Mysql查看相似的条件

Question

我有一个数据库，保存一些人的信息。

我发送以下查询：

(select 'NUMBER OF BRAND PARTNERS ACTIVE PROMOTERS' AS `NUMBER OF ACTIVE PROMOTERS`,count(*) AS `COUNT(*)` from `Registration_Summary` where promoter not like "bw%") 

和它的作品完美，但每当我创建一个视图来保存这个命令时，“凡子不喜欢‘BW％’”部分被完全忽略，为什么？！这么奇怪 ？！仿佛视图不能有“喜欢”的状态！

我的观点的命令是：

CREATE VIEW test AS 
(select 'NUMBER OF BRAND PARTNERS ACTIVE PROMOTERS' AS `NUMBER OF ACTIVE PROMOTERS`,count(*) AS `COUNT(*)` from `Registration_Summary` where promoter not like "bw%") 

---

我想我的问题应该是我如何创建基于一个“其中像”语句另一种观点认为一个看法？

Answer 1

我认为别的可能会出错。复制你的问题，如：

create table `Registration_Summary` (
    id integer primary key, 
    promoter varchar(200) 
); 

insert into `Registration_Summary` (id, promoter) values (1, 'bar1'); 
insert into `Registration_Summary` (id, promoter) values (2, 'bar2'); 
insert into `Registration_Summary` (id, promoter) values (3, 'bar3'); 
insert into `Registration_Summary` (id, promoter) values (4, 'foo1'); 
insert into `Registration_Summary` (id, promoter) values (5, 'foo1'); 
insert into `Registration_Summary` (id, promoter) values (6, 'foo1'); 
insert into `Registration_Summary` (id, promoter) values (7, 'foo1'); 
insert into `Registration_Summary` (id, promoter) values (8, 'afoo1'); 
insert into `Registration_Summary` (id, promoter) values (9, 'bfoo1'); 
insert into `Registration_Summary` (id, promoter) values (10, 'cfoo1'); 

从登记表

select * from `Registration_Summary` where promoter like 'foo%'; 

选择将产生如下输出

+----+----------+ 
| id | promoter | 
+----+----------+ 
| 4 | foo1  | 
| 5 | foo1  | 
| 6 | foo1  | 
| 7 | foo1  | 
+----+----------+ 

和

select * from `Registration_Summary` where promoter not like 'foo%'; 

会产生以下

+----+----------+ 
| id | promoter | 
+----+----------+ 
| 1 | bar1  | 
| 2 | bar2  | 
| 3 | bar3  | 
| 8 | afoo1 | 
| 9 | bfoo1 | 
| 10 | cfoo1 | 
+----+----------+ 

所以创建视图像你有

create or replace view testview as (select 'NUMBER OF BRAND PARTNERS ACTIVE PROMOTERS' AS `NUMBER OF ACTIVE PROMOTERS`,count(*) AS `COUNT(*)` from `Registration_Summary` where promoter not like 'foo%'); 

，然后从视图

SELECT * FROM testview; 

选择产生正确的输出是

+-------------------------------------------+------------+ 
| NUMBER OF ACTIVE PROMOTERS    | COUNT(*) | 
+-------------------------------------------+------------+ 
| NUMBER OF BRAND PARTNERS ACTIVE PROMOTERS |   6 | 
+-------------------------------------------+------------+ 

Answer 2

我想我知道什么是问题。我在另一个视图上发布where条件而不是表格，我认为这就是为什么我没有得到正确的答案。我对吗 ？

我还是没有解决问题，但是当我使用它的工作表中的SELECT命令创建视图，但是当我用它来从另一个视图中选择，它使用Where语句时给出了错误的答案。

Answer 3

回复旧线程是因为我遇到了可能是相同的问题。

DROP TABLE IF EXISTS like_test; 

CREATE TABLE like_test (
    id int(11) unsigned NOT NULL AUTO_INCREMENT COMMENT 'Unique ID' 
    ,textvalue varchar(64) COLLATE ucs2_bin NOT NULL 
    ,PRIMARY KEY (id) 
) ENGINE=InnoDB CHARSET=ucs2 COLLATE=ucs2_bin; 

INSERT INTO like_test (textvalue) 
VALUES 
('foofoo') 
,('foobar') 
,('barfoo') 
,('barbar'); 

DROP VIEW IF EXISTS view_like_test; 

CREATE VIEW view_like_test AS 
SELECT 
    id 
    ,textvalue 
FROM like_test 
WHERE textvalue NOT LIKE 'foo%'; 

SHOW CREATE VIEW view_like_test; 

MySQL的内部的mangling后，视图已经剥离LIKE条件：

CREATE 
    ALGORITHM = UNDEFINED 
    DEFINER = `symbols_mgr`@`%` 
    SQL SECURITY DEFINER 
VIEW `view_like_test` AS 
    select 
     `like_test`.`id` AS `id`, 
     `like_test`.`textvalue` AS `textvalue` 
    from 
     `like_test` 
    where 
     (not ((`like_test`.`textvalue` like ''))) 

的字符集/归类导致要被清空的LIKE条件 - 所以或许OP也有类似的表结构。

我在版本5.1 - 也许这从那时起停止成为一个问题。有一个快速去5。5等条件改变这一点：

(not ((`like_test`.`textvalue` like '\0\0\0f\0\0\0o\0\0\0o\0\0\0%'))) 

如果它是可行的，使得数据更加香草为MySQL（例如拉丁/ ASCII）可能是解决该问题的最简单的方法。或者你可以这样改变视图：

(not ((`like_test`.`textvalue` like 'foo%' collate utf8_bin))) 

...你在指定排序规则。