我试图使用case来计算特定文本何时包含varchar列。我一直试图破解它,但我有一个问题时,第二,第三种情况时,被添加。我使用的代码是:
SELECT *,
CASE
WHEN (Orders.Some_text LIKE "%test%" AND Orders.Some_text NOT LIKE "%found%") THEN 1
ELSE 0 END AS Test
FROM Orders;
现在我添加第二种情况:
SELECT *,
CASE
WHEN (Orders.Some_text LIKE "%test%" AND Orders.Some_text NOT LIKE "%found%") THEN 1
WHEN (Orders.Some_text LIKE "%test%" AND Orders.Some_text NOT LIKE "%missing%") THEN 1
ELSE 0 END AS Test
FROM Orders;
它在测试列中产生错误。
当单词测试被发现时,我想要的结果只是简单的1,并且它没有被发现、丢失或/和在测试过程中。
+------+--------------+------------+
| id | Some_text | Test |
+------+--------------+------------+
| 1 | test | 1 |
| 2 | test found | 0 |
| 3 | found test | 0 |
| 4 | test missing | 0 |
| 5 | missing test | 0 |
| 6 | test during | 0 |
| 7 | during test found | 0 |
| 8 | abc | 0 |
+------+--------------+------------+
复制数据集的代码:
CREATE TABLE Orders
(
id INT,
Some_text char(255));
insert into Orders values (1, "test");
insert into Orders values (2, "test found");
insert into Orders values (3, "found test");
insert into Orders values (4, "test missing");
insert into Orders values (5, "miss ing test");
insert into Orders values (6, "test during");
insert into Orders values (7, "during test found");
insert into Orders values (8, "abc");
2条答案
按热度按时间toe950271#
看来你得检查一下这两个词
found
以及missing
如果字段中没有,则将值设置为1。组合如下语句,它应该返回预期的输出。
输出
cidc1ykv2#
你的查询有冲突,
和
两者可能包含相同的结果。这在使用case循环时是不可接受的。