MySql的回顧六:子查詢(內查詢)

阿新 • • 發佈：2020-07-26

　　西北望鄉何處是，東南見月幾回圓。

月亮又慢悠悠的掛上了天空，趁著睡前夢囈，我就帶領各位可愛的讀者們探索MySql最後的子查詢部分。

說明:有些查詢結果出來結果截圖與題目要求不一樣會出現多餘的欄位是為了方便展示結果的可讀性。實際操作的讀者可以刪除SELECT後面多餘的欄位得到正確的結果。

#WHERE或HAVING後面
#1.標量子查詢(單行子查詢)
#2.列子查詢(多行子查詢)
#3.行子查詢(多列多行)
#特點:
#    ①子查詢放在小括號內
#    ②子查詢一般放在條件的右側
#    ③標量子查詢:一般搭配著單行操作符使用
#     單行操作符: > < >=  <= <> !-
#     列子查詢，一般搭配著多行操作符使用
#     IN,ANY 
/SOME(任意),ALL
#    ④子查詢的執行優先與主查詢執行,主查詢的條件用到了子查詢的結果。

#1.標量子查詢
#案例1:誰的工資比Abel高？
#①查詢Abel的工資
SELECT salary
FROM employees
WHERE last_name = 'Abel';

#②查詢員工的資訊，滿足Salary>①結果
SELECT *
FROM employees
WHERE salary>(SELECT salary FROM employees WHERE last_name='Abel');

#案例2.返回job_id與141號員工相同，salary比143號員工多的員工姓名，job_id,工資。
#①查141員工的job_id
SELECT job_id
FROM employees
WHERE employee_id 
='141';

#②查143員工的salary
SELECT salary
FROM employees
WHERE employee_id='143';

#③最後合併結果
SELECT CONCAT(last_name,first_name) AS 姓名,
job_id AS 工種編號,
salary AS 工資
FROM employees
WHERE job_id=(
    SELECT job_id
    FROM employees
    WHERE employee_id='141'
)
AND salary>(
    SELECT salary
    FROM employees
    WHERE employee_id='143' 

);

#案例3.返回公司工資最少的員工的last_name,job_id和salary。
SELECT MIN(salary)
FROM employees;

SELECT
last_name AS 姓,
salary AS 工資,
job_id AS 工種編號
FROM employees
WHERE salary=(
    SELECT MIN(salary)
    FROM employees
                 );

#案例4.查詢最低工資大於50號部門最低工資的部門id和其最低工資。
#①查50部門的最低工資
SELECT MIN(salary)
FROM employees
WHERE department_id=50;

#分組後，篩選條件①.【不用排除沒有部門的所以不篩選部門編號】
SELECT department_id AS 部門編號,
MIN(salary) AS 月薪
FROM employees
#WHERE department_id
GROUP BY department_id
HAVING 月薪>(
    SELECT MIN(salary)
    FROM employees
              );

#2.列子查詢(多行子查詢)
#返回多行
#使用多行比較操作符

#案例1.返回location_id是1400或1700的部門中的所有員工姓名。
#①查詢location_id是1400或1700的部門編號
SELECT DISTINCT department_id
FROM departments
WHERE location_id IN(1400,1700);

#②查詢員工姓名，要求部門號是①列表的某一個
SELECT CONCAT(last_name,first_name) AS 姓名
FROM employees
WHERE department_id IN (
    SELECT DISTINCT department_id
    FROM departments
    WHERE location_id IN(1400,1700)
    );

用ANY替代IN與上面同樣的結果
SELECT CONCAT(last_name,first_name) AS 姓名
FROM employees
WHERE department_id = ANY(
    SELECT DISTINCT department_id
    FROM departments
    WHERE location_id IN(1400,1700)
    );

