路漫漫其修遠兮，吾將上下而求索，又到了周末，我繼續(xù)帶各位看官學習回顧Mysql知識。

上次說到了流程控制函數(shù)，那就從流程控制函數(shù)來繼續(xù)學習吧！

#五.流程控制函數(shù)#1.if函數(shù):if else的效果IF(條件表達式,成立返回1,不成立返回2)#與Java三元運算相同SELECT IF(10>5,’大’,’小’);

SELECT last_name,commission_pct,IF(commission_pct IS NULL,’沒獎金呵呵’,’有獎金嘻嘻’) AS 備注FROM employees;

#2.CASE函數(shù)的使用一: swirch case的效果【等值判斷】回顧:switch(變量或表達式){case 常量1:語句1;break;...default:語句n;break;}mysql中case 要判斷的字段或表達式或變量when 常量1 then 要顯示的值1或者語句1;[語句要加分號，值不用加]when 常量2 then 要顯示的值2或者語句2;...slse要顯示的值n或者語句n;[默認值]end[結(jié)尾]case在SELECT后面相當于表達式用，后面不能放語句，只能是值.在后續(xù)的學習中，存儲過程與函數(shù)內(nèi)就可以單獨,用不搭配SLECT,就用語句.先是表達式的操作案例:查詢員工的工資,要求部門號=30，顯示的工資為1.1倍部門號=40，顯示的工資為1.2倍部門號=50，顯示的工資為1.3倍其他部門，顯示的工資為原工資

SELECT salary 原始工資,department_id,CASE department_idWHEN 30 THEN salary*1.1WHEN 40 THEN salary*1.2WHEN 50 THEN salary*1.3ELSE salaryEND AS 新工資FROM employees;

#2.CASE函數(shù)的使用二:類是于多重IF【區(qū)間判斷】回顧Java中:if(條件1){語句1;}else if(條件2){語句2;}...else{語句n;}

mysql中:casewhen 條件1 then 要顯示的值1或者語句[語句后面要加分號;]when 條件2 then 要顯示的值2或者語句[語句后面要加分號;]...else 要顯示的值n或語句nend案例:查詢員工的工資的情況如果工資>20000,顯示A級別如果工資>15000,顯示B級別如果工資>10000,顯示C級別否則,顯示D級別

==============流程函數(shù)到此結(jié)束，要想熟練運用還需要勤加練習。===============

提供幾道習題供讀者試試手！

#計算有幾種工資。SELECT COUNT(DISTINCT salary),COUNT(salary)FROM employees;

#5.count 函數(shù)的詳細介紹SELECT COUNT(salary) FROM employees;

#COUNT(統(tǒng)計所有列)SELECT COUNT(*) FROM employees;

#把表的行生成一個列每一列都是1。統(tǒng)計1的總數(shù)。count里面可以用任意常量值。SELECT COUNT(1) FROM employees;

#考慮到效率問題:#5.5之前都是MYISAM 下COUNT(*)最高，計數(shù)器直接返回#5.5之后默認都是INNODB下COUNT()與COUNT(1)都差不多， 比COUNT(字段)效率高，如果是字段要判斷字段是否為NULL。

#6.和分組函數(shù)一同查詢的字段有限制SELECT AVG(salary),employee_id FROM employees;

#1.查詢公司員工工資的最大值，最小值，平均值，總和。SELECT MAX(salary),MIN(salary),AVG(salary),SUM(salary)FROM employees;

SELECT MAX(salary) AS 最大值,MIN(salary) AS 最小值,ROUND(AVG(salary)) AS 最小值,SUM(salary) AS總和FROM employees;

#2.查詢員工表中最大入職時間和最小入職時間的相差天數(shù)。#DATEDIFF計算天數(shù)。DATEIFF(參數(shù)1-參數(shù)2)SELECT DATEDIFF(NOW(),(’1995-1-1’));

SELECT DATEDIFF(MAX(hiredate),MIN(hiredate)) AS DIFFRENCEFROM employees;

#3.查詢部門編號為90的員工個數(shù)。SELECT COUNT(*) AS個數(shù)FROM employeesWHERE department_id=90;

#進階5.分組查詢GROUP BY子句語法。可以使用GROUP BY子句將表中的數(shù)據(jù)分成若干組。語法:SELECT 分組函數(shù)，列(要求出現(xiàn)在GROUP BY的后面)FROM 表【W(wǎng)HERE篩選條件】GROP BY 分組的列表【ORDER BY 子句】注意: 查詢列表必須特殊,要求是分組函數(shù)和GROUP BY后出現(xiàn)的字段特點: 1.分組查詢中的篩選條件分為兩類

2.GROUP BY 子句支持單個字段分組，多個字段分組(多個字段之間用逗號隔開沒有順序要求)也支持表達式或函數(shù)分組(用的較少)3.也可以添加排序(排序放在整個分組查詢的最后)

#引入:查詢每個部門的平均工資，保留兩位小數(shù)。SELECT ROUND(AVG(salary),2) AS 平均工資FROM employees;

#簡單的分組查詢，添加分組前的篩選WHERE#案例1:查詢每個工種的最高工資。SELECT MAX(salary) AS 最高工資,job_id AS 工種編號FROM employeesGROUP BY job_id;

