使用内联输出计数带有jq 1.3的JSON数组的成员

Question

我有一个JSON文件：

[
  {
    "instance_AZs": [
        "us-east-1b",
        "us-east-1b",
        "us-east-1b",
        "us-east-1c",
        "us-east-1c",
        "us-east-1a",
        "us-east-1a",
        "us-east-1c",
        "us-east-1b",
        "us-east-1a",
        "us-east-1a",
        "us-east-1c",
        "us-east-1a",
        "us-east-1b",
        "us-east-1b",
        "us-east-1c",
        "us-east-1c",
        "us-east-1b"
    ],
    "ASGname": "myapp"
  }
]

期望的输出是：

myapp.us-east-1a.5
myapp.us-east-1b.7
myapp.us-east-1c.6

至于我可以使用jq 1.3语法，我编写了这段代码：

cat file.json | jq -r '.[] | .ASGname, (.instance_AZs | group_by(.) | map(. | length , (. | unique ) [] ))'

产生输出：

myapp
[
  5,
  "us-east-1a",
  7,
  "us-east-1b",
  6,
  "us-east-1c"
]

能否请您建议如何产生所需的输出？

Answer 1

这是一种方法：

jq -r '.[]|.ASGname as $g|.instance_AZs|group_by(.)[]|"\($g).\(.[0]).\(length)"' file.json

Answer 2

.[]
| .ASGname as $p
| reduce .instance_AZs[] as $k ({}; .[$k]+=1)
| keys_unsorted[] as $k
| "\($p).\($k).\(.[$k])"

将instance_AZs减少到一个对象，其中唯一的数组元素是数组中重复计数的关键，用它来产生所需的输出。

或者，如果你坚持按照你的方式做到这一点，将组分组为插值字符串而不是map，即参考@ hek2mgl的答案。