@L_197_5@现象

var PeopleLimit =make(map[String]map[String]int64)

func foo(arr1,arr2 String){
if x,ok := PeopleLimit[arr1][arr2];ok{
fmt.Println(" ok")
fmt.Println(X)
}else{
PeopleLimit[arr1] = make(map[String]int64)
fmt.Println(" not ok")
}

//添加新元素
PeopleLimit[arr1][arr2] = time.Now().Unix()
}

func main() {

foo("a","b")
foo("c","d")
foo("a","f")
fmt.Println(PeopleLimit)
return
}

按理应该要输出三个元素，但第一个被覆盖了，定位了很久才发现是二维map的判断问题：
执行foo("a","f")时
if x,ok := PeopleLimit[arr1][arr2];判断是二级key也不存在,但在执行第三次foo("a","f")时，@H_575_64@map[a][b]是存在，但@H_575_64@map[a][f]不存在，于是，进入else后重新@H_575_64@make将@H_575_64@map[a]清空了！！

修改后:

var PeopleLimit =make(map[String]map[String]int64)

func foo(arr1,ok := PeopleLimit[arr1];ok{
//修改点
if y,ok := x[arr2];ok{
fmt.Println(" ok")
fmt.Println(x,y)
}
}else{
PeopleLimit[arr1] = make(map[String]int64)
fmt.Println(" not ok")
}

//添加新元素
PeopleLimit[arr1][arr2] = time.Now().Unix()
}

func main() {

foo("a","b") foo("c","d") foo("a","f") fmt.Println(PeopleLimit) return}--------------------- 作者：yinnnnnnn 来源：CSDN 原文：https://blog.csdn.net/qq_35440678/article/details/83154780 版权声明：本文为博主原创文章，转载请附上博文链接！