; UNION_ALL＆＃39;作为分隔符，但我无法取得进展。

附录：我知道这是糟糕的架构设计，但我现在无法做任何事情。

Answer 1

如果我对您的理解正确：

select listagg('select ' || f.value, ' union all ') from table(flatten(input => parse_json(
                                                    '[1, ,77]'))) f;
+----------------------------------------------+
| LISTAGG('SELECT ' || F.VALUE, ' UNION ALL ') |
|----------------------------------------------|
| select 1 union all select 77                 |
+----------------------------------------------+
1 Row(s) produced. Time Elapsed: 0.739s

和：

select 1 union all select 77;
+----+
|  1 |
|----|
|  1 |
| 77 |
+----+
2 Row(s) produced. Time Elapsed: 0.638s

Answer 2

也许像这样：

  var grid = $("#tblPresentationsTable").data('kendoGrid')
                if(grid){
                    grid.destroy();
                    $("#tblPresentationsTable").empty()
                }


                $("#tblPresentationsTable tbody").html(tbodyPresentations)

                    $("#tblPresentationsTable").kendoGrid({
                        height: 550,
                        sortable: true,
                        toolbar: ["search"],
                        pageable: {
                            refresh: true,
                            pageSizes: true,
                            buttonCount: 5
                        }
                    });

Answer 3

在Snowflake中，此动态SQL：

with a as (
  select * from information_schema.tables 
  where table_catalog like 'CLASS%' and table_name = 'NAMES' and table_type = 'BASE TABLE'
),

b as (
  select *, 
    $$SELECT *, 'SCHEMA' as class FROM SCHEMA.names$$ as t,
    replace(t,'SCHEMA',lower(table_schema)) as sql,
  from a
)

select listagg(sql,'\nUNION ALL\n') within group (order by table_schema, table_catalog)
from b;

会产生：

SELECT *, 'class1' as class FROM class1.names
UNION ALL
SELECT *, 'class2' as class FROM class2.names
UNION ALL
etc.

$$$$$是字符串文字的单引号的替代方法。您也可以将单引号加倍来转义。

循环并将模式/表与UNION_ALL组合在一起

3 个答案: