for record_order in result_order: node_order = record_order['o'] customerID_value = node_order['customerID'] query_customer_id = "MATCH (u:Customer) WHERE u.customerID = $customer_id RETURN u" result_customer_id = graph.run(query_customer_id, parameters={"customer_id": f"{customerID_value}"}) for record_customer in result_customer_id: node_customer = record_customer['u'] relationship = Relationship(node_customer, customer_order_type, node_order) graph.create(relationship)
时间: 2024-03-15 14:42:03 浏览: 88
这段代码使用了查询结果 "result_order" 中的所有节点,并针对每个节点执行一些操作。具体来说,它查找了每个 "Order" 节点的 "customerID" 属性,并将其用作参数来查询匹配的 "Customer" 节点。然后,它使用 "Relationship" 类创建了一个在 "Customer" 节点和 "Order" 节点之间的关系,并使用 "graph.create" 方法在数据库中创建了这个关系。
这个循环的作用是将 "Customer" 节点与 "Order" 节点连接起来,以表示它们之间的关系。这种关系可以用来查询特定客户的订单历史记录,或者查询特定订单的客户信息。
相关问题
SELECT IFNULL( count( nt.id ), 0 ) total, IFNULL( count( ta.id ), 0 ) endtotal, cs.taskname,nt.id,nt.`name` FROM nk_xjsg_taskresult nt JOIN nk_xjsg_constask cs ON cs.id = nt.taskid AND cs.customerid = nt.customerid AND cs.pid = '40289e8c87fff7f80188042c15d50010' JOIN nk_xjsg_consplan nc ON nc.id = cs.pid AND nc.customerid = cs.customerid LEFT JOIN ( SELECT id, resultid, statusnum, statusname, endnum, ROW_NUMBER() over ( PARTITION BY resultid ORDER BY createdate DESC ) number FROM nk_xjsg_taskresult_date WHERE customerid = '40289e8c8803a391018803a6ef8f0000' ) ta ON ta.resultid = nt.id AND ta.number = 1 -- -- AND ta.endnum = nt.resultnum WHERE nt.customerid = '40289e8c8803a391018803a6ef8f0000' GROUP BY cs.taskname ,nt.id ,nt.`name`;我nk_xjsg_taskresult_date 有两天数据 为什么只查询出来一条 什么原因 怎么解决
根据你提供的 SQL 查询语句,可以看出使用了子查询来查询 `nk_xjsg_taskresult_date` 表中的数据,并使用了 `ROW_NUMBER()` 函数来获取每个 `resultid` 最新的一条数据。而在子查询中,使用了 `PARTITION BY resultid` 来对数据进行分组,所以每个 `resultid` 只会返回一条数据。
如果你想要查询 `nk_xjsg_taskresult_date` 表中所有的数据,可以将子查询中的 `PARTITION BY resultid` 移除,这样就不会对数据进行分组,从而返回所有的数据。修改后的 SQL 查询语句如下:
```
SELECT
IFNULL(count(nt.id), 0) total,
IFNULL(count(ta.id), 0) endtotal,
cs.taskname,
nt.id,
nt.`name`
FROM
nk_xjsg_taskresult nt
JOIN nk_xjsg_constask cs ON cs.id = nt.taskid
AND cs.customerid = nt.customerid
AND cs.pid = '40289e8c87fff7f80188042c15d50010'
JOIN nk_xjsg_consplan nc ON nc.id = cs.pid
AND nc.customerid = cs.customerid
LEFT JOIN (
SELECT
id,
resultid,
statusnum,
statusname,
endnum,
ROW_NUMBER() over (ORDER BY createdate DESC) number
FROM
nk_xjsg_taskresult_date
WHERE
customerid = '40289e8c8803a391018803a6ef8f0000'
) ta ON ta.resultid = nt.id
AND ta.number = 1
WHERE
nt.customerid = '40289e8c8803a391018803a6ef8f0000'
GROUP BY
cs.taskname,
nt.id,
nt.`name`;
```
这样就可以查询出 `nk_xjsg_taskresult_date` 表中的所有数据了。
SELECT IFNULL(COUNT(DISTINCT nt.id), 0) total, IFNULL(COUNT( DISTINCT ta.id), 0) endtotal, cs.taskname, nt.id, nt.name FROM nk_xjsg_taskresult nt JOIN nk_xjsg_constask cs ON cs.id = nt.taskid AND cs.customerid = nt.customerid AND cs.pid = '40289e8c87fff7f80188042c15d50010' JOIN nk_xjsg_consplan nc ON nc.id = cs.pid AND nc.customerid = cs.customerid LEFT JOIN ( SELECT id, resultid, statusnum, statusname, endnum FROM nk_xjsg_taskresult_date WHERE customerid = '40289e8c8803a391018803a6ef8f0000' ) ta ON ta.resultid = nt.id and ta.statusnum = '40' -- AND ta.number = 1 -- AND ta.endnum = nt.resultnum WHERE nt.customerid = '40289e8c8803a391018803a6ef8f0000' GROUP BY cs.taskname, nt.id, nt.name; 怎么让endtotal的值从5变成1
根据你提供的SQL查询语句,endtotal的值是通过COUNT函数统计的DISTINCT ta.id的数量,因此结果为5。如果你想让endtotal的值从5变成1,需要在COUNT函数中加入条件,只统计符合要求的数据。根据你的需求,可以在COUNT函数的参数中加入条件“ta.endnum = nt.resultnum”,确保只统计符合要求的数据。修改后的SQL查询语句如下:
```
SELECT IFNULL(COUNT(DISTINCT nt.id), 0) total, IFNULL(COUNT(DISTINCT CASE WHEN ta.endnum = nt.resultnum THEN ta.id ELSE NULL END), 0) endtotal, cs.taskname, nt.id, nt.name
FROM nk_xjsg_taskresult nt
JOIN nk_xjsg_constask cs ON cs.id = nt.taskid AND cs.customerid = nt.customerid AND cs.pid = '40289e8c87fff7f80188042c15d50010'
JOIN nk_xjsg_consplan nc ON nc.id = cs.pid AND nc.customerid = cs.customerid
LEFT JOIN (
SELECT id, resultid, statusnum, statusname, endnum
FROM nk_xjsg_taskresult_date
WHERE customerid = '40289e8c8803a391018803a6ef8f0000'
) ta ON ta.resultid = nt.id and ta.statusnum = '40' -- AND ta.number = 1
WHERE nt.customerid = '40289e8c8803a391018803a6ef8f0000'
GROUP BY cs.taskname, nt.id, nt.name;
```
修改后的查询语句中,将第二个COUNT函数的参数改为“DISTINCT CASE WHEN ta.endnum = nt.resultnum THEN ta.id ELSE NULL END”,这样只有符合条件的数据才会被统计。
阅读全文