[英]Go json.Unmarshall() works with single entity but not with slice
I'm using Go for a simple http client. 我正在将Go用于简单的http客户端。 Here's the entity I'm unmarshalling:
这是我要解组的实体:
type Message struct {
Id int64
Timestamp int64
Text string
Author User
LastEdited int64
}
type User struct {
Id int64
Name string
}
A single entity looks like this in JSON: 单个实体在JSON中如下所示:
{
"text": "hello, can you hear me?",
"timestamp": 1512964818565,
"author": {
"name": "andrea",
"id": 3
},
"lastEdited": null,
"id": 8
}
Go/json has no problem unmarshalling the single entity: Go / json可以解组单个实体没有问题:
var m Message
err = json.Unmarshal(body, &m)
if err != nil {
printerr(err.Error())
}
println(m.Text)
However, if the return of the endpoint is multiple entities: 但是,如果端点的返回是多个实体:
[
{
"text": "hello, can you hear me?",
"timestamp": 1512964800981,
"author": {
"name": "eleven",
"id": 4
},
"lastEdited": null,
"id": 7
}
]
And I change my corresponding Unmarshall to work on a slice of structs, Go throws an error: 然后,我将对应的Unmarshall更改为可在一片结构上工作,Go抛出错误:
var m []Message
err = json.Unmarshal(body, &m)
if err != nil {
printerr(err.Error()) // unexpected end of JSON input
}
for i := 0; i < len(m); i++ {
println(m[i].Text)
}
What gives? 是什么赋予了?
Works fine for me (try it on playground ), where are you getting the payload data from? 对我来说工作正常(在操场上尝试),您从哪里获取有效载荷数据? sounds like that's truncating it.
听起来好像被截断了。
package main
import (
"encoding/json"
"fmt"
)
type Message struct {
Id int64
Timestamp int64
Text string
Author User
LastEdited int64
}
type User struct {
Id int64
Name string
}
func main() {
body := []byte(`[
{
"text": "hello, can you hear me?",
"timestamp": 1512964800981,
"author": {
"name": "eleven",
"id": 4
},
"lastEdited": null,
"id": 7
}
]`)
var m []Message
err := json.Unmarshal(body, &m)
if err != nil {
fmt.Printf("error: %v") // unexpected end of JSON input
}
for i := 0; i < len(m); i++ {
fmt.Println(m[i].Text)
}
}
running it gives this output 运行它给出了这个输出
hello, can you hear me?
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.