Spring Boot + Vue3 前后端分离 实战wiki知识库系统

osc_43954056 2021-04-07 22:27:43
java Mysql Spring Boot javascript Vue.js


download:Spring Boot + Vue3 前后端分离 实战wiki知识库系统

通过一个界面简约、功能完整的知识库项目,带你快速掌握Spring Boot核心知识+ Vue全家桶全栈技能。在此过程中,打造一套通用的组件工具(可应用于其他项目),掌握前后端分离开发模式,搭建一个专属自己的、内容可灵活配置的知识库系统。

技术要求
具备Java、MySQL基础
具备HTML+CSS+JavaScript基础
环境参数
Spring Boot 2.4
Vue3
Vue CLI 4.5
函數操作
對條件字段做函數操作走不了索引。

select * from t1 where date =‘2019-05-21’;
優化:改成範圍查询

select * from t1 where c>=‘2019-05-21 00:00:00’ and c<=‘2019-05-21 23:59:59’;
隱式轉換
操作符與不同類型的操作對象一同運用時,就會發作類型轉換以使操作兼容。

select user_name,tele_phone from user_info where tele_phone =11111111111; / tele_phone varchar /
實践會做函數操作:

select user_name,tele_phone from user_info where cast(tele_phone as singed int) =11111111111;
優化:類型統一

select user_name,tele_phone from user_info where tele_phone =‘11111111111’;
含糊查询
通配符在前面

select * from t1 where a like ‘%1111%’;
優化:含糊查询必需包含條件字段前面的值

select * from t1 where a like ‘1111%’;
範圍查询
範圍查询數據量太多,需求回表,因而不走索引。

select * from t1 where b>=1 and b <=2000;
優化:降低單次查询範圍,分屢次查询。(實践可能速度沒得快太多,倡議走索引)

select from t1 where b>=1 and b <=1000;
show profiles;
±---------±-----------±-----------------------------------------+
| Query_ID | Duration | Query |
±---------±-----------±-----------------------------------------+
| 1 | 0.00534775 | select 
from t1 where b>=1 and b <=1000 |
| 2 | 0.00605625 | select * from t1 where b>=1 and b <=2000 |
±---------±-----------±-----------------------------------------+
2 rows in set, 1 warning (0.00 sec)
計算操作
即便是简單的計算

explain select * from t1 where b-1 =1000;
優化:將計算操作放在等號後面

explain select * from t1 where b =1000 + 1;

翻了很多题解,只能看懂这种解法,够直观够暴力。

class Solution {
public:
vector<string> restoreIpAddresses(string s) {
vector<string> res;

 for (int a = 1; a < 4; a ++ )
for (int b = 1; b < 4; b ++ )
for (int c = 1; c < 4; c ++ )
for (int d = 1; d < 4; d ++ ) //abcd分别表示四段ip地址长度
{
if (a + b + c + d == s.size()) //四段长度刚好
{
string s1 = s.substr(0, a); //分别截取四段ip地址
string s2 = s.substr(a, b);
string s3 = s.substr(a + b, c);
string s4 = s.substr(a + b + c);
if (check(s1) && check(s2) && check(s3) && check(s4))
{
string ip = s1 + '.' + s2 + '.' + s3 + '.' + s4;
res.push_back(ip);
}
}
}
return res;
}
bool check(string s) //判断ip地址每段的第一位不为0,或只有一位且该位为0
{
if (stoi(s) <= 255)
if (s[0] != '0' || (s[0] == '0' && s.size() == 1)) return true;
return false;
}
版权声明
本文为[osc_43954056]所创,转载请带上原文链接,感谢
https://my.oschina.net/u/5067889/blog/5011516

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