更改MySQL視圖中的select語句

我有一個叫wallcitiescouncils視圖，從這個視圖返回值的過程。我需要更快的respose，因此我想將此視圖更改爲Select語句，但目前沒有成功。更改MySQL視圖中的select語句

我需要的結果形成了兩個表，城市和議會，這是我的看法代碼：

CREATE OR REPLACE VIEW wallcitiescouncils AS 
SELECT 
    concat('p',cities.id) as id, 
    cities.name as name, 
    cities.country_id, 
    concat(cities.name,' (',councils.name,'), ',states.name,', ',countries.name) as fullname, 
    'City' as type 

FROM cities 
    JOIN countries ON (cities.country_id = countries.id) 
    JOIN states  ON (cities.state_id = states.id) 
    JOIN councils ON (cities.council_id = councils.id) 

UNION ALL 

SELECT 
    concat('c',councils.id) as id, 
    councils.name as name, 
    councils.country_id, 
    concat(councils.name,'(Council)') as fullname, 
    'Council' as type 

FROM councils

而這一次我的程序：

BEGIN 
    SELECT name as city, id as city_id, fullname 
    FROM wallcitiescouncils 
    WHERE country_id = _country_id AND name LIKE CONCAT(search , '%') ORDER BY name LIMIT _limit; 
END

我需要改變FROM wallcitiescouncils SELECT語句所以我不必調用視圖。

這是我的一個嘗試，但結果不正確（我沒有理事會的結果，並在'全名'總是得到相同的理事會名稱），我使用councils_id（只存在於城市表）嘗試知道，如果是一個城市或一個城市coucil：

SELECT 
    if(
     cities.council_id is null, 
     concat('c',councils.id), 
     concat('p',cities.id) 
     ) as id, 
    if(
     cities.council_id is null, 
     councils.name, 
     cities.name 
     ) as name, 

    if(
     cities.council_id is null, 
     councils.country_id, 
     cities.country_id 
     ) as country_id, 
    if(
     cities.council_id is null, 
     concat(councils.name,' (Municipio)'), 
     concat(cities.name,' (',councils.name,'), ',states.name,', ',countries.name) 
     ) as fullname 
FROM councils, cities 
    JOIN countries ON (country_id = countries.id) 
    JOIN states  ON (state_id = states.id) 
    /*JOIN councils ON (cities.council_id = councils.id)*/ 
where cities.id = 1 or councils.id = 1 limit 200

來源

2017-04-11 kikerrobles

爲什麼你認爲改變視圖爲select語句會做什麼工作得更快？你的查詢中有一個'LIKE'，你沒有發佈'EXPLAIN'的輸出，所以你是如何得出瓶頸在哪裏的？ – Mjh

[MySQL視圖作爲性能麻煩製造者（https://www.percona.com/blog/2007/08/12/mysql-view-as-performance-troublemaker/） – Devart

@Devart你連接一個10歲的文章。有一種確定MySQL瓶頸的標準方法，每個問題都會重複一遍。讓我們這樣做，並獲得正確的信息，而不是猜測。 – Mjh

嘗試此查詢firstly-

SELECT name as city, id as city_id, fullname FROM (
    SELECT 
    concat('p',cities.id) as id, 
    cities.name as name, 
    cities.country_id, 
    concat(cities.name,' (',councils.name,'), ',states.name,', ',countries.name) as fullname, 
    'City' as type 

    FROM cities 
    JOIN countries ON (cities.country_id = countries.id) 
    JOIN states  ON (cities.state_id = states.id) 
    JOIN councils ON (cities.council_id = councils.id) 

    UNION ALL 

    SELECT 
    concat('c',councils.id) as id, 
    councils.name as name, 
    councils.country_id, 
    concat(councils.name,'(Council)') as fullname, 
    'Council' as type 
    FROM councils 
) v 
WHERE 
    v.country_id = _country_id AND v.name LIKE CONCAT(search , '%') ORDER BY v.name LIMIT _limit;

來源

2017-04-11 15:46:26 Devart

更改MySQL視圖中的select語句

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