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

N: Number

A positive (or zero) Number specifying how many records to ignore

Return Value

Collection

Remarks

Return a collection when the first N entries in the object store are ignored. If it is requested to skip the LAST N entires rather than the first, 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 include the last 15th to 10th friend.

db.friends.orderBy('lastName').reverse().offset(10).limit(5);

Limitations

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

Performance Notes

If executed on simple queries, the native IDBCursor.advance() method will be used (fast execution). If advanced queries are used, the implementation have 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)

See Also

Collection.offset()

Collection.limit()