Mixing
Count relationships by types in Neo4j
1
I have many relationship types in the database. How do I count relationships by each type without using apoc?
neo4j cypher
edited Nov 21 '18 at 13:32
David Makogon
56.8k15106153
asked Nov 21 '18 at 12:02
Schatt95Schatt95
789
add a comment |
1
I have many relationship types in the database. How do I count relationships by each type without using apoc?
neo4j cypher
edited Nov 21 '18 at 13:32
David Makogon
56.8k15106153
asked Nov 21 '18 at 12:02
Schatt95Schatt95
789
add a comment |
1
1
1
I have many relationship types in the database. How do I count relationships by each type without using apoc?
neo4j cypher
edited Nov 21 '18 at 13:32
David Makogon
56.8k15106153
asked Nov 21 '18 at 12:02
Schatt95Schatt95
789
I have many relationship types in the database. How do I count relationships by each type without using apoc?
neo4j cypher
neo4j cypher
edited Nov 21 '18 at 13:32
David Makogon
56.8k15106153
asked Nov 21 '18 at 12:02
Schatt95Schatt95
789
edited Nov 21 '18 at 13:32
David Makogon
56.8k15106153
asked Nov 21 '18 at 12:02
Schatt95Schatt95
789
edited Nov 21 '18 at 13:32
David Makogon
56.8k15106153
edited Nov 21 '18 at 13:32
David Makogon
56.8k15106153
edited Nov 21 '18 at 13:32
David Makogon
56.8k15106153
56.8k15106153
asked Nov 21 '18 at 12:02
Schatt95Schatt95
789
asked Nov 21 '18 at 12:02
Schatt95Schatt95
789
asked Nov 21 '18 at 12:02
Schatt95Schatt95
789
789
add a comment |
add a comment |
1 Answer
1
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oldest
votes
1
Solution
MATCH ()-[relationship]-()
RETURN TYPE(relationship) AS type, COUNT(relationship) AS amount;
The first line specifies the pattern to define the relationship variable, which is used to determine type and amount in line two.
Example result
╒══════════════╤════════╕
│"type" │"amount"│
╞══════════════╪════════╡
│"BELONGS_TO" │1234567 │
├──────────────┼────────┤
│"CONTAINS" │432552 │
├──────────────┼────────┤
│"IS_PART_OF" │947227 │
├──────────────┼────────┤
│"HOLDS" │4 │
└──────────────┴────────┘
answered Nov 21 '18 at 12:48
ThirstForKnowledgeThirstForKnowledge
6651112
You should addORDER BY xyz DESC
– Guy Coder
Nov 21 '18 at 18:25
Thank you for the answer! It is fine unless you have duplicated relationship names between different nodes and you'd like to catch that. In such a case the following expression might be helpful:match (a)-[r]-(b) return distinct labels(a), labels(b), type(r) as type, count(r) as amount
– Schatt95
Nov 28 '18 at 7:20
add a comment |
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1 Answer
1
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Solution
MATCH ()-[relationship]-()
RETURN TYPE(relationship) AS type, COUNT(relationship) AS amount;
The first line specifies the pattern to define the relationship variable, which is used to determine type and amount in line two.
Example result
╒══════════════╤════════╕
│"type" │"amount"│
╞══════════════╪════════╡
│"BELONGS_TO" │1234567 │
├──────────────┼────────┤
│"CONTAINS" │432552 │
├──────────────┼────────┤
│"IS_PART_OF" │947227 │
├──────────────┼────────┤
│"HOLDS" │4 │
└──────────────┴────────┘
answered Nov 21 '18 at 12:48
ThirstForKnowledgeThirstForKnowledge
6651112
You should addORDER BY xyz DESC
– Guy Coder
Nov 21 '18 at 18:25
Thank you for the answer! It is fine unless you have duplicated relationship names between different nodes and you'd like to catch that. In such a case the following expression might be helpful:match (a)-[r]-(b) return distinct labels(a), labels(b), type(r) as type, count(r) as amount
– Schatt95
Nov 28 '18 at 7:20
add a comment |
1
Solution
MATCH ()-[relationship]-()
RETURN TYPE(relationship) AS type, COUNT(relationship) AS amount;
The first line specifies the pattern to define the relationship variable, which is used to determine type and amount in line two.
