Mysql查看相似的條件

Question

我有一個數據庫，保存一些人的信息。

我發送以下查詢：

(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，但每當我創建一個視圖來保存這個命令時，「凡子不喜歡‘BW％’」部分被完全忽略，爲什麼？！這麼奇怪 ？！彷彿視圖不能有「喜歡」的狀態！

我的觀點的命令是：

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%") 

---

我想我的問題應該是我如何創建基於一個「其中像」語句另一種觀點認爲一個看法？

Answer 1

我認爲別的可能會出錯。複製你的問題，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'); 

從登記表

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

選擇將產生如下輸出

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

和

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

會產生以下

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

所以創建視圖像你有

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%'); 

，然後從視圖

SELECT * FROM testview; 

選擇產生正確的輸出是

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

Answer 2

我想我知道什麼是問題。我在另一個視圖上發佈where條件而不是表格，我認爲這就是爲什麼我沒有得到正確的答案。我對嗎 ？

我還是沒有解決問題，但是當我使用它的工作表中的SELECT命令創建視圖，但是當我用它來從另一個視圖中選擇，它使用Where語句時給出了錯誤的答案。

Answer 3

回覆舊線程是因爲我遇到了可能是相同的問題。

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的內部的mangling後，視圖已經剝離LIKE條件：

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/歸類導致要被清空的LIKE條件 - 所以或許OP也有類似的表結構。

我在版本5.1 - 也許這從那時起停止成爲一個問題。有一個快速去5。5等條件改變這一點：

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

如果它是可行的，使得數據更加香草爲MySQL（例如拉丁/ ASCII）可能是解決該問題的最簡單的方法。或者你可以這樣改變視圖：

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

...你在指定排序規則。