我在json下面:
{
"deveui": "1000",
"vars": [
{ "name": "CfgNo",
"value":"255" },
{
"name":"Status","value":{
"lampS": "unknown(0xff)",
"lampL": "255"
}
},
{ "name" : "Volt", "value": "255" },
{
"name" : "gps", "value": {
"lat": "12.93",
"lon": "77.69"
}
},
{"name" : "status", "value":"up"},
{"name" : "last_status_change","value": 1503 }
]
}
预期输出为:
{[{ "deveui": "1000",
"CfgNo": "255",
"lampS": "unknown(0xff)",
"lampL": "255",
"Volt": "255",
"lat": "12.93",
"lon": "77.69",
"status": "up",
"last_status_change": 1503
}]}
是否可以转换为预期的输出格式。如果是这样,请帮我设计规格。
答案 0 :(得分:0)
它看起来应该很简单,但如果是地图或标量,Shift不能以不同的方式匹配。在这种情况下,"值"在你的输入中有时是一个标量,有时候是一个嵌套的地图。
如果您知道"名称"这将是多重价值,然后Jolt可以提供帮助。
规格
[
{
// Shift can't vary its behavior between
// the "value" key in the input being a
// scalar or a map.
// Thus the only way to do the desired flattening
// is if we know the names that will
// be multi-valued.
"operation": "shift",
"spec": {
// pass deveui thru
"deveui": "deveui",
"vars": {
"*": { // array index of vars[]
"name": {
// match value of name Status or gps
"Status|gps": {
// We know these entries have "value" that is map.
// For these, go back up the tree, 3 levels (0,1,2)
// and come back down to the "value" entry.
"@(2,value)": {
// for each key in the value, pass it thru
// to the output.
"*": "&"
}
},
"*": {
// for all other values of "name" that are not
// Status or gps, just go back up the tree
// and grab the scalear "value" and write it
// to the output at the matched value of name.
"@(2,value)": "&1"
}
}
}
}
}
}
]