SQL 与 Hive 技术总结

其他 2021-12-15 15:46:49 阅读次数: 0

1、

Hive 安装_w3cschool所有Hadoop的子项目，如Hive, Pig,和HBase，都需要Linux的操作系统。因此，需要安装Linux OS。以下是为Hive的安装执行的简单步骤：第1步：验证JAVA安装在Hive安装之前，Java必须在系统上已经安装。使用下面的命令来验证是否已经安装Java：$ java –_来自Hive 教程，w3cschool编程狮。https://www.w3cschool.cn/hive_manual/hive_install.html

2、

HIVE：窗口函数，用sql语句查询MySQL安装路径和版本_Jack_2085-CSDN博客SET NAMES utf8mb4;SET FOREIGN_KEY_CHECKS = 0;-- ------------------------------ Table structure for test_table-- ----------------------------DROP TABLE IF EXISTS `test_table`;CREATE TABLE `test_table` ( `id` bigint(6) NULL DEFAULT NULL, `prov...https://blog.csdn.net/weixin_54217632/article/details/120243482

3、

    3.1、根据条件查询连续登录N天的用户。

    3.2、查询连续登录天数最大的用户及天数。

表结构及数据：
DROP TABLE IF EXISTS `test5`;
CREATE TABLE `test5`  (
  `dt` varchar(10) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
  `user_id` varchar(50) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
  `age` int(11) NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of test5
-- ----------------------------
INSERT INTO `test5` VALUES ('2021-03-11', 'test_1', 23);
INSERT INTO `test5` VALUES ('2021-03-11', 'test_1', 23);
INSERT INTO `test5` VALUES ('2021-03-11', 'test_1', 23);
INSERT INTO `test5` VALUES ('2021-03-13', 'test_1', 23);
INSERT INTO `test5` VALUES ('2021-03-11', 'test_2', 19);
INSERT INTO `test5` VALUES ('2021-04-11', 'test_3', 39);
INSERT INTO `test5` VALUES ('2021-03-11', 'test_3', 39);
INSERT INTO `test5` VALUES ('2021-03-12', 'test_2', 19);
INSERT INTO `test5` VALUES ('2021-03-15', 'test_2', 19);
INSERT INTO `test5` VALUES ('2021-03-16', 'test_2', 19);
INSERT INTO `test5` VALUES ('2021-03-13', 'test_2', 19);
INSERT INTO `test5` VALUES ('2021-03-14', 'test_2', 19);

SET FOREIGN_KEY_CHECKS = 1;



------------------------------------------------ 下面是sql的实现语句



--  方法1 用case when then 方法


select distinct user_id 
from(
SELECT * ,
               CASE
               WHEN DATE_SUB(str_to_date(dt,'%Y-%m-%d'),INTERVAL 1 DAY) = str_to_date(@old,'%Y-%m-%d')  and @u_id=user_id and @old:=dt 
					THEN @size:=@size+1
					
               WHEN @old:=dt 
					THEN @size:=1 
               END
 AS tt, @u_id:=user_id
FROM (select * from test5 
group BY user_id,dt) t
ORDER BY user_id,dt) tb
where tb.tt = 2


-----------------------------------------------------------------



-- 求连续最多登录的天数

select MAX(tt) 
from(
SELECT * ,
               CASE
               WHEN DATE_SUB(str_to_date(dt,'%Y-%m-%d'),INTERVAL 1 DAY) = str_to_date(@old,'%Y-%m-%d')  and @u_id=user_id and @old:=dt 
					THEN @size:=@size+1
					
               WHEN @old:=dt 
					THEN @size:=1 
               END
 AS tt, @u_id:=user_id
FROM (select * from test5 
group BY user_id,dt) t
ORDER BY user_id,dt) tb


------------------------------------------------------------------

-- 查询连续登录系统天数，最多的人和一共连续登录了多少天。

select user_id, tt
from(
SELECT * ,
               CASE
               WHEN DATE_SUB(str_to_date(dt,'%Y-%m-%d'),INTERVAL 1 DAY) = str_to_date(@old,'%Y-%m-%d')  and @u_id=user_id and @old:=dt 
					THEN @size:=@size+1
					
