1、「select * from 表名; ”,—可查詢表中全部資料;
2、“select 欄位名稱from 表名;”,—可查詢表中指定欄位的資料;
3、“select distinct 欄位名稱from 表名;”,—可對錶中資料進行去重查詢。
4、“select 欄位名稱from 表名where 查詢條件;”,—可根據條件查詢表中指定欄位的資料;
#1)比較運算子:>, 95ec6993dc754240360e28e0de8de30a=, c4290fa39d2a3d6a61c5df07c46bf05e
查詢大於18歲的資訊
select * from students where age>18; select id, name,gender from students where age>18;查詢小於18歲的資訊
select * from students where age<18;查詢年齡為18歲的所有學生的名字
select * from students where age=18;
2)邏輯運算子:and,或, not
–18到28之間的學生資訊
select * from students where age>18_and age<28:–18歲以上的女性
select * from students where age>18 and gender="女"; select * from students where age>18 and gender=2;#–18以上或身高查過180(包含)以上
select * from students where age>18 or height>=180;不在18歲以上的女性這個範圍內的信息
select * from students where not (age>18 and gender=2);年齡不是小於或等於18並且是女性
select * from students where (not age<=18) and gender=2;
3)模糊查詢:like, rlike
% 替換1個或多個
_ 替換1個
查詢姓名中以“小”開始的名字select name from students where name="小"; select name from students where name like"小%";查詢姓名中有「小」所有的名字
select name from students whece name like "%小%";#查詢有2個字的名字
select name from students where name like "__";查詢有3個字的名字
select name from students where name like "__";查詢至少有2個字的名字select name from
students where name like "__%";rlike正則
查詢以周開始的姓名select name from students where name rlike "^周.*";查詢以周開始、倫結尾的姓名
select name from students where name rlike "^周.*伦$";
4)範圍查詢:in,not in,between…and,not between…and
查詢年齡為18、34的姓名
select name, age from students where age=18 or age=34; select name,age from students where age in (18,34);not in不非連續的範圍之內
年齡不是18、34歲之間的信息select name,age from students where age not in (18,34);between … and …表示在一個連續的範圍內
查詢年齡在18到34之間的信息select name,age from students where age between 18 and 34;not between … and …表示不在一個連續的範圍內
查詢年齡不在在18到34之間的信息select * from students where age not between 18 and 34;
空判斷
判空is null
查詢身高為空的資訊select *from students where height is null/NULL/Null;判非空is not null
select * from students where height is not null;
排序:order_by
–查詢年齡在18到34歲之間的男性,依照年齡從小到大排序
select * from students where (age between 18 and 34) and gender=1; select * from students where (age between 18 and 34) and gender=1 order by age; select * from students where (age between 18 and 34) and gender=1 order by age asc;查詢年齡在18到34歲之間的女性,身高從高到矮排序
select * from students where (age between18 and 34) and gender=2 order by height desc;order by多個字段
查詢年齡在18到34歲之間的女性,身高從高到矮排序,如果身高相同的情況下按照年齡從小到大排序select * from students where (age between 18 and 34) and gender=2 order by height desc,age asc;查詢年齡在18到34歲之間的女性,身高從高到矮排序,如果身高相同的情況下依照年齡從小到大排序,如果年齡也相同那麼依照id從大到小排序
select * from students where (age between 18 and 34) and gender=2 order by height desc,age asc, id desc;依照年齡從小到大、身高從高到矮的排序
select * from students order by age asc,height desc;
分組:group_by, group_concat():查詢內容, having
where :是對整個資料表資訊的判斷;
having:是對於分組後的資料進行判斷
–group by
依照性別分組,查詢所有的性別select gender from students group by gender;–計算每個性別中的人數
select gender, count(*) from students group by gender;where是在group by前面
–計算男性的人數select count(*) from students where gender='男';–group_concat(…)
查詢同種性別中的姓名select gender,group_concat(name) from students group by gender;having :having是在group by後面
查詢平均年齡超過30歲的性別,以及姓名select gender ,avg(age) from students group by gender having avg(age) > 30;查詢每種性別中的人數多於2個的信息
select gender,count(*) from students group by gender having count(*) > 2;– 查詢每組性別的平均年齡
select gender,avg(age) from students group by gender;
分頁:limit
limit start,count (start:表示从哪─个开始;count:表示数量) 即limit(第N页-1)*每个的个数,每页的个数; limit在使用的时候,要放在最后面.
