My problem on getting one row shown with two null values.
FROM tuotemerkki;
tmtunnus | tmnimi | maa
----------+----------+-------------
1 | McCee | Yhdysvallat
2 | KooTek | Italia
3 | Giardino | Italia
(3 rows)
FROM tuote;
ttunnus | tnimi | kuvaus | suositushinta | tmtunnus
---------+-----------------------+--------------------+---------------+----------
111 | Trimmeri TRCee | tehokas 4-tahtinen | 179.00 | 1
112 | Trimmerisiima Cee | laadukas siima | 6.99 | 1
113 | Moottorisaha MSCee RR | robusti ja raskas | 559.00 | 1
114 | Trimmerisiima Y | yleissiima | 3.99 | 2
115 | Lapio L | kevyt yleislapio | 23.95 | 2
(5 rows)
I need to get this selected with the NULL VALUE on giardino from tmnimi.
tmnimi | tnimi | kuvaus
----------+-----------------------+--------------------
McCee | Trimmeri TRCee | tehokas 4-tahtinen
McCee | Trimmerisiima Cee | laadukas siima
McCee | Moottorisaha MSCee RR | robusti ja raskas
KooTek | Trimmerisiima Y | yleissiima
KooTek | Lapio L | kevyt yleislapio
Giardino | |
(6 rows)
I get only this selected
SELECT tmnimi, tnimi, kuvaus
FROM tuote
CROSS JOIN tuotemerkki
WHERE tuote.tmtunnus = tuotemerkki.tmtunnus;
tmnimi | tnimi | kuvaus
--------+-----------------------+--------------------
McCee | Trimmeri TRCee | tehokas 4-tahtinen
McCee | Trimmerisiima Cee | laadukas siima
McCee | Moottorisaha MSCee RR | robusti ja raskas
KooTek | Trimmerisiima Y | yleissiima
KooTek | Lapio L | kevyt yleislapio
(5 rows)
Use a left join between the two tables:
SELECT
t1.tmnimi,
t2.tnimi,
t2.kuvaus
FROM tuotemerkki t1
LEFT JOIN tuote t2
ON t1.tmtunnus = t2.tmtunnus
When you left join from tuotemerkki to tuote, then every record in the former table is guaranteed to appear in the result set. Since the Giardino record does not match to anything in the tuote table, all the columns from that table would have a NULL value for the Giardino record.
Related
I have a Cypher query that shows the following output:
+----------------
| usid | count |
+----------------
| "000" | 1 |
| "000" | 0 |
| "000" | 0 |
| "001" | 1 |
| "001" | 1 |
| "001" | 0 |
| "002" | 2 |
| "002" | 2 |
| "002" | 0 |
| "003" | 4 |
| "003" | 2 |
| "003" | 2 |
| "004" | 4 |
| "004" | 4 |
| "004" | 4 |
+----------------
How can I get the below result with the condition SUM(count) <= 9.
+----------------
| usid | count |
+----------------
| "000" | 1 |
| "001" | 2 |
| "002" | 4 |
| "003" | 8 |
+----------------
Note: I have used the below query to get the 1st table data.
MATCH (us:USER)
WITH us
WHERE us.count <= 4
RETURN us.id as usid, us.count as count;
I don't know how you get your original data, so I will just use a WITH clause and assume the data is there:
// original data
WITH usid, count
// aggregate and filter
WITH usid, sum(count) as new_count
WHERE new_count <= 9
RETURN usid, new_count
Based on the updated question, the new query would look like:
MATCH (us:USER)
WHERE us.count <= 4
WITH us.id as usid, sum(us.count) as count
WHERE new_count <= 9
RETURN usid, count
˙˙˙
How to combine distinct and max within join table below?
Table_details_usage
UID | VE_NO | START_MILEAGE | END_MILEAGE
------------------------------------------------
1 | ASD | 410000 | 410500
2 | JWQ | 212000 | 212350
3 | WYS | 521000 | 521150
4 | JWQ | 212360 | 212400
5 | ASD | 410520 | 410600
Table_service_schedule
SID | VE_NO | SV_ONMILEAGE | SV_NEXTMILEAGE
------------------------------------------------
1 | ASD | 400010 | 410010
2 | JWQ | 212120 | 222120
3 | WYS | 511950 | 521950
4 | JWQ | 212300 | 222300
5 | ASD | 410510 | 420510
How to get display as below (only max value)?
