相互计数的搜索列表(第二次尝试) [英] Search list with mutual count (2nd try)

问题描述

我创建了新的数据集来解释我想要的结果.这是和this 和 这个等等.

解决方案

正如我在评论中所说，尝试同时查询连接和断开连接的节点似乎不是一个好主意.

如果您只想要连接的节点，请尝试以下查询:

START me=node:node_auto_index(UserName = 'manish')匹配我-[:friends]-mf-[:friends]-other, me-[r?]-otherWHERE other.UserName!=~'(?i)dh.*'RETURN DISTINCT ID(other), r.ApprovalStatus AS 状态, count(mf) AS 相互, ID(me)ORDER BY 相互 DESC , 状态 ASC

请注意，我不得不在匹配子句中添加另一个模式，因为您的批准状态是在(我)和(其他)之间，而不是在(我)和(共同朋友)之间，这就是您所做的！

这将返回您预期答案的前 3 行并忽略 dhansukh.

I have created fresh dataset to explain my desired result. and here is the link

Or you can trigger this command using cypher.

create 
(_6  {UserName:"dhansukh", UserProfileID:'1000', EMailID:'[e-mail]'}),
(_5  {UserName:"dhruv", UserProfileID:'516', EMailID:'[e-mail]'}),
(_4  {UserName:"dharmistha", UserProfileID:'5262', EMailID:'[e-mail]'}),
(_3  {UserName:"dinesh", UserProfileID:'995', EMailID:'[e-mail]'}),
(_2  {UserName:"dharmesh", UserProfileID:'502', EMailID:'[e-mail]'}),
(_1  {UserName:"manish", UserProfileID:'1', EMailID:'[e-mail]'}),
_1-[:friends {ApprovalStatus: 1} ]->_2,
_1-[:friends {ApprovalStatus: 1} ]->_3,
_1-[:friends {ApprovalStatus: 2} ]->_5,
_2-[:friends {ApprovalStatus: 1} ]->_3,
_2-[:friends {ApprovalStatus: 1} ]->_5,
_3-[:friends {ApprovalStatus: 1} ]->_4

Now I am trying following query, but it is not given me my expected result.

START me=node:node_auto_index(UserProfileID = '1'), other=node(*) 
MATCH pMutualFriends=me-[r?:friends]-mf-[r1:friends]-other 
WHERE other.UserName =~ '(?i)dh.*' AND other.UserProfileID <> 1 
RETURN other.UserProfileID, other.UserName, r.ApprovalStatus, COUNT(pMutualFriends) AS mutualCount

In the above result set, I have get duplicate records, (due to ApprovalStatus), If I remove ? from relationship, it just shows linked node only, but I want all nodes started with 'dh'. node 6 also missing, don't know why? mutual count also showing wrong in some case. Only that node should be consider in mutual count which is has ApprovalStatus = 1. like login node (eg. node 1) and search nodes both have relationship's property ApprovalStatus = 1.

EDIT : My expected result set :

UserProfileID  UserName     ApprovalStatus  MutualCount 
-------------  --------     --------------  ----------- 
502            dharmesh     1               2           (node 3 & 5 )
516            dhruv        2               1           (node 2)
5262           dharmistha   null            1           (node 3) 
1000           dhansukh     null            0               

EDIT : I am updating image for clear understanding.

I am suffering from this issue last 20-25 days, and not getting proper solution, I don't know where is the problem. I have already post this problem many times on stackoverflow. here is the link, and this and this and many more.

解决方案

As I said in my comment, trying to query for connected and disconnected nodes at the same time doesn't seem to be a good idea.

If you want only connected nodes, try the following query :

START me=node:node_auto_index(UserName = 'manish') 
MATCH me-[:friends]-mf-[:friends]-other, me-[r?]-other 
WHERE other.UserName! =~ '(?i)dh.*' 
RETURN DISTINCT ID(other), r.ApprovalStatus AS status, count(mf) AS mutual, ID(me) 
ORDER BY mutual DESC , status ASC

Please note that I had to add another pattern in the match clause, because your approval status is between (me) and (other), and not between (me) and (mutual friend), which is what you were doing!

This will return the first 3 lines of your expected answer and ignores dhansukh.

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