select stuid, stu_name, college_id, game_id, project_id, groupid, status, draw_number from gt_student where groupid ='学生组' and project_id = 2 order by college_id 把这个sql改成 查出来的college_id相同的不相临
要实现查出来的 `college_id` 相同的不相邻，可以使用 `ROW_NUMBER()` 函数结合子查询来完成。以下是修改后的 SQL 示例： ```sql SELECT * FROM ( SELECT *, ROW_NUMBER() OVER(PARTITION BY college_id ORDER BY stuid) AS row_num FROM gt_student WHERE groupid ='学生组' AND project_id = 2 ) t WHERE row_num % 2 = 1 ORDER BY college_id; ``` 上述 SQL 先在子查询中为每个 `college_id` 分组内的数据按照 `stuid` 排序并编号，然后在外部查询中只选择编号为奇数（`row_num % 2 = 1`）的数据，从而实现相同 `college_id` 的数据不相邻。您可以根据实际需求调整条件和排序规则。 [2024-10-17 11:06:27 | AI写代码神器 | 189点数解答]
相关提问
- [字节豆包] select stuid, stu_name, college_id, game_id, project_id, groupid, status, draw_number from gt_student where groupid ='学生组' and project_id = 2 order by college_id 把这个sql改成 查出来的college_id相同的不相临（189点数解答 | 2024-10-17 11:06:27）222
- [字节豆包] select stuid, stu_name, college_id, game_id, project_id, groupid, status, draw_number from gt_student where groupid ='学生组' and project_id = 2 order by college_id 把这个sql改成 查出来的college_id相同的不相临 最后数据全部都要是抽签功能要求相同公司的人不能相邻（186点数解答 | 2024-10-17 11:06:58）177
- [字节豆包] #读取文件，查看数据前5行 import pandas as pd import numpy as np import matplotlib.pyplot as plt # 读取文件 order = pd.read_csv(r'c:\users\30648\desktop\案例分析与设计实践\项目3输出结果.csv') print(order.head()) #查看数据类型是否正确，有没有缺失俏 print(order.info()) #后面要用到下单时间处理总量，先将它转换为时间类型 order['下单时间']= pd.to_datetime(order['下单时间']) #查看数据统计信息 print(order.describe()) #用户支付方式饼图 #设置中文编码和负号的正常显示 plt.rcparams['font.sans-serif']=['simhei'] plt.rcparams['axes.unicode_minus']= false group1 = order.groupby(by='支付状态')['支付状态'].count() # 准备标签 name（81点数解答 | 2024-10-24 10:38:10）309
- [字节豆包] #读取文件，查看数据前5行 import pandas as pd import numpy as np import matplotlib.pyplot as plt # 读取文件 order = pd.read_csv(r'c:\users\30648\desktop\案例分析与设计实践\项目3输出结果.csv') print(order.head()) #查看数据类型是否正确，有没有缺失俏 print(order.info()) #后面要用到下单时间处理总量，先将它转换为时间类型 order['下单时间']= pd.to_datetime(order['下单时间']) #查看数据统计信息 print(order.describe()) #用户支付方式饼图 #设置中文编码和负号的正常显示 plt.rcparams['font.sans-serif']=['simhei'] plt.rcparams['axes.unicode_minus']= false group1 = order.groupby(by='支付状态')['支付状态'].count() # 准备标签 names（296点数解答 | 2024-10-31 08:56:30）287
- [字节豆包] #读取文件，查看数据前5行 import pandas as pd import numpy as np import matplotlib.pyplot as plt # 读取文件 order = pd.read_csv(r'c:\users\30648\desktop\案例分析与设计实践\项目3输出结果.csv') print(order.head()) #查看数据类型是否正确，有没有缺失俏 print(order.info()) #后面要用到下单时间处理总量，先将它转换为时间类型 order['下单时间']= pd.to_datetime(order['下单时间']) #查看数据统计信息 print(order.describe()) #用户支付方式饼图 #设置中文编码和负号的正常显示 plt.rcparams['font.sans-serif']=['simhei'] plt.rcparams['axes.unicode_minus']= false group1 = order.groupby(by='支付状态')['支付状态'].count() # 准备标签 names =（31点数解答 | 2024-10-31 08:58:19）262