               WHEN @old:=dt 
					THEN @size:=1 
               END
 AS tt, @u_id:=user_id
FROM (select * from test5 
group BY user_id,dt) t
ORDER BY user_id,dt) tb
where tb.tt = 
(
select MAX(tt) 
from(
SELECT * ,
               CASE
               WHEN DATE_SUB(str_to_date(dt,'%Y-%m-%d'),INTERVAL 1 DAY) = str_to_date(@old,'%Y-%m-%d')  and @u_id=user_id and @old:=dt 
					THEN @size:=@size+1
					
               WHEN @old:=dt 
					THEN @size:=1 
               END
 AS tt, @u_id:=user_id
FROM (select * from test5 
group BY user_id,dt) t
ORDER BY user_id,dt) tb)

4、查询用户连续登录系统


DROP TABLE IF EXISTS `test_login_time`;
CREATE TABLE `test_login_time`  (
  `id` int(0) NOT NULL,
  `user_name` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_as_ci NULL DEFAULT NULL,
  `login_time` datetime(0) NULL DEFAULT NULL,
  PRIMARY KEY (`id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_0900_as_ci ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of test_login_time
-- ----------------------------
INSERT INTO `test_login_time` VALUES (1, 'jack', '2021-11-17 10:15:37');
INSERT INTO `test_login_time` VALUES (2, 'jack', '2021-11-18 10:15:37');
INSERT INTO `test_login_time` VALUES (3, 'jack', '2021-11-19 10:15:37');
INSERT INTO `test_login_time` VALUES (4, 'jack', '2021-11-20 10:15:37');
INSERT INTO `test_login_time` VALUES (5, 'jack', '2021-11-21 10:15:37');
INSERT INTO `test_login_time` VALUES (6, 'jack', '2021-11-22 10:15:37');
INSERT INTO `test_login_time` VALUES (7, 'jack', '2021-11-16 10:15:37');
INSERT INTO `test_login_time` VALUES (8, 'jack', '2021-11-15 10:15:37');
INSERT INTO `test_login_time` VALUES (9, 'lucy', '2021-12-01 10:15:37');
INSERT INTO `test_login_time` VALUES (10, 'lucy', '2021-12-02 10:15:37');
INSERT INTO `test_login_time` VALUES (11, 'lucy', '2021-12-03 10:15:37');
INSERT INTO `test_login_time` VALUES (12, 'lucy', '2021-12-04 10:15:37');
INSERT INTO `test_login_time` VALUES (13, 'lucy', '2021-12-06 10:15:37');
INSERT INTO `test_login_time` VALUES (14, 'lucy', '2021-12-08 10:15:37');
INSERT INTO `test_login_time` VALUES (15, 'lucy', '2021-12-09 10:15:37');
INSERT INTO `test_login_time` VALUES (16, 'lucy', '2021-12-10 10:15:37');
INSERT INTO `test_login_time` VALUES (17, 'mark', '2022-02-01 10:15:37');
INSERT INTO `test_login_time` VALUES (18, 'mark', '2022-02-02 10:15:37');
INSERT INTO `test_login_time` VALUES (19, 'mark', '2022-02-03 10:15:37');
INSERT INTO `test_login_time` VALUES (20, 'mark', '2022-02-04 10:15:37');
INSERT INTO `test_login_time` VALUES (21, 'mark', '2022-03-06 10:15:37');
INSERT INTO `test_login_time` VALUES (22, 'mark', '2022-06-08 10:15:37');
INSERT INTO `test_login_time` VALUES (23, 'mark', '2022-05-09 10:15:37');
INSERT INTO `test_login_time` VALUES (24, 'mark', '2022-05-10 10:15:37');
INSERT INTO `test_login_time` VALUES (25, 'mark', '2023-08-10 10:15:37');
INSERT INTO `test_login_time` VALUES (26, 'mayun', '2099-09-15 10:15:37');
INSERT INTO `test_login_time` VALUES (27, 'mayun', '2099-09-15 10:15:37');
INSERT INTO `test_login_time` VALUES (28, 'mayun', '2099-09-15 10:15:37');
INSERT INTO `test_login_time` VALUES (30, 'mayun', '2099-09-16 10:15:37');
INSERT INTO `test_login_time` VALUES (31, 'mayun', '5999-08-18 10:15:37');

SET FOREIGN_KEY_CHECKS = 1;

