英文:
Re-using Redigo connection instead of recreating it every time
问题
连接到Redigo并在函数内部操作数据就像涂抹黄油一样容易，但当你需要重复使用连接时，问题就来了，显然是出于性能/实用性的原因。
像这样在函数内部操作是可以的：
func main() {client, err := redis.Dial("tcp", ":6379")if err != nil {panic(err)}defer client.Close()client.Do("GET", "test:1")}
但是将其放在外部就不行：
var Client = redis.Dial("tcp", ":6379")defer Client.Close()func main() {Client.Do("GET", "test:1")}
会返回以下错误：
./main.go:1: multiple-value redis.Dial() in single-value context./main.go:2: non-declaration statement outside function body
我尝试将连接放在const（常量）中，将defer放在主函数中，但都没有奏效。
这是一个更大的问题，因为我有许多其他函数需要与Redis通信，但每次重新创建Redis连接似乎很愚蠢。
Redigo API只展示了如何创建一个Dial实例，但没有进一步解释如何重复使用它。
也许你在我的讲话中迷失了，但我想在这里提供一些背景信息，所以我的明确而简明的问题是：你如何重复使用（而不是每次重新创建）Redigo连接？
英文:
Connecting to Redigo and manipulating data inside a function is easy like butter, but the problem comes when you have to re-use its connection, obviously for performance/practicality reasons.
Doing it inside a function like this works:
func main() {client, err := redis.Dial("tcp", ":6379")if err != nil {panic(err)}defer client.Close()client.Do("GET", "test:1")}
But bringing it outside doesn't:
var Client = redis.Dial("tcp", ":6379")defer Client.Close()func main() {Client.Do("GET", "test:1")}
With the following error(s) returned:
./main.go:1: multiple-value redis.Dial() in single-value context./main.go:2: non-declaration statement outside function body
I've tried putting the connection as a const(ant), putting defer inside the main function to my dismay not working too.
This is an even bigger concern as I have many other functions that have to communicate to Redis, but recreating the connection to Redis everytime seems silly.
The Redigo API just shows how to create a Dial instance but doesn't go further by explaining how to re-use it.
You may've been lost in my talk, but I wanted to put a bit of context here, so my clear and concise question is: How do you go about re-using (not recreating everytime) a Redigo connection?
答案1
得分: 21
最好的方法是使用Pools，这里简要介绍了它的用法：Redigo Pools。
全局变量最终不会重用连接，所以我最终使用了类似下面的代码（使用了之前提到的Pools）：
func newPool() *redis.Pool {return &redis.Pool{MaxIdle: 80,MaxActive: 12000, // 最大连接数Dial: func() (redis.Conn, error) {c, err := redis.Dial("tcp", ":6379")if err != nil {panic(err.Error())}return c, err},}}var pool = newPool()func main() {c := pool.Get()defer c.Close()test, _ := c.Do("HGETALL", "test:1")fmt.Println(test)}如果你想在另一个函数中重用这个连接池，可以这样做：```gofunc test() {c := pool.Get()defer c.Close()test2, _ := c.Do("HGETALL", "test:2")fmt.Println(test2)}
英文:
The best way turned out to be using Pools, which are briefly documented here: Redigo Pools.
A global variable won't eventually reuse a connection, so I ended up with something like this (using Pools as noted before):
func newPool() *redis.Pool {return &redis.Pool{MaxIdle: 80,MaxActive: 12000, // max number of connectionsDial: func() (redis.Conn, error) {c, err := redis.Dial("tcp", ":6379")if err != nil {panic(err.Error())}return c, err},}}var pool = newPool()func main() {c := pool.Get()defer c.Close()test,_:=c.Do("HGETALL", "test:1")fmt.Println(test)}
If for example you want to reuse a pool inside another function you do it like this:
func test() {c := pool.Get()defer c.Close()test2,_:=c.Do("HGETALL", "test:2")fmt.Println(test2)}
答案2
得分: -1
redis.Dial()方法返回客户端错误。要修复它，你应该将以下代码替换为：
var Client, _ = redis.Dial("tcp", ":6379")
英文:
The redis.Dial() method returns client error. To fix it, you should replace:
var Client = redis.Dial("tcp", ":6379")
with:
var Client, _ = redis.Dial("tcp", ":6379")
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