Atomic Property Updates
Atomic procedures handle the concurrency, it’s add a lock to the resource.
If two users access to the same resource at the same time, with the parameter times
(default value 5) we can determine how many time retry to modify the resource, until the lock is release.
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adds the number to the value of the property |
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subtracts the number to the value of the property |
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concatenate the string to the property |
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inserts the object in the chosen position of the array |
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remove from the array the element to the position selected |
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update the property with the result of the expression |
CREATE (p:Person {name:'Tom',age: 40})
CREATE (p:Person {name:'Will',age: 35})
CREATE (p:Person {name:'David', children: ['Anne','Sam','Paul']})
CREATE (p:Person {name:'John', cars: ['Class A','X3','Focus']})
CREATE (p:Person {name:'Ryan', salary1:1800, salary2:1500})
age
for Tom
:MATCH (n:Person {name:'Tom'})
CALL apoc.atomic.add(n,'age',10,5)
YIELD oldValue, newValue
RETURN n
n |
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{"name":"Tom","age":50} |
age
for Tom
:MATCH (n:Person {name:'Tom'})
CALL apoc.atomic.subtract(n,'age',10,5)
YIELD oldValue, newValue
RETURN n
n |
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{"name":"Tom","age":40} |
iam
to the name
property for Will
:MATCH (p:Person {name:'Will',age: 35})
CALL apoc.atomic.concat(p,"name",'iam',5)
YIELD newValue
RETURN p
p |
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{"name":"William","age":35} |
Mary
in position 2 of children
list:MATCH (p:Person {name:'David'})
CALL apoc.atomic.insert(p,'children',2,'Mary',5)
YIELD newValue
RETURN p
p |
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{"name":"David","children":["Anne","Sam","Mary","Paul"]} |
X3
, which is at position 1, from the array cars
MATCH (p:Person {name:'John'})
CALL apoc.atomic.remove(p,'cars',1,5)
YIELD newValue
RETURN p
p |
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{"name":"John","cars":["Class A","Focus"]} |
salary1
with the result of an expression:MATCH (p:Person {name:'Ryan'})
CALL apoc.atomic.update(p,'salary1','n.salary1*3 + n.salary2',5)
YIELD newValue
RETURN p
The expression always have to be referenced with the n.
that refers to the entity passed as the first parameter.
If we rename our node/rel (as in the example above) we have anyway to refer to it in the expression as n
.
p |
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{"name":"Ryan","salary1":6900,"salary2":1500} |