#案例.返回location_id不是1400或1700的部門中的所有員工姓名。
SELECT CONCAT(last_name,first_name) AS 姓名
FROM employees
WHERE department_id NOT IN(
    SELECT DISTINCT department_id
    FROM departments
    WHERE location_id IN(1400,1700)
);
==============================
SELECT CONCAT(last_name,first_name) AS 姓名
FROM employees
WHERE department_id <> ALL(
    SELECT DISTINCT department_id
    FROM departments
    WHERE location_id IN(1400,1700)
);

#案例2.返回其他工種中比job_id為IT_PROG部門任意一工資低的員工工號,
#   姓名,job_id以及salary
#①把IT_PROG部門中的工資查出來
SELECT DISTINCT salary
FROM employees
WHERE job_id='IT_PROG';

#②把不是IT_PROG部門資訊查出來
SELECT *
FROM employees
WHERE job_id != 'IT_PROG';

#③合併①與②在員工表中查出來
SELECT employee_id AS 員工編號,
CONCAT(last_name,first_name) AS 姓名,
job_id AS 工種編號,
salary AS 工資
FROM employees
WHERE job_id != 'IT_PROG'
AND salary<ANY(
    SELECT salary
    FROM employees
    WHERE job_id='IT_PROG'
       );

用MAX代替ANY與上面同樣的效果
SELECT employee_id AS 員工編號,
CONCAT(last_name,first_name) AS 姓名,
job_id AS 工種編號,
salary AS 工資
FROM employees
WHERE job_id <> 'IT_PROG'
AND salary<(
    SELECT MAX(salary)
    FROM employees
    WHERE job_id='IT_PROG'
       );

#案例3.返回其他部門中比job_id為‘IT_PROG’部門所有工資都低的員工
#的員工號，姓名，job_id以及salary。
#①先把IT_PROG部門的工資查出來。
SELECT DISTINCT salary
FROM employees
WHERE job_id='IT_PROG';

SELECT    employee_id AS 員工號,
CONCAT(last_name,first_name) AS 姓名,
job_id AS 工種編號,
salary AS 工資
FROM employees
WHERE salary<ALL(
    SELECT DISTINCT salary
    FROM employees
    WHERE job_id='IT_PROG'
)
    AND job_id <> 'IT_PROG';
=============================
MIN替代ALL
SELECT    employee_id AS 員工號,
CONCAT(last_name,first_name) AS 姓名,
job_id AS 工種編號,
salary AS 工資
FROM employees
WHERE salary<(
　　SELECT MIN(salary)
　　FROM employees
　　WHERE job_id='IT_PROG'
)
    AND job_id <> 'IT_PROG';

#3.行子查詢(結果集一行多列或者多行多列)
#案例1.查詢員工編號最小並且工資最高的員工資訊.引入
SELECT MIN(employee_id)
FROM employees;
=================
SELECT MAX(salary)
FROM employees;

SELECT *
FROM employees
WHERE employee_id = (
    SELECT MIN(employee_id)
    FROM employees
)
AND salary = (
    SELECT MAX(salary)
    FROM employees
                 );

這種查詢結果使用虛擬欄位，單行操作符必須一致可以使用。查出來與上面同樣的效果。
SELECT *
FROM employees
WHERE (employee_id,salary)=(
    SELECT MIN(employee_id),
    MAX(salary)
    FROM employees
                           );

#二.SELECT子查詢
#僅僅支援標量子查詢，結果是一行一列
#案例1.查詢每個部門的員工個數
SELECT d.*,(SELECT COUNT(*) FROM employees)
FROM departments d;

新增條件
SELECT d.*,(SELECT COUNT(*)
FROM employees e
WHERE e.department_id=d.department_id
) AS 個數
FROM departments d;

#案例2.查詢員工號=102的部門名。
SELECT department_name
FROM departments;
==============
SELECT employee_id
FROM employees
WHERE employee_id = 102;

SELECT employee_id,
(
    SELECT department_name
    FROM departments d
    WHERE e.department_id=d.department_id
)
FROM employees e
WHERE employee_id=102;

