json - 在 Go 中合并动态数据结构

Question

我有这个传入的有效负载，我无法更改。

{
"source": "some random source",
"table": "hosts_table",
"data": [
    ["address", "id", "services_with_info"],
    ["0.0.0.1", 1111, [
            ["service_3", "is very cool", 1],
            ["service_4", "is very cool", 2]
        ]
    ],
    ["0.0.0.2", 2222, [
            ["service_3", "is very cool", 3],
            ["service_4", "is very cool", 4]
        ]
    ]
]}

我需要获取“数据”的第一个索引并创建一个新的 JSON 对象，看起来像这样......

"data": [{
"address": "0.0.0.1", 
"id": 1111, 
"services_with_info":[
    {
        "service_name": "service_1", 
        "service_message": "is very cool", 
        "service_id": 1
    }, 
    {...}
]}, 
{...}]

然后从中构建一个[]Host，数据结构为 5k“hosts”长。我能够将其映射到结构，但需要先将其转换为这种格式。我了解如何解码 JSON，但前提是我可以将有效负载转换为上述内容。

Answer 1

您可以为此使用 json.Unmarshal 并根据您的条件解析数据。我只是为 "data" 做这件事，你也可以为 "services_with_info"

做同样的事情

b := []byte(`{
            "source": "some random source",
            "table": "hosts_table",
            "data": [
                ["address", "id", "services_with_info"],
                ["0.0.0.1", 1111, [
                        ["service_3", "is very cool", 1],
                        ["service_4", "is very cool", 2]
                    ]
                ],
                ["0.0.0.2", 2222, [
                        ["service_3", "is very cool", 3],
                        ["service_4", "is very cool", 4]
                    ]
                ]
            ]}`)
var f interface{}
err := json.Unmarshal(b, &f)
if err != nil {
    fmt.Println(err)
    return
}
m := f.(map[string]interface{})
result := make(map[string]interface{})
results := make(map[string][]map[string]interface{})
for k, v := range m {
    if k == "data" {
        s := v.([]interface{})
        header := make([]interface{}, 3)
        for i, u := range s {
            if i == 0 {
                header = u.([]interface{})
            } else {
                row := u.([]interface{})
                for j, k := range header {
                    result[k.(string)] = row[j]
                }
                results["data"] = append(results["data"], result)
            }
        }
    }
}
fmt.Println(results)

这里的"results"是需要的"data"。