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三表多表查询 左连接 太常见了 数据量如果过大导致查询速度不及预期（4、5秒之内）应当优化
三表多表查询
左连接
太常见了
数据量如果过大导致查询速度不及预期（4、5秒之内）应当优化
select u.code,u.username, c.*
from ad_call_record r
left join lego_user.user u on r.user_id = u.id
left join ad_callback_content c on r.ref_id = c.id
win10老是自动更新，彻底禁止自动更新
Java两个数字字符串相加、两个数字字符串相乘（含%）
Java两个数字字符串相加
//两个数字字符串相加
String a = "75.2";
String b = "77.5504000";
BigDecimal c = new BigDecimal(a);
BigDecimal add = c.add(new BigDecimal(b));
System.out.println(add.toString());
//如果要保留两位小数
System.out.println("=====" + add.setScale(2, BigDecimal.ROUND_HALF_UP));
两个数字字符串相乘（含%）
//两个数字字符串相乘
String a = "75.233";
String b = "50";//50%
BigDecimal c = new BigDecimal(a);
BigDecimal multiply = c.multiply(new BigDecimal(b)).divide(new BigDecimal(100));
System.out.println(multiply.toString());
//如果要保留两位小数
System.out.println("=====" + multiply.setScale(2, BigDecimal.ROUND_HALF_UP));
首单付费用户总数sql
select DISTINCT lo.place_user_id AS '首单付费用户总数' from lego_order.order lo where DATE_FORMAT(lo.pay_time,'%Y-%m-%d') = '2021-12-01'
and (lo.pay_environment is null or lo.pay_environment != 'Sandbox')
and lo.status = 2 and lo.place_user_type = 1
and lo.pay_amount > 0 and lo.pay_type not in ('SSCM')
and lo.platform = 'sscm'
and lo.place_user_id not in (
select place_user_id from lego_order.order lo where DATE_FORMAT(lo.pay_time,'%Y-%m-%d') <'2021-12-01' and lo.platform = 'sscm'
and (lo.pay_environment is null or lo.pay_environment != 'Sandbox')
and lo.status = 2 and lo.place_user_type = 1
and lo.pay_amount > 0 and lo.pay_type not in ('SSCM')
and lo.platform = 'sscm'
)#19
@GetMapping要使用@RequestParam(“userId“)/@PathVariable(name = “userCode“)
@GetMapping要使用@RequestParam(“userId”)/@PathVariable(name = “userCode”)
否则会报错
@PostMaping---->@RequestBody(对象)
常见数据库表后缀
copy结尾的是备份
flow结尾的是流水
rel结尾的是中间关联表
statistics结尾的是数据统计表
log结尾的是日志表
config结尾的是配置表
for循环遍历数组、集合（含代码）
1、泛型数组遍历出来，集合用ArrayList,遍历格式为(foreach)：
for(类（将所有类型带过来） 循环变量名称（自定义一个e）: 要被遍历的对象){
}2、for(元素的数据类型（int这些，包装类型integer也可） 变量自定义一个 : Collection集合or数组){
}
3、上代码：People.java
package com.equals;public class People {
private Integer id;
private String name;
private String post;
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getPost() {
return post;
}
public void setPost(String post) {
this.post = post;
}
public People(String name,String post){//
this.id = id;
this.name = name;
this.post = post;
}
@Override public String toString() {
return "People{" + "id=" + id + ", name='" + name + '\'' + ", post='" + post + '\'' + '}';
}}
Test.java
package com.equals;import java.util.ArrayList;import java.util.Collection;public class test{ public static void main(String[] args) {
ArrayList<People> people = new ArrayList<>();
people.add(new People("zhoonghau","php"));
people.add(new People("zhoonghauzhognhuada","java1000")); people.add(new People("zhoonghau3","php1111"));
for (People e:people ) {
System.out.println("People{" + "id=" + e.getId() + ", name='" + e.getName() + '\'' + ", post='" + e.getPost() + '\'' + '}'); }
String[] arr = new String[]{"学生","职场人","管理员"}; Integer[] arr2 = new Integer[]{1,2,23};
for (Integer w:arr2 ) {
System.out.print(w+" ");
}
System.out.println();
Collection<String> collection = new ArrayList<>(); collection.add("jiyu "+"java");
collection.add("xiaaohuadonghe");
collection.add("zhongtongsida");
for (String z:collection){
System.out.println(z);
} }}
4、结果
People{id=null, name=‘zhoonghau’, post=‘php’}People{id=null, name=‘zhoonghauzhognhuada’, post=‘java1000’}People{id=null, name=‘zhoonghau3’, post=‘php1111’}1 2 23 jiyu javaxiaaohuadonghezhongtongsida
访问字符串
下面函数get_str2返回数组str的首地址，而从内存的堆区申请字符串的空间，可返回全部的
#include <stdio.h> char *get_str2(){ char str2[] = {"testing local pointer"}; return str2;
} char *get(){ char str; str = (char)malloc(100); if(!str) return NULL; strcpy(str,"testing local pointer"); return str; } int main(){ char *p; int i; char *m; int j; char a[] = {"testing local pointer"};
printf("第一种方法输出的是：\n"); p = get_str2();
for(i=0;*(p+i);i++)
putchar(*(p+i));
printf("\n");
m = get(); printf("第二种方法输出的是：\n"); for(j=0;(m+j);j++) putchar((m+j)); printf("\n");
}
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