#三.FROM 後面
注意:將子查詢結果充當一張表，要求必須起別名
#案例:查詢每個部門的平均工資等級。
SELECT ROUND(AVG(salary),2),department_id
FROM employees
GROUP BY department_id;

SELECT e.平均工資,j.grade_level
FROM job_grades AS j
,(
SELECT ROUND(AVG(salary),2) AS 平均工資,department_id
    FROM employees
    GROUP BY department_id
) AS e
WHERE e.平均工資 BETWEEN j.lowest_sal AND j.highest_sal;

#1999語法，老師答案
SELECT e.*,j.grade_level
FROM (
　　SELECT ROUND(AVG(salary),2) AS 平均工資,department_id
　　FROM employees
　　GROUP BY department_id
) AS e
INNER JOIN job_grades j
ON e.平均工資 BETWEEN j.lowest_sal AND j.highest_sal;

#四.EXISTS後面(相關子查詢)
語法:EXISTS(完整的查詢語句)
備註:完整的查詢語句可以是一行一列，可以使一行多列
注意:先走外查詢，然後根據某個欄位的值再去過濾
EXISTS 判斷(布林型別)值存不存在,結果只有兩種:1有，0沒有
#引入
SELECT EXISTS(SELECT employee_id FROM employees);

查詢工資3W的員工資訊
SELECT EXISTS(SELECT * FROM employees WHERE salary=30000);

#案例引入.查詢員工名和部門名
#查員工名與部門編號
SELECT first_name,department_id
FROM employees
WHERE department_id;

#查部門名
SELECT department_name
FROM departments;

#查員工名與部門名
SELECT e.first_name,d.department_name
FROM employees e
INNER JOIN (     SELECT department_name,department_id
    FROM departments
) AS d
ON e.department_id=d.department_id;

#案例1..查有員工的部門名
SELECT department_name
FROM departments d
WHERE EXISTS(
    SELECT *
    FROM employees e
    WHERE d.department_id=e.department_id
                );

使用IN代替EXISTS，同樣是上面的結果
SELECT department_name
FROM departments d
WHERE d.department_id IN(
    SELECT department_id
    FROM employees
                    );

#案例2.查詢沒有女朋友的男神資訊
#IN方法
SELECT *
FROM boys bo
WHERE bo.id  NOT IN(
    SELECT boyfriend_id
    FROM beauty be
);
===============
#EXISTS方法
SELECT *
FROM boys bo
WHERE NOT EXISTS(
    SELECT boyfriend_id
    FROM beauty be
    WHERE bo.id=be.boyfriend_id
);

進階9:聯合查詢
UNION 聯合 合併:將多條查詢語句的結果合併成一個結果。
語法:
    查詢語句1
    UNION
    查詢語句2
    UNION
    ...

應用場景:
要查詢的結果來自於多個表，且多個表沒有直接的連線關係，
但查詢資訊一致時。
網頁搜尋內容，內容從不同的表中檢索聯合起來返回給使用者。

特點:
1.要求多條查詢語句的查詢列數是一致的。
2.要求多條查詢語句的查詢的每一列的型別和順序最好一致。
3.使用UNION關鍵字預設去重，如果使用UNION ALL全部展示，包含重複項

#引入案例1.:查詢部門編號>90或者郵箱包含A的員工資訊
SELECT * FROM employees
WHERE email LIKE '%a%' OR department_id>90;
聯合查詢
SELECT * FROM employees WHERE email LIKE '%a%'
UNION
SELECT * FROM employees WHERE department_id>90;

感謝能認真讀到這裡的夥伴們，MySql查詢部分結束，相信螢幕前的你照著我部落格裡的模板可以完成一些簡單的SQL查詢語句，SQL既然學了，以後還是要多練習一下，SQL1992與1999語法在主流的關係型資料庫都是通用的。後續我會繼續進行對MySql的知識進行擴充套件，感興趣的同志互相關注一唄！o(^▽^)o