--   ---------------------------------------------------

--  方法1、


-- 繁琐版本

select *,
         CASE
               WHEN  lag_t2=1 
					THEN @size:=@size+1
					ELSE
               @size:=1 
               END
 AS tt,
 
-- if方法也行
 if(lag_t2=1,@size_2:=@size_2+1,@size_2:=1) as tt_2
from(
select *,
TIMESTAMPDIFF(DAY,lead_t1,login_time) as lag_t2
from(
select *,
lag(t.login_time, 1) over (PARTITION by t.user_name order by t.login_time) as lead_t1
from test_login_time t 
) ta)tab;








******************************************************************
-- - 简单版本

-- 从来没有连续登录的用户，和连续n次登录的用户。
select *,
 if(ta.tg=1 and ta.t_name=1,@size:=@size+1,@size:=1) as t_num
from(
select *,lead(left(login_time,10), 1) over () = 
     DATE_ADD(left(login_time,10),INTERVAL 1 day) as tg ,
	  lead(user_name, 1) over () = user_name  as t_name
	  from test_login_time t 
ORDER BY t.user_name,t.login_time) ta;

------------------------------------------------------------ 上面sql的升级版本

select *,@size,
 if(ta.tg=1 and ta.t_name=1,@size:=@size+1,@size:=1) as t_num
from(
select *,lead(left(login_time,10), 1) over () = 
     DATE_ADD(left(login_time,10),INTERVAL 1 day) as tg ,
	  lead(user_name, 1) over () = user_name  as t_name
	  from test_login_time t 
ORDER BY t.user_name,t.login_time) ta;

------------------------------------------- 下面是个要解决的问题。

-- 为什么下面的sql语句，@size变量不重新计算。
select *,
 if(lead(left(login_time,10), 1) over () = 
     DATE_ADD(left(login_time,10),INTERVAL 1 day),
	  @size:=@size+1,
	  @size:=1) as t_num
from(
select * from test_login_time t 
ORDER BY t.user_name,t.login_time) ta

5、求各科学生成绩第二名的学生

    

   


DROP TABLE IF EXISTS `km_cj`;
CREATE TABLE `km_cj`  (
  `km` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_as_ci NULL DEFAULT NULL,
  `user_id` int(0) NOT NULL,
  `cj` bigint(0) NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_0900_as_ci ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of km_cj
-- ----------------------------
INSERT INTO `km_cj` VALUES ('jack', 8, 1);
INSERT INTO `km_cj` VALUES ('jack', 7, 2);
INSERT INTO `km_cj` VALUES ('jack', 1, 3);
INSERT INTO `km_cj` VALUES ('jack', 2, 4);
INSERT INTO `km_cj` VALUES ('jack', 3, 5);
INSERT INTO `km_cj` VALUES ('jack', 4, 6);
INSERT INTO `km_cj` VALUES ('jack', 5, 7);
INSERT INTO `km_cj` VALUES ('jack', 6, 8);
INSERT INTO `km_cj` VALUES ('lucy', 9, 1);
INSERT INTO `km_cj` VALUES ('lucy', 10, 2);
INSERT INTO `km_cj` VALUES ('lucy', 11, 3);
INSERT INTO `km_cj` VALUES ('lucy', 12, 4);
INSERT INTO `km_cj` VALUES ('lucy', 13, 1);
INSERT INTO `km_cj` VALUES ('lucy', 14, 1);
INSERT INTO `km_cj` VALUES ('lucy', 15, 2);
INSERT INTO `km_cj` VALUES ('lucy', 16, 3);
INSERT INTO `km_cj` VALUES ('mark', 17, 1);
INSERT INTO `km_cj` VALUES ('mark', 18, 2);
INSERT INTO `km_cj` VALUES ('mark', 19, 3);
INSERT INTO `km_cj` VALUES ('mark', 20, 4);
INSERT INTO `km_cj` VALUES ('mark', 21, 1);
INSERT INTO `km_cj` VALUES ('mark', 23, 1);
INSERT INTO `km_cj` VALUES ('mark', 24, 2);
INSERT INTO `km_cj` VALUES ('mark', 22, 1);
INSERT INTO `km_cj` VALUES ('mark', 25, 1);
INSERT INTO `km_cj` VALUES ('mayun', 26, 1);
INSERT INTO `km_cj` VALUES ('mayun', 27, 1);
INSERT INTO `km_cj` VALUES ('mayun', 28, 2);
INSERT INTO `km_cj` VALUES ('mayun', 30, 1);
INSERT INTO `km_cj` VALUES ('mayun', 31, 1);

