jQuery中的SQL Join模拟 [英] Analogue for sql join in jq

问题描述

我有以下json:

[
  {"id": "1", "type": "folder", "title": "folder-1"},
  {"id": "2", "type": "folder", "title": "folder-2"},
  {"id": "3", "type": "item", "title": "item-1", "folder": "1"},
  {"id": "4", "type": "item", "title": "item-2", "folder": "2"},
  {"id": "5", "type": "item", "title": "item-3"}
]

基本上，我需要使用jq生成此输出，这类似于sql join的结果:

Basically, I need to produce this output using jq, which is similar to the result of sql join:

[
  {"type": "item", "title": "item-1", "folder": "folder-1"},
  {"type": "item", "title": "item-2", "folder": "folder-2"},
  {"type": "item", "title": "item-3"}
]

有什么想法吗?

推荐答案

尝试使用此过滤器:

map(
    (select(.type=="item") | { key: .folder, value: { type, title } }),
    (select(.type=="folder") | { key: .id, value: { folder: .title } })
)
| group_by(.key)
| map(
    (map(select(.key != null) | .value) | add)
    // map(.value)[]
)

您必须将其分解为几个步骤.

You'll have to break this out into steps.

1. 获取项目和文件夹，并为每个项目和文件夹获取您感兴趣的值，并为其分配要关联的密钥.

1. Get the items and folders and for each, take the values you're interested in and assign it a key to associate with.

map(
    (select(.type=="item") | { key: .folder, value: { type, title } }),
    (select(.type=="folder") | { key: .id, value: { folder: .title } })
)

按键将所有分组

Group all by the key

| group_by(.key)

然后将包含键(文件夹)的值与其他值组合

Then combine the values that have keys (folders) and the value otherwise

| map(
    (map(select(.key != null) | .value) | add)
    // map(.value)[]
)

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