Table.offset()

Returns a Collection (ordered by primary key), where the first N items in the object store are ignored.

Syntax

db.[table].offset(N)

Parameters

Return Value

Collection

Remarks

Returns a collection where the first N entries in the object store are ignored. If it is requested to skip the LAST N entries rather than the FIRST N entries, then this method can be used in combination with the Collection.reverse() method.

This method is equivalent to:

db.[table].toCollection().offset(N)

or:

db.[table].orderBy(':id').offset(N)

Sample

This sample will sort friends by lastName and output to the console the 11th to the 15th friend from the end.

db.friends.orderBy('lastName')
    .reverse()
    .offset(10)
    .limit(5)
    .toArray()
    .then(function (results) {
        console.log (JSON.stringify(results));
    });

Limitations

In combination with the or() method, the offset() method makes no sense since the sort order of the result will be undefined (this is because or() works on multiple different indexes in parallel). Instead, use sortBy() and then slice the resulting array from the requested offset.

Performance Notes

If executed on simple queries, then the native IDBCursor.advance() method will be used (fast execution). If advanced queries are used, then the implementation has to execute a query to iterate all items and ignore N items using a JS filter.

Examples where offset() will be fast

db.[table].offset(N)
db.[table].where(index).equals(value).offset(N)
db.[table].where(index).above(value).offset(N)
db.[table].where(index).below(value).offset(N)
db.[table].where(index).between(value).offset(N)
db.[table].where(index).startsWith(value).offset(N)

Examples where offset() will have to iterate the collection:

db.[table].where(index).equalsIgnoreCase(value).offset(N)
db.[table].where(index).startsWithIgnoreCase(value).offset(N)
db.[table].where(index).anyOf(valueArray).offset(N)
db.[table].where(index).above(value).and(filterFunction).offset(N)