SELECT u.users_username, u.givenname, u.familyname, u.studentassent, u.parentconsent, u.birthdate, u.gender
FROM users AS u
JOIN classes_users as c
ON c.users_username = u.users_username
JOIN classes_users as x
ON c.classes_id = x.classes_id
WHERE x.users_username = "johnny" AND x.role = "teacher"
SELECT u.users_username, u.givenname, u.familyname, u.studentassent, u.parentconsent, u.birthdate, u.gender
FROM users AS u
WHERE u.users_username
IN (
SELECT c.users_username
FROM classes_users as c
JOIN classes_users as x
ON c.classes_id = x.classes_id
WHERE x.users_username = "johnny" AND x.role = "teacher"
)
我想第一個是更好,但我仍然在學習如何寫出更好的SQL語句是不是發生了什麼,所有的內部明確,使一個更好的說法比在這種情況下的其他。
如果有更好的寫作方式比我寫的兩種方式,請讓我知道。謝謝。編輯: 有老師和學生。通過查看classes_users表可以找到他們在任何班級中作爲學生或教師的位置。我想要做的是,當給予用戶時,找到他是老師的班級,然後返回這些班級中的所有學生。
這裏是我的DB模式:
-- -----------------------------------------------------
-- Table `kcptech`.`users`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `kcptech`.`users` (
`users_username` VARCHAR(63) NOT NULL ,
`password` VARCHAR(255) NULL DEFAULT NULL ,
`salt` VARCHAR(127) NULL DEFAULT NULL ,
`givenname` VARCHAR(96) NULL DEFAULT NULL ,
`familyname` VARCHAR(128) NULL DEFAULT NULL ,
`privileges` TINYINT NULL DEFAULT NULL ,
`studentassent` TINYINT(1) UNSIGNED NULL DEFAULT NULL ,
`parentconsent` TINYINT(1) UNSIGNED NULL DEFAULT NULL ,
`birthdate` DATE NULL DEFAULT NULL ,
`gender` VARCHAR(1) NULL DEFAULT NULL ,
`registration` TIMESTAMP NULL DEFAULT NULL ,
PRIMARY KEY (`users_username`) ,
UNIQUE INDEX `uname_UNIQUE` (`users_username` ASC))
ENGINE = InnoDB;
-- -----------------------------------------------------
-- Table `kcptech`.`classes`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `kcptech`.`classes` (
`classes_id` INT UNSIGNED NOT NULL AUTO_INCREMENT ,
`course` VARCHAR(127) NULL ,
`period` VARCHAR(31) NULL DEFAULT '' ,
`organization` VARCHAR(127) NULL DEFAULT '' ,
PRIMARY KEY (`classes_id`) ,
UNIQUE INDEX `id_UNIQUE` (`classes_id` ASC))
ENGINE = InnoDB;
-- -----------------------------------------------------
-- Table `kcptech`.`classes_users`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `kcptech`.`classes_users` (
`classes_id` INT UNSIGNED NOT NULL ,
`users_username` VARCHAR(64) NOT NULL ,
`role` VARCHAR(12) NOT NULL ,
PRIMARY KEY (`classes_id`, `users_username`) ,
INDEX `fk_class_users__users_users_username` (`users_username` ASC) ,
INDEX `fk_class_users__class_class_id` (`classes_id` ASC) ,
CONSTRAINT `fk_class_users__users_users_username`
FOREIGN KEY (`users_username`)
REFERENCES `kcptech`.`users` (`users_username`)
ON DELETE NO ACTION
ON UPDATE NO ACTION,
CONSTRAINT `fk_class_users__class_class_id`
FOREIGN KEY (`classes_id`)
REFERENCES `kcptech`.`classes` (`classes_id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
請發佈您的架構並描述您要實現的目標。如果我理解正確,那麼您試圖爲名爲johnny並且有老師角色的用戶提供信息?我懷疑我們可以提出一些模式改進。此外,JOIN將總是擊敗這樣的情況的子查詢。此外,這篇文章可能更適合codereview.stackexchange.com – Corbin
我已經添加了一個數據庫模式以清晰起見。 – user594044