Kotlin: How apply all list elements to the object?

user_12125755 0
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I have such kotlin code. I didn't write it very elegantly, but it works correctly. How can I write this more beautifully using kotlin tools?

I need to apply all elements from the filters list to the query object.

fun create(filters: List<FilterFactory.Filter>, searchQuery: SearchQuery): Query {
    var query = Query()
    filters.forEach { filter -> query = filter.apply(query) }

    return query
}
kotlin refactoring
2 Answers

You can refactor the implementation of the function by using fold here:

fun create(filters: List<FilterFactory.Filter>): Query =
    filters.fold(Query()) { q, f -> f.apply(q) }

Note that your initial code didn't make use of the searchQuery parameter, so I removed it.

As for the API, you might want to use an extension function instead, with the toX() convention for the name. For instance:

fun List<FilterFactory.Filter>.toQuery(): Query =
    fold(Query()) { q, f -> f.apply(q) }

toQuery() on a list of filters might be confusing to some, though. Maybe a better approach would be an extension on Query itself:

fun Query.withFilters(filters: List<FilterFactory.Filter>): Query =
    filters.fold(this) { q, f -> f.apply(q) }

Then to use it you'll need to provide the initial Query instance on the call site:

val filteredQuery = Query().withFilters(filters)
user_15408180

Another thing you could do is a simple apply call to configure the object you've created:

fun create(filters: List<FilterFactory.Filter>, searchQuery: SearchQuery): Query {
    return Query().apply { 
        filters.forEach { filter -> filter.apply(this) }
    }
}

which lends itself nicely to the single-expression style

fun create(filters: List<FilterFactory.Filter>, searchQuery: SearchQuery) =
    Query().apply { 
        filters.forEach { filter -> filter.apply(this) }
    }

(I'm assuming you're using searchQuery in there somewhere, e.g. the Query constructor)

user_135982220
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