Get Max value from Table_service_schedule (SV_NEXTMILEAGE) and Get Max value from Table_details_usage (END_MILEAGE)
SID | VE_NO | SV_NEXTMILEAGE | END_MILEAGE
--------------------------------------------
5 | ASD | 420510 | 410600
4 | JWQ | 222300 | 212400
3 | WYS | 521950 | 521150
Something in the lines of:
SELECT
SID,
VE_NO,
SV_NEXTMILEAGE,
(select max(END_MILEAGE) from Table_details_usage d where d.VE_NO = s.VE_NO) END_MILEAGE
FROM Table_service_schedule s
WHERE SID = (SELECT max(SID) FROM Table_service_schedule s2 WHERE s2.VE_NO = s.VE_NO)
Probably could need to change the direct value ov SV_NEXTMILEAGE to max as well if the id:s aren't in order...
I have a tree structure like node(1)->node(2)->node(3). I have name as an property used to retrieve a node.
Given a node say node(3), i wanna retrieve node(1).
Query tried :
MATCH (p:Node)-[:HAS*]->(c:Node) WHERE c.name = "node 3" RETURN p LIMIT 5
But, not able to get node 1.
Your query will not only return "node 1", but it should at least include one path containing it. It's possible to filter the paths to only get the one traversing all the way to the root, however:
MATCH (c:Node {name: "node 3"})<-[:HAS*0..]-(p:Node)
// The root does not have any incoming relationship
WHERE NOT (p)<-[:HAS]-()
RETURN p
Note the use of the 0 length, which matches all cases, including the one where the start node is the root.
Fun fact: even if you have an index on Node:name, it won't be used (unless you're using Neo4j 3.1, where it seems to be fixed since 3.1 Beta2 at least) and you have to explicitly specify it.
MATCH (c:Node {name: "node 3"})<-[:HAS*0..]-(p:Node)
USING INDEX c:Node(name)
WHERE NOT (p)<-[:HAS]-()
RETURN p
Using PROFILE on the first query (with a numerical id property instead of name):
+-----------------------+----------------+------+---------+-------------------------+----------------------+
| Operator | Estimated Rows | Rows | DB Hits | Variables | Other |
+-----------------------+----------------+------+---------+-------------------------+----------------------+
| +ProduceResults | 0 | 1 | 0 | p | p |
| | +----------------+------+---------+-------------------------+----------------------+
| +AntiSemiApply | 0 | 1 | 0 | anon[23], c -- p | |
| |\ +----------------+------+---------+-------------------------+----------------------+
| | +Expand(All) | 1 | 0 | 3 | anon[58], anon[67] -- p | (p)<-[:HAS]-() |
| | | +----------------+------+---------+-------------------------+----------------------+
| | +Argument | 1 | 3 | 0 | p | |
| | +----------------+------+---------+-------------------------+----------------------+
| +Filter | 1 | 3 | 3 | anon[23], c, p | p:Node |
| | +----------------+------+---------+-------------------------+----------------------+
| +VarLengthExpand(All) | 1 | 3 | 5 | anon[23], p -- c | (c)<-[:HAS*]-(p) |
| | +----------------+------+---------+-------------------------+----------------------+
| +Filter | 1 | 1 | 3 | c | c.id == { AUTOINT0} |
| | +----------------+------+---------+-------------------------+----------------------+
| +NodeByLabelScan | 3 | 3 | 4 | c | :Node |
+-----------------------+----------------+------+---------+-------------------------+----------------------+
Total database accesses: 18
and on the second one:
+-----------------------+----------------+------+---------+-------------------------+------------------+
| Operator | Estimated Rows | Rows | DB Hits | Variables | Other |
+-----------------------+----------------+------+---------+-------------------------+------------------+
| +ProduceResults | 0 | 1 | 0 | p | p |
| | +----------------+------+---------+-------------------------+------------------+
| +AntiSemiApply | 0 | 1 | 0 | anon[23], c -- p | |
| |\ +----------------+------+---------+-------------------------+------------------+
| | +Expand(All) | 1 | 0 | 3 | anon[81], anon[90] -- p | (p)<-[:HAS]-() |
| | | +----------------+------+---------+-------------------------+------------------+
| | +Argument | 1 | 3 | 0 | p | |
| | +----------------+------+---------+-------------------------+------------------+
| +Filter | 1 | 3 | 3 | anon[23], c, p | p:Node |
| | +----------------+------+---------+-------------------------+------------------+
| +VarLengthExpand(All) | 1 | 3 | 5 | anon[23], p -- c | (c)<-[:HAS*]-(p) |
| | +----------------+------+---------+-------------------------+------------------+
| +NodeUniqueIndexSeek | 1 | 1 | 2 | c | :Node(id) |
+-----------------------+----------------+------+---------+-------------------------+------------------+
Total database accesses: 13
Is it possible to do union of two queries (from the same entity) in core data? In SQL speak, if entity is called t, then consider that T has following data:
+------+------+------+
| x | y | z |
+------+------+------+
| 1 | 11 | 2 |
| 1 | 12 | 3 |
| 2 | 11 | 1 |
| 3 | 12 | 3 |
Then I am trying to run the following query (using core data - not SQLite)
select x, y, sum(z)
from t
group by 1, 2
union
select x, 1 as y, sum(z)
from t
group by 1, 2
order by x, y, 1
;
+------+------+--------+
| x | y | sum(z) |
+------+------+--------+
| 1 | 1 | 5 |
| 1 | 11 | 2 |
| 1 | 12 | 3 |
| 2 | 1 | 1 |
| 2 | 11 | 1 |
| 3 | 1 | 3 |
| 3 | 12 | 3 |
+------+------+--------+
7 rows in set (0.00 sec)
Is it possible?
