@Brenigan , "&" on the RHS returns the value found at the specified path position. When used inside "[]" the meaning is the same and the returned value is used as an index of that array. For example, take this expression: "@1": "outer[&4].inner[&2].t", The different matches for "&" expressions are shown below: [{
"operation": "shift",
"spec": {
"events": {
"*": { <-- &4 matches the position of the item in the events array (e.g. 0, 1, 2)
"transports": { <-- &3 matches "transports"
"*": { <-- &2 matches the position of the item in the transports array (e.g. 0, 1)
"*": { <-- &1 matches the same as &0
"@1": "outer[&4].inner[&2].t", <-- &0 matches the value of the item in the transports array (e.g. PUSH, SMS)
"@(3,eventTypeCode)": "outer[&4].inner[&2].etc"
}
}
}
}
},
"*": "&"
}
}] So, the expression "outer[&4].inner[&2].t" will create an "outer" array that has as many elements as the "events" array. Each element of the "outer" array will have an "inner" attribute whose value will be another array. This array will have as many elements as the respective "transports" array, each one with the "t" and "etc" attributes. Hope this helps. Cheers, André
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