duanmei1850 2016-06-11 07:51
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Golang检查接口类型是否为nil

When unmarshalling a string of json, using Golang's json.Unmarshal() function, I always unmarshal the string into a map[string]interface{} - I'm not sure weather there is a more optimal way of decoding json.

To get to the point:

Sometimes the json's unmarshalled type is nil, not string (or int etc.). This always throws a panic, saying:

interface conversion: interface is nil, not int

How can I avoid this panic or check if the interface's type is nil or not?

Example:

Here is an example of my problem in action: https://play.golang.org/p/0ATzXBbdoS

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  • dtmjl4427 2016-06-11 08:18
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    Check is the key exist instead of letting it panic.

    func keyExists(decoded map[string]interface{}, key string) {
        val, ok := decoded[key]
        return ok && val != nil
    }
    
    func main() {
    
    jsonText := `{
            "name": "Jimmy",
            "age":  23
        }`
    
        var decoded map[string]interface{}
    
        if err := json.Unmarshal([]byte(jsonText), &decoded); err != nil {
            fmt.Println(err)
            os.Exit(0)
        }
    
        if keyExists(decoded, "name") {
            fmt.Println(decoded["name"].(string))
        }
        if keyExists(decoded, "age") {
            fmt.Println(decoded["age"].(float64))
        }
        if keyExists(decoded, "gender") {
            fmt.Println(decoded["gender"].(int))
        }
    }
    

    Also, this is far from being optimal if you know what your json will look like. As specified in the documentation, you can unmarshal it directly into a struct:

    type Human struct {
        Name string
        Age int
        Gender int
    }
    
    func main() {
        jsonText := `{
            "name": "Jimmy",
            "age":  23
        }`
    
        decoded := Human{}
    
        if err := json.Unmarshal([]byte(jsonText), &decoded); err != nil {
            fmt.Println(err)
            os.Exit(0)
        }
    
        fmt.Println(decoded.Name)
        fmt.Println(decoded.Age)
        fmt.Println(decoded.Gender)
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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