SET FOREIGN_KEY_CHECKS = 1;



--   -----------------------------------------------------------------




-- -- 用窗口函数，求各科成绩第二名的学生

select * from (
select *,
DENSE_RANK() over (PARTITION by km ORDER BY cj desc) as d_r_num
from km_cj t) ta
where ta.d_r_num=2



-- 用普通方法求各科成绩第二名的学生
select tb_three.km, tb_three.user_id,tb_three.cj
from 
(select t_tu.km as km_two, max(t_tu.cj)  as max_cj_two from 
km_cj t_tu
where t_tu.user_id not in(
select tb.user_id
from
(select km as km_one,max(cj) as max_cj from km_cj t
group by  km) ta
LEFT JOIN  km_cj  tb
on  ta.km_one = tb.km and ta.max_cj = tb.cj)
GROUP BY t_tu.km) t_zj
LEFT JOIN km_cj  tb_three
on t_zj.km_two=tb_three.km and t_zj.max_cj_two = tb_three.cj

6、求连续3个相同的数。


DROP TABLE IF EXISTS `km_cj`;
CREATE TABLE `km_cj`  (
  `km` varchar(255) ,
  `user_id` int(0),
  `cj` bigint(0)
) 


INSERT INTO `km_cj` VALUES ('语文', 8, 1);
INSERT INTO `km_cj` VALUES ('语文', 7, 2);
INSERT INTO `km_cj` VALUES ('语文', 1, 3);
INSERT INTO `km_cj` VALUES ('语文', 2, 4);
INSERT INTO `km_cj` VALUES ('语文', 3, 5);
INSERT INTO `km_cj` VALUES ('语文', 4, 6);
INSERT INTO `km_cj` VALUES ('语文', 5, 7);
INSERT INTO `km_cj` VALUES ('语文', 6, 8);
INSERT INTO `km_cj` VALUES ('数学', 9, 1);
INSERT INTO `km_cj` VALUES ('数学', 10, 2);
INSERT INTO `km_cj` VALUES ('数学', 11, 3);
INSERT INTO `km_cj` VALUES ('数学', 12, 4);
INSERT INTO `km_cj` VALUES ('数学', 13, 1);
INSERT INTO `km_cj` VALUES ('数学', 14, 1);
INSERT INTO `km_cj` VALUES ('数学', 15, 2);
INSERT INTO `km_cj` VALUES ('数学', 16, 3);
INSERT INTO `km_cj` VALUES ('英语', 17, 1);
INSERT INTO `km_cj` VALUES ('英语', 18, 2);
INSERT INTO `km_cj` VALUES ('英语', 19, 3);
INSERT INTO `km_cj` VALUES ('英语', 20, 4);
INSERT INTO `km_cj` VALUES ('英语', 21, 1);
INSERT INTO `km_cj` VALUES ('英语', 23, 1);
INSERT INTO `km_cj` VALUES ('英语', 24, 2);
INSERT INTO `km_cj` VALUES ('英语', 22, 1);
INSERT INTO `km_cj` VALUES ('英语', 25, 1);
INSERT INTO `km_cj` VALUES ('体育', 26, 1);
INSERT INTO `km_cj` VALUES ('体育', 27, 1);
INSERT INTO `km_cj` VALUES ('体育', 28, 2);
INSERT INTO `km_cj` VALUES ('体育', 30, 1);
INSERT INTO `km_cj` VALUES ('体育', 31, 1);





--  -----------------------------------------------------------


---- 求连续3个相同的数。
select DISTINCT t_num
from  
(select id,num as t_num,
LEAD(num,1) over () as t1,
LEAD(num,2) over () as t2
from test_num) ta
where ta.t_num = ta.t1 and
       ta.t_num = ta.t2

7、关于学生成绩排名的所有问题查询总结：


DROP TABLE IF EXISTS `sc`;
CREATE TABLE `sc`  (
  `s_id` int(255) ,
  `s_g` varchar(10) ,
  `s_score` int(4)
) ;

