New answers tagged

1

If I understand your question correctly, you want to filter a list based on the next item in the list instead of the item itself, i.e. in pseudocode: if(items[i] == stop && items[i + 1] == start) return items[i]; This question focuses on how to observe the previous item, but it's essentially doing the same thing and can be tailored to your ...


1

Excellent analysis. Writing an object graph is notoriously difficult because it can involve a lot of foreign key lookups and these are hard to implement without incurring a lot of waits. To make a dent in the performance, you need to widen your approach. Few ideas to consider: Blasting all the new data into a temporary table and reconcile with the main ...


Top 50 recent answers are included