英文:
ERROR: need type assertion
问题
我以为我已经断言过了（就我学习Go的程度而言），但我一直收到这个错误消息：cannot use readBack["SomePIN"] (type interface {}) as type string in argument to c.String: need type assertion
以下是我的代码（这段代码片段来自一个请求处理函数，我正在使用Echo Web框架和Tiedot NoSQL数据库）：
// 要获取查询结果文档，只需按照Tiedot的readme.md中所述进行读取
for id := range queryResult {
readBack, err := aCollection.Read(id)
if err != nil {
panic(err)
}
if readBack["OtherID"] == otherID {
if _, ok := readBack["SomePIN"].(string); ok {
return c.String(http.StatusOK, readBack["SomePIN"])
}
}
}
请注意，这是一个代码片段，可能需要根据你的实际情况进行适当的调整。
英文:
I thought I have asserted (as far as I've learnt Go), but I keep getting this error
cannot use readBack["SomePIN"] (type interface {}) as type string in argument to c.String: need type assertion
Here is my code (this snippet is from a Request Handler function and I'm using Echo Web framework and Tiedot NoSQL database)
// To get query result document, simply
// read it [as stated in the Tiedot readme.md]
for id := range queryResult {
readBack, err := aCollection.Read(id)
if err != nil {
panic(err)
}
if readBack["OtherID"] == otherID {
if _, ok := readBack["SomePIN"].(string); ok {
return c.String(http.StatusOK, readBack["SomePIN"])
}
}
}
答案1
得分: 6
你正在将readBack["SomePIN"]断言为字符串 - 在if语句中。然而，这对readBack["SomePIN"]没有任何改变 - 它仍然是一个interface{}类型。在Go语言中，类型是不会改变的。以下是可以工作的代码示例：
for id := range queryResult {
readBack, err := aCollection.Read(id)
if err != nil {
panic(err)
}
if readBack["OtherID"] == otherID {
if somePIN, ok := readBack["SomePIN"].(string); ok {
return c.String(http.StatusOK, somePIN)
}
}
}
你之前丢弃了类型断言的字符串值，但是你需要它。所以保留它，作为somePIN，然后使用它。
最后注意一点 - 使用value, ok = interfaceVal.(type)的语法是一个好的实践。如果interfaceVal不是字符串类型，你将得到value = ""和ok = false。如果你从类型断言中省略了ok值，并且interfaceVal不是字符串类型，程序将会panic。
英文:
You are asserting readBack["SomePIN"] as a string - in the if statement. That doesn't make any change to readBack["SomePIN"], however - it's still an interface{}. In Go, nothing ever changes type. Here's what will work:
for id := range queryResult {
readBack, err := aCollection.Read(id)
if err != nil {
panic(err)
}
if readBack["OtherID"] == otherID {
if somePIN, ok := readBack["SomePIN"].(string); ok {
return c.String(http.StatusOK, somePIN)
}
}
}
You were tossing the string value from your type assertion, but you want it. So keep it, as somePIN, and then use it.
Final note - using the value, ok = interfaceVal.(type) syntax is a good practice. If interfaceVal turns out to be a non-string, you'll get value = "" and ok = false. If you eliminate the ok value from the type assertion and interfaceVal is a non-string, the program will panic.
答案2
得分: 3
看起来你正在将其转换为具体类型并丢弃转换结果，我认为这样应该可以工作：
if somePinString, ok := readBack["SomePIN"].(string); ok {
return c.String(http.StatusOK, somePinString)
}
请注意，这是一段Go语言代码。
英文:
It looks like your converting to a concrete type and throwing away the conversion, I think this should work:
if somePinString, ok := readBack["SomePIN"].(string); ok {
return c.String(http.StatusOK, somePinString)
}
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