How to dynamically create a struct with one less property? - go

Question

Is there a way to copy a generic struct (i.e. a struct whose property names are unknown) and skip a single, known property?
Here is what I know:
The parameter to my function--I will call the parameter myData-- is of type interface{}.
myData is a struct.
myData has a known property path.
myData has anywhere from 0 to 6 or so other properties, none of which are known a priori.
Once I remove that path property, then the “leftover” is one of say 30 possible struct types.
So I want to strip path out of myData (or more accurately make a copy that omits path) so that various bits of generated code that try to coerce the struct to one of its possible types will be able to succeed.
I have found examples of copying a struct by reflection, but they typically create an empty struct of the same underlying type, then fill it in. So is it even possible to delete a property as I have outlined...?

Answer 1

You can use reflect.StructOf to dynamically create structs from a list of fields.
package main
import (
"fmt"
"reflect"
)
type A struct {
Foo string
Bar int
Baz bool // to be skipped
}
type B struct {
Foo string
Bar int
}
func main() {
av := reflect.ValueOf(A{"hello", 123, true})
fields := make([]reflect.StructField, 0)
values := make([]reflect.Value, 0)
for i := 0; i < av.NumField(); i++ {
f := av.Type().Field(i)
if f.Name != "Baz" {
fields = append(fields, f)
values = append(values, av.Field(i))
}
}
typ := reflect.StructOf(fields)
val := reflect.New(typ).Elem()
for i := 0; i < len(fields); i++ {
val.Field(i).Set(values[i])
}
btyp := reflect.TypeOf(B{})
bval := val.Convert(btyp)
b, ok := bval.Interface().(B)
fmt.Println(b, ok)
}