Question

如何使用jq的reduce函数来处理多种操作？

给出一个像这样的模型：

{
  "usage" : {
    "os:linux" : {
      "2018_11_18" : {
        "amount" : 601046342,
        "credits" : 99783
      },
      "2018_10_18" : {
        "amount" : 839785106,
        "credits" : 173121
      }
    }
  }
}

和JQ文件：

reduce .usage."os:linux"[] as $item (
{"credits":0,"minutes":0}; 
."credits" += $item.credits
)

我得到：

{
  "credits": 272904,
  "minutes": 0
}

很酷，但是我想同时介绍两个领域，例如：

reduce .usage."os:linux"[] as $item (
{"credits":0,"minutes":0}; 
."credits" += $item.credits,
."minutes" += $item.amount /1000/60
)

分号会导致错误，逗号通过，但仅处理最后一个任务，即

{
  "credits": 0,
  "minutes": 24013.85746666667
}

Answer 1

总有明确的方法：

reduce .usage."os:linux"[] as $item (
  {"credits": 0, "minutes": 0 }; 
  {
    "credits": (.credits + $item.credits),
    "minutes": (.minutes + $item.amount / 60000) 
  }
)

Answer 2

或更简洁：

reduce .usage."os:linux"[] as $item (
  {}; 
  .credits += $item.credits
  | .minutes += ($item.amount / 60000) )

或者...

或将简化形式抽象为通用工具功能sigma/1：

def sigma(s): reduce s as $s (null; . + $s);

.usage."os:linux"
| {credits: sigma(.[].credits),
   minutes: (sigma(.[].amount) / 60000) }

用jq reduce汇总多个值

2 个答案:

或者...