我正在尝试基于嵌套的Json创建单个json输出:
{
"Id" : "1",
"items" : [
{"item_name" : "shirt","value" : 10},
{"item_name" : "dress","value" : 20},
{"item_name" : "ice cream","value" : 30}
]
}
我的预期输出是:
[
{
"id": "1",
"item_name": "shirt",
"value": 10,
"index_position": 0
},
{
"id": "1",
"item_name": "dress",
"value": 20,
"index_position": 1
},
{
"id": "1",
"item_name": "ice cream",
"value": 30,
"index_position": 2
}
]
我能够提取的唯一输出是此代码段:
https://jqplay.org/s/G6mYAI47LE
迭代内部数组并同时获取外部对象数据的最佳方法是什么?
答案 0 :(得分:1)
Q.E.D。如下:
.Id as $Id
| .items
| [ range(0; length) as $index_position
| {$Id} + .[$index_position] + {$index_position} ]
此处简洁(或至少具有DRY风格)的关键在于jq表达式{$x}
扩展为{"x": $x}
。