他のデータからデータを探すmysql

Question

「impedgato」テーブルの「cognome」フィールドから始まる「sede」テーブルから「via」フィールドを見つける方法は、まったく分かりません。

CREATE TABLE IF NOT EXISTS `mydb`.`impiegato` (
    `codice` VARCHAR(45) NOT NULL, 
    `cognome` VARCHAR(45), 
    `nomedipart` VARCHAR(45), 
    `sede` VARCHAR(45), 
    `data` VARCHAR(45), 
    PRIMARY KEY (`codice`)) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `mydb`.`sede` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `mydb`.`sede` (
    `cittaSede` VARCHAR(45) NOT NULL, 
    `via` VARCHAR(45) NULL, 
    `cap` VARCHAR(45) NULL, 
    PRIMARY KEY (`cittaSede`)) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `mydb`.`dipartimento` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `mydb`.`dipartimento` (
    `nome` VARCHAR(45) NOT NULL, 
    `cittaSede` VARCHAR(45) NOT NULL, 
    `Telefono` VARCHAR(45), 
    `codDirett` VARCHAR(45), 
    INDEX `fk_dipartimento_idx` (`codDirett` ASC), 
    PRIMARY KEY (`nome`), 
    INDEX `fk_sede_idx` (`cittaSede` ASC), 
    CONSTRAINT `fk_dipartimento` 
    FOREIGN KEY (`codDirett`) 
    REFERENCES `mydb`.`impiegato` (`codice`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION, 
    CONSTRAINT `fk_sede` 
    FOREIGN KEY (`cittaSede`) 
    REFERENCES `mydb`.`sede` (`cittaSede`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `mydb`.`progetto` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `mydb`.`progetto` (
    `nomeProgetto` VARCHAR(45) NOT NULL, 
    `budget` VARCHAR(45), 
    PRIMARY KEY (`nomeProgetto`)) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `mydb`.`partecipazione` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `mydb`.`partecipazione` (
    `codImpieg` VARCHAR(45) NOT NULL, 
    `progetto` VARCHAR(45), 
    PRIMARY KEY (`codImpieg`), 
    INDEX `fk_progetto_idx` (`progetto` ASC), 
    CONSTRAINT `fk_impiegato` 
    FOREIGN KEY (`codImpieg`) 
    REFERENCES `mydb`.`impiegato` (`codice`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION, 
    CONSTRAINT `fk_progetto` 
    FOREIGN KEY (`progetto`) 
    REFERENCES `mydb`.`progetto` (`nomeProgetto`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB; 

Answer 1

あなたは同じsede値でsedeの行と希望cognomeでimpiegato行を一致させる必要があり、そしてそれはjoinsは何をすべきかです：で、「経由」のデータで何

select via 
from sede s 
join impiegato i 
on  s.sede = i.sede 
where i.cognome = 'desiredCognome'