限制查詢出來的資料數量
select *from students where gender=1 limit 2;查詢前5個資料
select* from students limit 0,5;查詢id6-10(包含)的書序
select * from students limit 5,5;每頁顯示2個,第1個頁
select * from students limit 0,2;每頁顯示2個,第2個頁
select * from students limit 2,2;每頁顯示2個,第3個頁面
select * from students limit 4,2;每頁顯示2個,第4個頁面
select * from students limit 6,2;每頁顯示2個,顯示第6頁的資訊,依照年齡從小到大排序
select * from students order by age asc limit 10,2;– 如果重新排序了,那麼會顯示第一頁
select * from students where gender=2 order by height des limit 0,2;
5)聚合函數:count(), max() , min(), sum(), avg(), round()
聚合函數
-總數-- count
-查詢男性有多少人,女性有多少人select count(*) from students where gender=1; select count(*) as 男性人数 from students where gender=1; select count(*) as 女性人数 from students where gender=2;-最大值-最小值
– max --min
一查詢最大的年齡select max (age) from students;–查詢女性的最高身高
select max (height) from students where gender=2;-求和
–sum
-計算所有人的年齡總和select sum ( age) from students;–平均值
– avg
–計算平均年齡select avg(age) from students;–計算平均年齡
select sum ( age) / count(* ) from students;–四舍五入round ( 123.23 ,_1)保留1位小数
–计算所有人的平均年龄,保留2位小数select round (sum(age)/count(*),2) from students; select round ( sum(age)/count(*),3) from students ;–计算男性的平均身高保留2位小数
select round(avg (height),2) from students where gender=1; select name,round(avg(height),2) from students where gender=1;
6)连接查询 :inner join, left join, right join
inner join
select … from 表 A inner join表B;select * from students inner join classes;查询有能够对应班级的学生以及班级信息
select * from students inner join classes on students.cls_id=classes.id;按照要求显示姓名、班级
select students.*, classes.name from students inner join classes on students.cls_id=classes.id; select students.name,classes.name from students inner join classes on students.cls_id=classes.id;给数据表起名字
select s.name,c.name from students as s inner join classes as c on s.cls_id=c.id;查询有能够对应班级的学生以及班级信息,显示学生的所有信息,只显示班级名称
select s.*,c.name from students as s inner join classes as c on s.cls_id=c.id;在以上的查询中,将班级姓名显示在第1列
select c.name,s.* from students as s inner join classes as c on s.cls_id=c.id;查询有能够对应班级的学生以及班级信息,按照班级进行排序
select c.xxx s.xxx from student as s inner join clssses as c on … order by …;select c.name,s.* from students as s inner join classes as c on s.cls_id=c.id order by c.name;当时同一个班级的时候,按照学生的id进行从小到大排序
select c.name,s.* from students as s inner join classes as c on s.cls_id=c.id order by c.name,s.id;left join
查询每位学生对应的班级信息select * from students as s left join classes as c on s.cls_id=c.id;查询没有对应班级信息的学生
– select … from xxx as s left join xxx as c on… where …
– select … from xxx as s left join xxx as c on… . … having …select * from students as s left join classes as c on s.cls_id=c.id having c.id is null; select * from students as s left join classes as c on s.cls_id=c.id where c.id is null;
left join是按照左边的表为基准和右边的表进行查询,查到就显示,查不到就显示为null
补充
查询所有字段:select * from 表名;
查询指定字段:select 列1,列2,... from 表名;
使用 as 给字段起别名: select 字段 as 名字.... from 表名;
查询某个表的某个字段:select 表名.字段 .... from表名;
可以通过 as 给表起别名: select 别名.字段 .... from 表名 as 别名;
消除重复行: distinct 字段
注意:WHERE子句中是不能用聚集函数作为条件表达式的!
二、总结
(1)命令:select * from a26d98d33123a70024fa8ba5642906c6;
(2)命令:select 7d63c9f8ce32ff1ae4915f31bebead74 from a26d98d33123a70024fa8ba5642906c6;
命令:select distinct 7d63c9f8ce32ff1ae4915f31bebead74 from a26d98d33123a70024fa8ba5642906c6
升序:asc
降序:desc
降序排列命令:select dfb08b8fe164362ddbbdb0005b7a6721 from a26d98d33123a70024fa8ba5642906c6 order by dfb08b8fe164362ddbbdb0005b7a6721 desc
不加desc一般默认为升序排列
命令:select 162131b8562e13ae131483a4c9d25bb6, Sum(score) from a26d98d33123a70024fa8ba5642906c6 group by 162131b8562e13ae131483a4c9d25bb6
假设现在又有一个学生成绩表(result)。要求查询一个学生的总成绩。我们根据学号将他们分为了不同的组。
命令:
select id, Sum(score) from result group by id;
现在有两个表学生表(stu)和成绩表(result)。
当连接运算符为“=”时,为等值连接查询。
现在要查询年龄小于20岁学生的不及格成绩。
select stu.id,score from stu,result where stu.id = result.id and age < 20 and score < 60;
等值查询效率太低
①语法
select f1,f2,f3,.... from table1 left/right outer join table2 on 条件;
②左外连接查询,例如
select a.id,score from (select id,age from stu where age < 20) a (过滤左表信息) left join (select id, score from result where score < 60) b (过滤右表信息) on a.id = b.id;
左外连接就是左表过滤的结果必须全部存在。右表中如果没有与左表过滤出来的数据匹配的记录,那么就会出现NULL值
③右外连接查询,例如
select a.id,score from (select id,age from stu where age < 20) a (过滤左表信息) right join (select id, score from result where score < 60) b (过滤右表信息) on a.id = b.id;
右外连接就是左表过滤的结果必须全部存在
①语法
select f1,f2,f3,.... from table1 inter join table2 on 条件;
②例如
select a.id,score from (select id,age from stu where age < 20) a (过滤左表信息) inner join (select id, score from result where score < 60) b (过滤右表信息) on a.id = b.id;
在图书表(t_book)和图书类别表(t_bookType)中
①.union
使用union关键字是,数据库系统会将所有的查询结果合并到一起,然后去掉相同的记录;
select id from t_book union select id from t_bookType;
②.union all
使用union all,不会去除掉重复的记录;
select id from t_book union all select id from t_bookType;
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