I have two tables in my Access-database. They look something like this:
访问数据库中有两个表。它们看起来是这样的:
Table1
+--------------+----------+----------+----------+
| Kabelnummer | Column1 | Column2 | Column3 |
+--------------+----------+----------+----------+
| 1 | x | x | x |
+--------------+----------+----------+----------+
| 2 | x | x | x |
+--------------+----------+----------+----------+
| 3 | x | x | x |
+--------------+----------+----------+----------+
| 4 | x | x | x |
+--------------+----------+----------+----------+
table2
+--------------+----------+----------+----------+
| Kabelnummer | Column1 | Column2 | Column3 |
+--------------+----------+----------+----------+
| 1 | x | x | x |
+--------------+----------+----------+----------+
| 2 | x | x | x |
+--------------+----------+----------+----------+
| 3 | x | x | x |
+--------------+----------+----------+----------+
| 4 | x | x | x |
+--------------+----------+----------+----------+
I need a query that gives me 1 table with the data from table1 added to the data from table2:
我需要一个查询，将表1的数据添加到表2的数据中，得到1个表:
TableTotal
+--------------+----------+----------+----------+
| Kabelnummer | Column1 | Column2 | Column3 |
+--------------+----------+----------+----------+
| 1 | x | x | x |
+--------------+----------+----------+----------+
| 2 | x | x | x |
+--------------+----------+----------+----------+
| 3 | x | x | x |
+--------------+----------+----------+----------+
| 4 | x | x | x |
+--------------+----------+----------+----------+
| 1 | x | x | x |
+--------------+----------+----------+----------+
| 2 | x | x | x |
+--------------+----------+----------+----------+
| 3 | x | x | x |
+--------------+----------+----------+----------+
| 4 | x | x | x |
+--------------+----------+----------+----------+
The names "Column1", "Column2" and "Column3" are the same in both tables
两个表中的“Column1”、“Column2”和“Column3”的名称是相同的
2 个解决方案
#1
7
SELECT *
FROM Table1
UNION
SELECT *
FROM table2;
#2
0
The question asks for non-distinct values while the current answers provide distinct values. The method below provides non-distinct values such that
这个问题要求非明显的值，而当前的答案提供不同的值。下面的方法提供了非不同的值
SELECT *
FROM Table1
UNION ALL
SELECT *
FROM table2;
which is often more efficient than the union method, particularly with large data sets (not having to compute the distinct).
这通常比union方法更有效，特别是对于大型数据集(不需要计算不同的数据集)。
#1
7
SELECT *
FROM Table1
UNION
SELECT *
FROM table2;
#2
0
The question asks for non-distinct values while the current answers provide distinct values. The method below provides non-distinct values such that
这个问题要求非明显的值，而当前的答案提供不同的值。下面的方法提供了非不同的值
SELECT *
FROM Table1
UNION ALL
SELECT *
FROM table2;
which is often more efficient than the union method, particularly with large data sets (not having to compute the distinct).
这通常比union方法更有效，特别是对于大型数据集(不需要计算不同的数据集)。