Thanks!
I have 2 mysql databases. (1) authorID, authorFirst, authorLast. (2) courseID, courseName, courseAuthor1, courseAuthor2, courseAuthor3, courseAuthor4. How to join 2 columns from authors database into one OR ALL 4 of the courseAuthor columns in the courses database?
authors
authorID | authorFirst | authorLast
1 | firstA | lastA
2 | firstB | lastB
3 | firstC | lastC
4 | firstD | lastC
courses
courseID | courseName | courseAuthor1 | courseAuthor2 | courseAuthor3 | courseAuthor4
1 | course1 | 2 | 3 | NULL | NULL
2 | course 2 | 1 | 4 | 3 | NULL
3 | course 33 | 4 | 1 | 2 | NULL
4 | course 4 | 3 | NULL | NULL | NULL
I Want Results to look like this:
courseID | courseName | courseAuthor1 | courseAuthor2 | courseAuthor3 | courseAuthor4
1 | course1 | firstB LastB | firstC LastC | NULL | NULL
2 | course 2 | firstA LastA | firstD LastD | firstC LastC | NULL
3 | course 33 | firstD LastD | firstA LastA | firstB LastB | NULL
4 | course 4 | firstC LastC | NULL | NULL | NULL
I don't even know where to start. Here is my code:
SELECT * FROM courses JOIN authors ON authors.authorID IN
(courses.courseAuthor1, courses.courseAuthor2, courses.courseAuthor3,
courses.courseAuthor4)
Which shows:
results | courseID | courseName | courseAuthor1 | courseAuthor2 | courseAuthor3 | courseAuthor4 | authorID | authorFirst | authorLast
1 | 1 | course1 | 2 | 3 | NULL | NULL | 2 | firstB | lastB
2 | 1 | course1 | 2 | 3 | NULL | NULL | 3 | firstC | lastC
3 | 2 | course 2 | 1 | 4 | 3 | NULL | 1 | firstA | lastA
4 | 2 | course 2 | 1 | 4 | 3 | NULL | 4 | firstD | lastD
5 | 2 | course 2 | 1 | 4 | 3 | NULL | 3 | firstC | lastC
6 | 3 | course 33 | 4 | 1 | 2 | NULL | 4 | firstD | lastD
7 | 3 | course 33 | 4 | 1 | 2 | NULL | 1 | firstA | lastA
8 | 3 | course 33 | 4 | 1 | 2 | NULL | 2 | firstB | lastB
9 | 4 | course 4 | 3 | NULL | NULL | NULL | 3 | firstC | lastC
Thanks So Much!
It took me a while, but i've answered my question. Here it is:
SELECT c1.courseID, c1.courseName,
CONCAT(a1.authorFirst,' ',a1.authorLast) AS aOne,
CONCAT(a2.authorFirst,' ',a2.authorLast) AS aTwo,
CONCAT(a3.authorFirst,' ',a3.authorLast) AS aThree,
CONCAT(a4.authorFirst,' ',a4.authorLast) AS aFour
FROM courses c1
LEFT JOIN authors a1 ON c1.courseAuthor1 = a1.authorID
LEFT JOIN authors a2 ON c1.courseAuthor2 = a2.authorID
LEFT JOIN authors a3 ON c1.courseAuthor3 = a3.authorID
LEFT JOIN authors a4 ON c1.courseAuthor4 = a4.authorID