滤波过滤一个暗号查询结果

Question

你好，这里是我当前的查询，我想“再过滤器”：

START movie = node(*) 
MATCH user-[:LIKE]->category-[:SIMILAR*0..3]-()<-[:TAGGED]->movie 
WHERE user.name = "current_user" 

WITH DISTINCT movie, user, category 

RETURN user.name, category.name, ID(movie), movie.name 
ORDER BY movie.name; 

http://console.neo4j.org/r/u19iim

这是它看起来像当前查询后：

+--------------+----------------+-----------+-------------------------+ 
| user.name | category.name | ID(movie) | movie.name    | 
+--------------+----------------+-----------+-------------------------+ 
| current_user | c    | 14  | movie_c_and_d_and_e  | 
| current_user | d    | 14  | movie_c_and_d_and_e  | 
| current_user | e    | 14  | movie_c_and_d_and_e  | 
| current_user | a    | 9   | movie_of_a_and_b_and_b1 | 
| current_user | b    | 9   | movie_of_a_and_b_and_b1 | 
| current_user | b    | 10  | movie_of_b2_first  | 
| current_user | b    | 11  | movie_of_b2_second  | 
| current_user | c    | 12  | movie_of_c    | 
| current_user | d    | 13  | movie_of_d_and_e  | 
| current_user | e    | 13  | movie_of_d_and_e  | 
+--------------+----------------+-----------+-------------------------+ 

我想GROUP BY COUNT(sugg) AS category_count提取此：

+--------------+----------------+-----------+-------------------------+ 
| user.name | category_count | ID(movie) | movie.name    | 
+--------------+----------------+-----------+-------------------------+ 
| current_user | 3    | 14  | movie_c_and_d_and_e  | 
| current_user | 2    | 9   | movie_of_a_and_b_and_b1 | 
| current_user | 2    | 13  | movie_of_d_and_e  | 
| current_user | 1    | 10  | movie_of_b2_first  | 
| current_user | 1    | 11  | movie_of_b2_second  | 
| current_user | 1    | 12  | movie_of_c    | 
+--------------+----------------+-----------+-------------------------+ 

我该如何做到这一点？

类似的问题： - how to have two aggregation in cypher query in neo4j?

更新
这里的工作结果（有示范：http://tinyurl.com/cywlycc）：

START movie = node(*) 
MATCH user-[:LIKE]->category-[:SIMILAR*0..3]-()<-[:TAGGED]->movie 
WHERE user.name = "current_user" 
WITH DISTINCT movie, category WITH COUNT(movie) AS category_count, movie, collect(category.name) as categorized 
RETURN category_count, ID(movie), movie.name, categorized 
ORDER BY category_count DESC; 

Answer 1

START movie = node(*) 
MATCH user-[:LIKE]->category-[:SIMILAR*0..3]-()<-[:TAGGED]->movie 
WHERE user.name = "current_user" 
WITH DISTINCT movie, user, category 
RETURN user.name, count(category.name) as category_count, ID(movie), movie.name 
ORDER BY category_count desc, movie.name asc 

http://console.neo4j.org/r/69rfkn