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题解 | #确定最佳顾客的另一种方式（二）#

select c.cust_name, sum(o2.item_price * o2.quantity ) as total_price from Customers c inner join Orders o1 on c.cust_id = o1.cust_id inner join...

2024-11-26

0 57

购买prod_id 为 BR01 的产品的所有顾客的电子邮件

select c.cust_email from Customers c inner join Orders o1 on c.cust_id = o1.cust_id inner join OrderItems o2 on o1.order_num = o2.order_num wh...

2024-11-26

0 62

题解 | #插入记录（三）#

REPLACE INTO examination_info (exam_id,tag,difficulty,duration,release_time) values (9003, 'SQL', 'hard', 90, '2021-01-01 00:00:00') 应该使用'REPLACE INT...

2024-11-06

0 54

题解 | #返回每个顾客不同订单的总金额#

select t2.cust_id,sum(t1.item_price*t1.quantity) as total_ordered from OrderItems t1 left join Orders t2 on t1.order_num = t2.order_num group by t2...

2024-11-05

0 58

题解 | #统计每个单词出现的个数#

#!/bin/bash awk -F" " '{for(i=1;i<=NF;i++)a[$i]++}END{for(i in a)print i,a[i]}' nowcoder.txt | sort -k2

2024-09-09

1 113

-查询某个IP的详细访问情况#

#!/bin/bash cat nowcoder.txt | grep "192.168.1.22" | awk '{print $7}' | sort | uniq -c | awk '{print $1 " " $2 }'

2024-09-09

0 115

题解 | #netstat练习1-查看各个状态的连接数#

#!/bin/bash cat nowcoder.txt | grep "tcp"|awk '{a[$6]++}END{for(i in a)print i,a[i]}'| sort -rn -k2

2024-09-08

0 125

题解 | #找出字符串中第一个只出现一次的字符#

#include <iostream> #include<unordered_map> using namespace std; int main() { string s; unordered_map<int, char>st; cin...

2023-10-09

0 260

题解 | #字符串排序#

#include <iostream> #include <string> #include<vector> #include<algorithm> using namespace std; int main() { vector<s...

2023-10-09

0 278

题解 | #数字颠倒#

#include <iostream> #include<algorithm> using namespace std; int main() { string s; cin>>s; reverse(s.begin(), s....

2023-10-09

0 283