#案例2:查詢每個位置上的部門個數(shù)。SELECT COUNT(*) AS 總數(shù),location_idFROM departmentsGROUP BY location_id;

#添加篩選條件#查詢郵箱中包含a字符的，平均工資保留兩位小數(shù)，每個部門的平均工資SELECT email,ROUND(AVG(salary),2),department_idFROM employeesWHERE email LIKE ’%a%’GROUP BY department_id;

#案例2:查詢每個領導手下員工有獎金的的最高工資SELECT MAX(salary),manager_idFROM employeesWHERE commission_pct IS NOT NULLGROUP BY manager_id;

添加復雜的篩選,添加分組后的篩選HAVING#案例1：查詢哪個部門的員工個數(shù)>2#①查詢每個部門的員工個數(shù)SELECT COUNT(*),department_idFROM employeesGROUP BY department_id;

#②根據(jù)①的結(jié)果，查詢哪個部門的員工個數(shù)>2SELECT COUNT(*) AS 總數(shù),department_id AS 部門編號FROM employeesGROUP BY department_idHAVING COUNT(*)>2;

#案例2.查詢每個工種有獎金的員工的最高工資>12000的工種編號和最高工資。#①查詢每個工種有獎金的員工的最高工資。#原始表能篩選的就放在FROM 后面用WHERE。

SELECT job_id AS 員工編號,MAX(salary) AS 最高工資FROM employeesWHERE commission_pct IS NOT NULLGROUP BY job_id;

#②根據(jù)①結(jié)果繼續(xù)篩選，最高工資>12000。SELECT job_id AS 員工編號,MAX(salary) AS 最高工資FROM employeesWHERE commission_pct IS NOT NULLGROUP BY job_idHAVING MAX(salary)>12000;

#案例3.查詢領導編號>102的每個領導手下的最低工資>5000的領導編號是哪個,以及其最低工資。#①查詢領導編號>102的每個領導手下的最低工資SELECT MIN(salary) AS 最低工資,manager_id AS 領導編號FROM employeesWHERE manager_id > 102GROUP BY manager_id;

#②最低工資大于5000的。SELECT MIN(salary) AS 最低工資,manager_id AS 領導編號FROM employeesWHERE manager_id > 102GROUP BY manager_idHAVING 最低工資 > 5000;

#案例:按員工姓名的長度分組,查詢每一組的員工個數(shù)，篩選員工個數(shù)>5的有哪些#MYSQL內(nèi)GROUP BY與HAVING后面支持別名，WHERE不支持別名。但ORACLE數(shù)據(jù)庫的GROUP BY與HAVING是不支持別名的。SELECT LENGTH(CONCAT(last_name,first_name)) AS 姓名長度,COUNT(*) AS 個數(shù)FROM employeesGROUP BY 姓名長度HAVING 姓名長度 > 5;

#按多個字段分組#案例:查詢每個部門工種的員工的平均工資(保留兩位小數(shù))。SELECT ROUND(AVG(salary),2) AS 平均工資,department_id AS 部門,job_id AS 工種FROM employeesGROUP BY 部門,工種;

#添加排序#案例:查詢每個部門每個工種的員工的平均工資,并且按平均工資的高低顯示。SELECT ROUND(AVG(salary),2) AS 平均工資,department_id AS 部門,job_id AS 工種FROM employeesGROUP BY 工種,部門ORDER BY 平均工資 DESC;

#案例:查詢部門不能為空的，每個部門每個工種的員工的平均工資,并且按平均工資的高低顯示。SELECT ROUND(AVG(salary),2) AS 平均工資,department_id AS 部門,job_id AS 工種FROM employeesWHERE department_id IS NOT NULLGROUP BY 工種,部門ORDER BY 平均工資 DESC;

#案例:查詢部門不能為空的，每個部門每個工種的員工的平均工資高于10000的,并且按平均工資的高低顯示。SELECT ROUND(AVG(salary),2) AS 平均工資,department_id AS 部門,job_id AS 工種FROM employeesWHERE department_id IS NOT NULLGROUP BY 工種,部門HAVING 平均工資 > 10000ORDER BY 平均工資 DESC;

加強練習:

#1.查詢各job_id的員工工資的最大值，最小值，平均值，總和，并按job_id升序。SELECT job_id,MAX(salary),MIN(salary),AVG(salary),SUM(salary)FROM employeesGROUP BY job_idORDER BY job_id ASC;

#2.查詢員工最高工資和最低工資的差距(DIFFERENCE)。SELECT MAX(salary) - MIN(salary) AS DIFFERENCEFROM employees;=======================================================SELECT MAX(salary) AS 最高,MIN(salary) AS 最低,MAX(salary)-MIN(salary) AS DIFFERENCEFROM employees;

#3.查詢各個管理者手下員工的最低工資,其中最低工資不能低于6000,沒有管理者的員工不能計算在內(nèi)。SELECT MIN(salary),CONCAT(last_name,first_name),manager_idFROM employeesWHERE manager_id IS NOT NULLGROUP BY manager_idHAVING MIN(salary)>6000;

到此結(jié)束，MySql的統(tǒng)計，分組查詢到此結(jié)束。如果沒有感覺的看官可以自己手動練習一下。

夏天的太陽總是那么亮的刺眼，但多沐浴一下陽光也補鈣，想不到吧！o(^?^)o

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