我寫了一個查詢,并從出生日期開始,我計算出一個人的年齡,然后使用AS年齡創建一個年齡字段.

我的問題是,是否有可能再次匹配那個年齡段？

像這樣的東西,

SELECT `candidates`.`candidate_id`,

`candidates`.`first_name`,

`candidates`.`surname`,

`candidates`.`DOB`,

`candidates`.`gender`,

DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(`candidates`.`DOB`, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(`candidates`.`DOB`, '00-%m-%d')) AS `age`

FROM `candidates`

WHERE `age` <= 20

任何幫助將不勝感激？

—–所以我把WHERE更改為HAVING,這是我的完整查詢——

SELECT `candidates`.`candidate_id`,

`candidates`.`first_name`,

`candidates`.`surname`,

`candidates`.`DOB`,

`candidates`.`gender`,

`candidates`.`talent`,

`candidates`.`location`,

`candidates`.`availability`,

`candidate_assets`.`url`,

`candidate_assets`.`asset_size`

FROM `candidates`

LEFT JOIN `candidate_assets` ON `candidate_assets`.`candidates_candidate_id` = `C`.`candidate_id`

WHERE `C`.`availability` = 'yes'

AND C.talent = "actor"

AND C.skill = "accents"

AND C.gender = "male"

AND HAVING DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(`candidates`.`DOB`, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(`candidates`.`DOB`, '00-%m-%d')) <= 69

AND HAVING DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(`candidates`.`DOB`, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(`candidates`.`DOB`, '00-%m-%d')) <=19

AND HAVING DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(`candidates`.`DOB`, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(`candidates`.`DOB`, '00-%m-%d')) <=49

我收到以下錯誤,

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘HAVING DATE_FORMAT(NOW(), ‘%Y’) – DATE_FORMAT(candidates.DOB, ‘%Y’) – (DATE_’ at line 15

對于我的生活,我不知道它是什么.

解決方法:

一種方法是將原始查詢包裝為子查詢,并將WHERE子句移動到外部查詢：

SELECT * FROM (

SELECT `candidates`.`candidate_id`, -- this is the original query

`candidates`.`first_name`,

`candidates`.`surname`,

`candidates`.`DOB`,

`candidates`.`gender`,

DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(`candidates`.`DOB`, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(`candidates`.`DOB`, '00-%m-%d')) AS `age`

FROM `candidates` ) as innertable

WHERE `age` <= 20 -- this is now part of the outer query

澄清：這可以在MySQL 5中運行

注意：這假設原始查詢有效

標簽：mysql,sql

來源： https://codeday.me/bug/20190530/1185794.html

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