Example result
╒══════════════╤════════╕
│"type" │"amount"│
╞══════════════╪════════╡
│"BELONGS_TO" │1234567 │
├──────────────┼────────┤
│"CONTAINS" │432552 │
├──────────────┼────────┤
│"IS_PART_OF" │947227 │
├──────────────┼────────┤
│"HOLDS" │4 │
└──────────────┴────────┘
answered Nov 21 '18 at 12:48
ThirstForKnowledgeThirstForKnowledge
6651112
You should addORDER BY xyz DESC
– Guy Coder
Nov 21 '18 at 18:25
Thank you for the answer! It is fine unless you have duplicated relationship names between different nodes and you'd like to catch that. In such a case the following expression might be helpful:match (a)-[r]-(b) return distinct labels(a), labels(b), type(r) as type, count(r) as amount
– Schatt95
Nov 28 '18 at 7:20
add a comment |
1
1
1
Solution
MATCH ()-[relationship]-()
RETURN TYPE(relationship) AS type, COUNT(relationship) AS amount;
The first line specifies the pattern to define the relationship variable, which is used to determine type and amount in line two.
Example result
╒══════════════╤════════╕
│"type" │"amount"│
╞══════════════╪════════╡
│"BELONGS_TO" │1234567 │
├──────────────┼────────┤
│"CONTAINS" │432552 │
├──────────────┼────────┤
│"IS_PART_OF" │947227 │
├──────────────┼────────┤
│"HOLDS" │4 │
└──────────────┴────────┘
answered Nov 21 '18 at 12:48
ThirstForKnowledgeThirstForKnowledge
6651112
Solution
MATCH ()-[relationship]-()
RETURN TYPE(relationship) AS type, COUNT(relationship) AS amount;
The first line specifies the pattern to define the relationship variable, which is used to determine type and amount in line two.
Example result
╒══════════════╤════════╕
│"type" │"amount"│
╞══════════════╪════════╡
│"BELONGS_TO" │1234567 │
├──────────────┼────────┤
│"CONTAINS" │432552 │
├──────────────┼────────┤
│"IS_PART_OF" │947227 │
├──────────────┼────────┤
│"HOLDS" │4 │
└──────────────┴────────┘
answered Nov 21 '18 at 12:48
ThirstForKnowledgeThirstForKnowledge
6651112
edited Nov 21 '18 at 13:15
answered Nov 21 '18 at 12:48
ThirstForKnowledgeThirstForKnowledge
6651112
answered Nov 21 '18 at 12:48
ThirstForKnowledgeThirstForKnowledge
6651112
answered Nov 21 '18 at 12:48
ThirstForKnowledgeThirstForKnowledge
6651112
6651112
You should addORDER BY xyz DESC
– Guy Coder
Nov 21 '18 at 18:25
Thank you for the answer! It is fine unless you have duplicated relationship names between different nodes and you'd like to catch that. In such a case the following expression might be helpful:match (a)-[r]-(b) return distinct labels(a), labels(b), type(r) as type, count(r) as amount
– Schatt95
Nov 28 '18 at 7:20
add a comment |
You should addORDER BY xyz DESC
– Guy Coder
Nov 21 '18 at 18:25
Thank you for the answer! It is fine unless you have duplicated relationship names between different nodes and you'd like to catch that. In such a case the following expression might be helpful:match (a)-[r]-(b) return distinct labels(a), labels(b), type(r) as type, count(r) as amount
– Schatt95
Nov 28 '18 at 7:20
You should add
ORDER BY xyz DESC– Guy Coder
Nov 21 '18 at 18:25
You should add
ORDER BY xyz DESC– Guy Coder
Nov 21 '18 at 18:25
Thank you for the answer! It is fine unless you have duplicated relationship names between different nodes and you'd like to catch that. In such a case the following expression might be helpful:
match (a)-[r]-(b) return distinct labels(a), labels(b), type(r) as type, count(r) as amount– Schatt95
Nov 28 '18 at 7:20
Thank you for the answer! It is fine unless you have duplicated relationship names between different nodes and you'd like to catch that. In such a case the following expression might be helpful:
match (a)-[r]-(b) return distinct labels(a), labels(b), type(r) as type, count(r) as amount– Schatt95
Nov 28 '18 at 7:20
add a comment |
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