golang 复杂json Unmarshal后如何取值

2019-09-20 10:55:26 407 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

weixin_38048154 墨衍会员 · AI 创作全网分发关注 + 加入社区

告别Keil/IAR：在VSCode中用CMake+Clangd打造丝滑的STM32开发环境（附完整配置）本文详细介绍了如何在VSCode中使用CMake和Clangd搭建高效的STM32开发环境，替代传统的Keil/IAR工具。通过完整的配置指南和快速开发技巧，帮助开发者实现轻量级、高度可定制的现代化开发流程，显著提升STM32项目的开发效率。阅读详情

```
package main

import "fmt"
import "encoding/json"

func main() {
json_list_str := `[{"ID":"7fab2946-9d9c-39d0-387e-a0239cf95351","Node":"consul-4","Address":"192.168.15.113","Datacenter":"dc1","TaggedAddresses":{"lan":"192.168.15.113","wan":"192.168.15.113"},"NodeMeta":{"consul-network-segment":""},"ServiceKind":"","ServiceID":"web4","ServiceName":"web4","ServiceTags":["master"],"ServiceAddress":"192.168.15.113","ServiceWeights":{"Passing":1,"Warning":1},"ServiceMeta":{},"ServicePort":8082,"ServiceEnableTagOverride":false,"ServiceProxyDestination":"","ServiceProxy":{},"ServiceConnect":{},"CreateIndex":2974,"ModifyIndex":2974}]`

var result []interface{}
err := json.Unmarshal([]byte(json_list_str), &result)
fmt.Println(result)

//todo 如何将“Address”取出？
}
```

...全文

2024年淘宝平板电脑销售数据分析本项目的目标是从淘宝电商平台上自动化爬取商品信息，并通过数据分析帮助了解价格、销量等相关指标，为商业决策提供支持。使用Selenium实现数据爬取，结合Python的数据处理与可视化工具，进一步分析和展示结果。阅读详情

weixin_38048154

其他技术讨论专区 479 成员 790,916 内容

+ 关注 + 加入社区

切换为时间正序

weixin_38079044 2019-09-20

打赏
举报

``` var result map[string]interface{} err := json.Unmarshal([]byte(json_list_str), &result) fmt.Println(err, result) ``` 提示报错了。 json: cannot unmarshal array into Go value of type map[string]interface {} map[] 在线代码执行链接 https://www.bytelang.com/o/s/c/c9YRSqkXPFs=

weixin_38087157 2019-09-20

打赏
举报

非常感谢支持！最终这样解决问题了。 ``` package main import "fmt" import "encoding/json" func main() { json_list_str := `[{"ID":"7fab2946-9d9c-39d0-387e-a0239cf95351","Node":"consul-4","Address":"192.168.15.113","Datacenter":"dc1","TaggedAddresses":{"lan":"192.168.15.113","wan":"192.168.15.113"},"NodeMeta":{"consul-network-segment":""},"ServiceKind":"","ServiceID":"web4","ServiceName":"web4","ServiceTags":["master"],"ServiceAddress":"192.168.15.113","ServiceWeights":{"Passing":1,"Warning":1},"ServiceMeta":{},"ServicePort":8082,"ServiceEnableTagOverride":false,"ServiceProxyDestination":"","ServiceProxy":{},"ServiceConnect":{},"CreateIndex":2974,"ModifyIndex":2974}]` var result []map[string]interface{} err := json.Unmarshal([]byte(json_list_str), &result) fmt.Println(err, result) // 取出result变量里面的"Address"字段。 fmt.Println(err, result[0]["Address"]) // output // <nil> [map[ServiceKind: ServiceTags:[master] ServicePort:8082 ServiceProxyDestination: TaggedAddresses:map[lan:192.168.15.113 wan:192.168.15.113] ServiceAddress:192.168.15.113 ServiceEnableTagOverride:false ServiceProxy:map[] ServiceConnect:map[] ModifyIndex:2974 ID:7fab2946-9d9c-39d0-387e-a0239cf95351 Datacenter:dc1 ServiceID:web4 ServiceName:web4 ServiceMeta:map[] CreateIndex:2974 Node:consul-4 Address:192.168.15.113 NodeMeta:map[consul-network-segment:] ServiceWeights:map[Passing:1 Warning:1]]] // <nil> 192.168.15.113 } ```

weixin_38096743 2019-09-20

打赏
举报

某些复杂的json也可以考虑这个包：https://github.com/tidwall/gjson

weixin_38052215 2019-09-20

打赏
举报

map[string]interface{}

weixin_38060006 2019-09-20

打赏
举报

json数据里面的"Address":"192.168.15.113"，需要如何取出来？

golang 复杂json Unmarshal后如何取值

相关推荐