-- ----------------------------
-- Records of sc
-- ----------------------------
INSERT INTO `sc` VALUES (1, 's1', 79);
INSERT INTO `sc` VALUES (2, 's1', 79);
INSERT INTO `sc` VALUES (3, 's1', 59);
INSERT INTO `sc` VALUES (4, 's2', 81);
INSERT INTO `sc` VALUES (5, 's2', 73);
INSERT INTO `sc` VALUES (6, 's3', 82);
INSERT INTO `sc` VALUES (7, 's3', 82);
INSERT INTO `sc` VALUES (8, 's3', 91);




--  ------------------------------------




select * from  sc;
-- s_id  学生id
-- s_g   科目
-- s_score 成绩


-- 实现窗口函数：ROW_NUMBER()
select s_id,s_g,s_score,row_num
from(
select *,
if(s_g=@s_g,@num:=@num+1,@num:=1) as row_num, @s_g:=s_g
from(
select * from sc,(select @num:=null) r
GROUP BY s_g,s_id,s_score
ORDER BY s_g,s_score desc) ta)tab;



-- 实现窗口函数：DENSE_RANK()
select s_id,s_g,s_score,row_num
from(
select *,
if(s_g=@s_g and s_score!=@s_score,@num:=@num+1,if(s_g=@s_g and s_score=@s_score,@num,@num:=1)) as row_num, @s_g:=s_g,@s_score:=s_score
from(
select * from sc,(select @num:=1) r
GROUP BY s_g,s_id,s_score
ORDER BY s_g,s_score desc) ta)tab



--  -------------------问题总结：

-- -- 第一次执行的时候，为什么，row_num 都等于1那？是Navicat的问题，还是sql语句的问题。



-- 一条sql语句 实现窗口函数：ROW_NUMBER(),DENSE_RANK()和RANK()
select 
s_id_ten as s_id,s_g_ten as s_g,s_score_ten as s_score,
row_num_one_ten as ROW_NUMBER ,row_num_ten as DENSE_RANK,
z_j_ban as rank
from (
select *,
if(row_num_one_ten=row_num_ten or @row_num_ten=row_num_ten,row_num_ten,row_num_one_ten) as z_j_ban,@row_num_ten:=row_num_ten
from(
select s_id as s_id_ten,s_g as s_g_ten,s_score as s_score_ten,row_num_one as row_num_one_ten,row_num as row_num_ten
from(
select *,
if(s_g=@s_g_one,@num_one:=@num_one+1,@num_one:=1) as row_num_one,@s_g_one:=s_g,
if(s_g=@s_g and s_score!=@s_score,@num:=@num+1,if(s_g=@s_g and s_score=@s_score,@num,@num:=1)) as row_num, @s_g:=s_g,@s_score:=s_score
from(
select * from sc,(select @num_one:=null,@num:=1) r
GROUP BY s_g,s_id,s_score
ORDER BY s_g,s_score desc) ta)tab)tabc)tabcd

8、在没有分组的情况下，实现rank()函数


DROP TABLE IF EXISTS `score`;
CREATE TABLE `score`  (
  `id` int(11),
  `name` varchar(255),
  `score` int(11) 
);


INSERT INTO `score` VALUES (1, '001', 100);
INSERT INTO `score` VALUES (2, '002', 99);
INSERT INTO `score` VALUES (3, '003', 99);
INSERT INTO `score` VALUES (4, '004', 96);
INSERT INTO `score` VALUES (5, '005', 95);
INSERT INTO `score` VALUES (6, '006', 95);
INSERT INTO `score` VALUES (7, '007', 65);




--  --------------------------------------- 上面是表及数据


select * from score;


-- 用 case when then  else end 的实现过程
select 
tmp.id,tmp.name,tmp.score,
-- 顺序一直在变大
@j:=@j+1 as j,
-- 只有在前后二次排序值不同时才会使用顺序号
@k:=(case when @pre_score=tmp.score then @k else @j end) as rank,
@pre_score:=tmp.score as pre_score
from 
(
-- 成绩排序
select * from score order by score desc
) tmp,
-- @k 表示最终的排名(相同值时序号相同） 
-- @j 表示顺序排名 
-- @pre_score上一次排序值
(select @k :=0,@j:=0, @pre_score:=0) sdcore



