<img src="https://s2.loli.net/2025/03/07/2wJs973gNcFAdEm.jpg" >


感谢建议。目前可以使用逗号分割备份多库，例如：db1,db2,db3

别来江浙沪，大同小异，全是人造景观。

@gpt5 <-- 人家在反串你们看不出来，还认真了（ doge

Q: How many "r" comes in the "rrrrrrabbrritrr"


A:

Thinking
To solve this problem, we need to count the number of "bbrritrr" in the given string. Let's break it down step by step:

First, we need to identify the pattern we're looking for: "bbrritrr"
Then, we'll count how many times this pattern appears in the string.
Let's count:

"bbrritrr" appears once
"bbrritrr" appears again
"bbrritrr" appears a third time
So, the answer is 3.

Reflection: Wait, I made a mistake in my counting. Let me double-check the string:

"bbrritrrbbrritrrbbrritrr"

I was correct in my initial count. There are indeed 3 occurrences of "bbrritrr" in the string.

Therefore, the correct answer is 3.

Response
There are 3 "bbrritrr" in the given string.

灯头灯尾的绿灯，有时候安全性不如灯中的红灯。

https://fast.ex.noerr.eu.org/t/689751 看看这个帖子，来自 2020 年的感悟。

没感觉到。加个微信吧，下次感到头晕直接 v 我

收收 python 、js 、php 味。go 的正规业务里，都要得先定义 struct ，不到万不得已不用 map[string]any

@mangoDB 不搞了

昨夜见军帖，可汗...

根据我的观察，flutter 还是用 vscode 写的人多

没汉化？你的 java 是基于易语言引擎特供版吗？

@body007 这样是不行的，function type 是 unique ，不能通过断言比较。

建议用 interface 的实现来判断，如果强行判断的话，那本质是比较 function signature 。拿到 test 和 fn 的 type 然后比较即可。

```go

type fn func(ctx context.Context) []string

func test(ctx context.Context) []string {
return []string{"1"}
}

func main() {
if matchFunctionSignature(reflect.TypeOf(test), reflect.TypeOf((fn)(nil))) {
fmt.Println(true)
} else {
fmt.Println(false)
}
}

func matchFunctionSignature(a, b reflect.Type) bool {
if a.Kind() != reflect.Func || b.Kind() != reflect.Func {
return false
}

if a.NumIn() != b.NumIn() || a.NumOut() != b.NumOut() {
return false
}

for i := 0; i < a.NumIn(); i++ {
if a.In(i) != b.In(i) {
return false
}
}

for i := 0; i < a.NumOut(); i++ {
if a.Out(i) != b.Out(i) {
return false
}
}
return true
}

```

quick_sort = lambda arr: [] if not arr else quick_sort([val for val in arr[1:] if val <= arr[0]]) + [arr[0]] + quick_sort([val for val in arr[1:] if val > arr[0]])