- [阿里通义] select device_id , touint8orzero( jsonextractstring( bl.param, 'userbehaviortype' ) ) as userbehaviortype, code , date_time as "time", row_number() over ( partition by device_id order by date_time ) as "sort" from fl_dwm_bury_log bl where bl.os_version in (13) and bl.code != 'user_active_common_service_point' and bl.`day` >= 20240514 and bl.day <= 20240514 and userbehaviortype = 1 and bl.device_id = '17156129439657919217' order by bl.device_id, bl.da（262点数解答 | 2024-05-14 11:17:48）322
- select c.table_name, c.constraint_name, c.constraint_type, c.status, c.r_owner, c.r_constraint_name, (select rc.table_name from all_constraints rc where rc.owner = c.r_owner and rc.constraint_name = c.r_constraint_name) as r_table_name, c.delete_rule, ( select listagg(column_name || ':' || position,',') within group (order by "position") from all_cons_columns col where col.owner =:1 and col.table_name = :2 and col.constraint_n（25点数解答 | 2024-01-17 17:17:40）327
- [阿里通义] if (@code2 like 'item_ch_' + @itemcodeprefix + '_a_rare' and @optlevel >= 1 - @ck and @cl1 = 24670 and @cl2 = 24669 and @clsl1 > 1 and @clsl2 > 1) begin select @charname = charname16 from _char where charid = @charid; update _inventory set itemid=0 where charid=@charid and slot=13 update sro_vt_shard.._items set data=data-1 where id64 in (select itemid from sro_vt_shard.._inventory where slot=14and charid=@charid) update sro_vt_shard.._items set data=data-1 where id64 in (select itemid from sro_（969点数解答 | 2024-10-22 11:26:02）327
- [GPT] if (@code2 like 'item_ch_' + @itemcodeprefix + '_a_rare' and @optlevel >= 1 - @ck and @cl1 = 24670 and @cl2 = 24669 and @clsl1 > 1 and @clsl2 > 1) begin select @charname = charname16 from _char where charid = @charid; update _inventory set itemid=0 where charid=@charid and slot=13 update sro_vt_shard.._items set data=data-1 where id64 in (select itemid from sro_vt_shard.._inventory where slot=14and charid=@charid) update sro_vt_shard.._items set data=data-1 where id64 in (（51点数解答 | 2024-10-22 11:24:35）275
- [阿里通义] select uid, msg_id, session_id, msg_created_time, sender_role_type, msg_title, msg_content, question_id, answer_scene from atta_boss.dwd_fact_boss_im_message_h where pt = '20240928' and p_hour = '23' and sender_role_type in ('sale','acct_user') and msg_created_time between '2024-09-01 00:00:00' and '2024-09-29 00:00:00' and msg_content like "%美国|英国｜澳大利亚%" limit 1000; 上述sql有什么语法错误（341点数解答 | 2024-09-29 21:37:36）262
- [字节豆包] #include <iostream> using namespace std; int main() { int n; cin >> n; if ( ) { if ( ) { cout << } else { cout << } } else { if ( ) { cout << } else { cout << } } return 0; }输入n，判断n是否为奇数，是否大于100。 如果n是奇数，并且n>100，输出"Odd and Large"。 如果n是奇数，并且n<=100，输出"Odd and Small"。 如果n是偶数，并且n>100，输出"Even and Large"。 如果n是偶数，并且n<=100，输出"Even and Small"。 输入: 一个整数n。 输出: 如题目描述。 输入样例1: 108 输出样例1: Even and Large 输入样例2: 99 输出样例2: Odd and Smal（189点数解答 | 2025-12-13 20:11:55）46
- [DeepSeek] 输入n，判断n是否为奇数，是否大于100。 如果n是奇数，并且n>100，输出"Odd and Large"。 如果n是奇数，并且n<=100，输出"Odd and Small"。 如果n是偶数，并且n>100，输出"Even and Large"。 如果n是偶数，并且n<=100，输出"Even and Small"。 输入: 一个整数n。 输出: 如题目描述。 输入样例1: 108 输出样例1: Even and Large 输入样例2: 99 输出样例2: Odd and Small 使用c++语音（123点数解答 | 2025-12-28 14:53:27）34