-- 用 if  的实现过程
select 
tmp.id,tmp.name,tmp.score,
-- 顺序一直在变大
@j:=@j+1 as j,
-- 只有在前后二次排序值不同时才会使用顺序号
@k:=if(@pre_score=tmp.score ,@k ,@j) as rank,
@pre_score:=tmp.score as pre_score_ed
from 
(
-- 成绩排序
select * from score order by score desc
) tmp,
-- @k 表示最终的排名(相同值时序号相同） 
-- @j 表示顺序排名 
-- @pre_score上一次排序值
(select @k :=0,@j:=0, @pre_score:=0) sdcore

9、简单版本：留存率的问题。

-- --- ----数据库表


DROP TABLE IF EXISTS `15.17_user_login`;
CREATE TABLE `15.17_user_login`  (
  `uid` varchar(255) ,
  `login_time` varchar(255)
) ;

INSERT INTO `15.17_user_login` VALUES ('1', '2019/1/1 6:00');
INSERT INTO `15.17_user_login` VALUES ('2', '2019/1/1 10:00');
INSERT INTO `15.17_user_login` VALUES ('3', '2019/1/1 19:00');
INSERT INTO `15.17_user_login` VALUES ('1', '2019/1/2 10:00');
INSERT INTO `15.17_user_login` VALUES ('2', '2019/1/2 9:00');
INSERT INTO `15.17_user_login` VALUES ('3', '2019/1/2 14:00');
INSERT INTO `15.17_user_login` VALUES ('1', '2019/1/3 8:00');
INSERT INTO `15.17_user_login` VALUES ('2', '2019/1/9 14:00');
INSERT INTO `15.17_user_login` VALUES ('3', '2019/1/9 10:00');
INSERT INTO `15.17_user_login` VALUES ('3', '2019/1/9 15:00');




--  -------------------



select(CASE 
	WHEN left(login_time,8)="2019/1/2" THEN
		"次日留存率"
	WHEN left(login_time,8)="2019/1/3" THEN
		"3日留存率"
	WHEN left(login_time,8)="2019/1/9" THEN
		"9日留存率"
	ELSE
		"其他时间留存率"
END 
) as len_u, count(uid) as id_num from (select * from `15.17_user_login` GROUP BY uid,left(login_time,8)) r
GROUP BY (
CASE 
	WHEN left(login_time,8)="2019/1/2" THEN
		"次日留存率"
	WHEN left(login_time,8)="2019/1/3" THEN
		"3日留存率"
	WHEN left(login_time,8)="2019/1/9" THEN
		"9日留存率"
	ELSE
		"其他时间留存率"
END 
)

10、

11、

12、

13、

14、

15、

16、

17、

18、

19、

20、

21、

22、

23、

24、

25、

26、

27、

28、

29、

30、

31、

32、

33、

34、

35、

36、

37、

38、

39、

40、

41、

42、

43、

44、

45、

46、

47、

48、

49、

50.

猜你喜欢

转载自blog.csdn.net/weixin_54217632/article/details/120819965
今日推荐
火速冲上 GitHub 热榜 —— 开源编程语言、框架哪有这么可爱？
北京人形机器人创新中心发布全球首个纯电驱拟人奔跑的全尺寸人形机器人“天工”
LFOSSA 源来如此公开课 | 掌握云原生未来：CNCF 认证全面攻略与备考秘籍
国产云输入法——仅华为无云端数据上传安全问题
开源日报 | 工业开源项目OGG 1.0；姐姐，你要和我一起配置火狐吗；苹果AI遥遥落后？Fedora 40
开放签电子签章：停止新增，优化体验，前进更进（五一假期前工作）
开源日报 | 中学生开源前端动画引擎；全球首个Llama3 8B中文版开源模型；联想电脑恐出局；Linus讽刺AI炒作
周排行
浏览器对同一域名进行请求的最大并发连接数
React Hook之自定义Hook
【转】MyBatis缓存机制
-Java-泛型
自动化测试常用脚本-发送邮件
LeetCode#859: Buddy Strings
java、Python处理字符串
第二篇の博客
Hadoop伪分布式环境安装
SQL Server进阶（十一）临时表、表变量
每日归档
更多
2024-04-27(56)
2024-04-26(39)
2024-04-25(22)
2024-04-24(36)
2024-04-23(26)
2024-04-22(39)
2024-04-21(0)
2024-04-20(6)
2024-04-19(5)
2024-04-18(0)

代码天地 © 2018 codetd.com

境外服务器   最新电视剧2022