Given
鉴于
[{
"objects": [{
"key": "value"
},{
"key": "value"
}]
}, {
"objects": [{
"key": "value"
}, {
"key": "value"
}]
}]
How do I generate
我怎么生成
[{
"objects": [{
"id": 0,
"key": "value"
},{
"id": 1,
"key": "value"
}]
}, {
"objects": [{
"id": 2,
"key": "value"
}, {
"id": 3,
"key": "value"
}]
}]
Using jq?
使用金桥?
I tried to use this one, but id
s are all 0
:
我试着用这个,但是id都是0:
jq '[(-1) as $i | .[] | {objects: [.objects[] | {id: ($i + 1 as $i | $i), key}]}]'
2 个解决方案
#1
1
The key to a simple solution here is to break the problem down into easy pieces. This can be accomplished by defining a helper function, addId/1
. Once that is done, the rest is straightforward:
一个简单解决方案的关键是将问题分解成简单的部分。这可以通过定义辅助函数addId/1来实现。一旦这样做了,剩下的就很简单了:
# starting at start, add {id: ID} to each object in the input array
def addId(start):
reduce .[] as $o
([];
length as $l
| .[length] = ($o | (.id = start + $l)));
reduce .[] as $o
( {start: -1, answer: []};
(.start + 1) as $next
| .answer += [$o | (.objects |= addId($next))]
| .start += ($o.objects | length) )
| .answer
#2
0
Inspired by @peak answer, I came up with this solution. Not much difference, just shorter way to generate IDs and opt for foreach
instead of reduce
since there is intermediate result involved.
受@peak回答的启发,我想出了这个解决方案。没有太大的差别,只是更短的生成id的方法,并选择foreach而不是reduce,因为其中包含了中间结果。
def addIdsStartWith($start):
[to_entries | map((.value.id = .key + $start) | .value)];
[foreach .[] as $set (
{start: 0};
.set = $set |
.start as $start | .set.objects |= addIdsStartWith($start) |
.start += ($set.objects | length);
.set
)]
#1
1
The key to a simple solution here is to break the problem down into easy pieces. This can be accomplished by defining a helper function, addId/1
. Once that is done, the rest is straightforward:
一个简单解决方案的关键是将问题分解成简单的部分。这可以通过定义辅助函数addId/1来实现。一旦这样做了,剩下的就很简单了:
# starting at start, add {id: ID} to each object in the input array
def addId(start):
reduce .[] as $o
([];
length as $l
| .[length] = ($o | (.id = start + $l)));
reduce .[] as $o
( {start: -1, answer: []};
(.start + 1) as $next
| .answer += [$o | (.objects |= addId($next))]
| .start += ($o.objects | length) )
| .answer
#2
0
Inspired by @peak answer, I came up with this solution. Not much difference, just shorter way to generate IDs and opt for foreach
instead of reduce
since there is intermediate result involved.
受@peak回答的启发,我想出了这个解决方案。没有太大的差别,只是更短的生成id的方法,并选择foreach而不是reduce,因为其中包含了中间结果。
def addIdsStartWith($start):
[to_entries | map((.value.id = .key + $start) | .value)];
[foreach .[] as $set (
{start: 0};
.set = $set |
.start as $start | .set.objects |= addIdsStartWith($start) |
.start += ($set.objects | length);
.set
)]