[TINKERPOP-1482] has(x).has(y) chains should be has(x.and(y))

• Type: Improvement
• Status: Closed
• Priority: Major
• Resolution: Fixed
• Affects Version/s: 3.2.2
• Fix Version/s: 3.2.4
• Component/s: process
• Assignee: Marko A. Rodriguez
• Reporter: Marko A. Rodriguez
• Created: 29/Sep/16 16:43
• Updated: 16/Nov/16 13:37
• Resolved: 16/Nov/16 13:37

Description

Lots of work has been done to inline filters for 3.2.3. A final application should be to inline has.has into has(x.and). If this is done, providers may have significant breaking changes as they will need to search HasContainers beyond P.eq() to walking an and/or-tree. However, the benefit of this grouping is two fold:

1. Its a clean, consistent composition.
2. It reduces the number of HasContainers and HasSteps in areas where provider "fold-ins" are not being used.