How to update information in an existing node instead of creating a new one using Dgraph?
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I am writing a Golang application using Dgraph for persisting objects. From the documentation, I can infer that a new UID and hence a new node is created everytime I mutate an object/run the code.
Is there a way to update the same node data instead for creating a new node?
I tried changing the UID to use "_:name" for the UID field but even this creates a new node everytime the application is run. I wish to be able to update the existing node if it is already present in the DB instead of creating a new node for it.
dgraph dgraph-dgo
asked Oct 11 at 4:53
hemanik
155313
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up vote
1
down vote
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I am writing a Golang application using Dgraph for persisting objects. From the documentation, I can infer that a new UID and hence a new node is created everytime I mutate an object/run the code.
Is there a way to update the same node data instead for creating a new node?
I tried changing the UID to use "_:name" for the UID field but even this creates a new node everytime the application is run. I wish to be able to update the existing node if it is already present in the DB instead of creating a new node for it.
dgraph dgraph-dgo
asked Oct 11 at 4:53
hemanik
155313
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am writing a Golang application using Dgraph for persisting objects. From the documentation, I can infer that a new UID and hence a new node is created everytime I mutate an object/run the code.
Is there a way to update the same node data instead for creating a new node?
I tried changing the UID to use "_:name" for the UID field but even this creates a new node everytime the application is run. I wish to be able to update the existing node if it is already present in the DB instead of creating a new node for it.
dgraph dgraph-dgo
asked Oct 11 at 4:53
hemanik
155313
I am writing a Golang application using Dgraph for persisting objects. From the documentation, I can infer that a new UID and hence a new node is created everytime I mutate an object/run the code.
Is there a way to update the same node data instead for creating a new node?
I tried changing the UID to use "_:name" for the UID field but even this creates a new node everytime the application is run. I wish to be able to update the existing node if it is already present in the DB instead of creating a new node for it.
dgraph dgraph-dgo
dgraph dgraph-dgo
asked Oct 11 at 4:53
hemanik
155313
asked Oct 11 at 4:53
hemanik
155313
asked Oct 11 at 4:53
hemanik
155313
asked Oct 11 at 4:53
hemanik
155313
asked Oct 11 at 4:53
hemanik
155313
155313
add a comment |
add a comment |
1 Answer
1
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2
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Unfortunately the docs aren't very beginner friendly yet :/
To modify / mutate existing data you have to run a set operation and supply a rdf-triple like <uid> <predicate> "value" / <objectYouWantToModify> <attributeYouWantToModify> "quotedStringValue". If it is not an attribute but an edge, the value has to be another <uid>.
The full mutation would be for example
{
set {
<0x2> <name> "modified-name" .
}
}
The . terminates the sequence and there is an optional fourth parameter you can use to also assign a label.
Check https://www.w3.org/TR/n-quads/ for further details.
answered Nov 11 at 8:24
TommyF
1,0651726
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Unfortunately the docs aren't very beginner friendly yet :/
To modify / mutate existing data you have to run a set operation and supply a rdf-triple like <uid> <predicate> "value" / <objectYouWantToModify> <attributeYouWantToModify> "quotedStringValue". If it is not an attribute but an edge, the value has to be another <uid>.
The full mutation would be for example
{
set {
<0x2> <name> "modified-name" .
}
}
The . terminates the sequence and there is an optional fourth parameter you can use to also assign a label.
Check https://www.w3.org/TR/n-quads/ for further details.
answered Nov 11 at 8:24
TommyF
1,0651726
add a comment |
up vote
2
down vote
Unfortunately the docs aren't very beginner friendly yet :/
To modify / mutate existing data you have to run a set operation and supply a rdf-triple like <uid> <predicate> "value" / <objectYouWantToModify> <attributeYouWantToModify> "quotedStringValue". If it is not an attribute but an edge, the value has to be another <uid>.
The full mutation would be for example
{
set {
<0x2> <name> "modified-name" .
}
}
The . terminates the sequence and there is an optional fourth parameter you can use to also assign a label.
Check https://www.w3.org/TR/n-quads/ for further details.
answered Nov 11 at 8:24
TommyF
1,0651726
add a comment |
up vote
2
down vote
up vote
2
down vote
Unfortunately the docs aren't very beginner friendly yet :/
To modify / mutate existing data you have to run a set operation and supply a rdf-triple like <uid> <predicate> "value" / <objectYouWantToModify> <attributeYouWantToModify> "quotedStringValue". If it is not an attribute but an edge, the value has to be another <uid>.
The full mutation would be for example
{
set {
<0x2> <name> "modified-name" .
}
}
The . terminates the sequence and there is an optional fourth parameter you can use to also assign a label.
Check https://www.w3.org/TR/n-quads/ for further details.
answered Nov 11 at 8:24
TommyF
1,0651726
Unfortunately the docs aren't very beginner friendly yet :/
To modify / mutate existing data you have to run a set operation and supply a rdf-triple like <uid> <predicate> "value" / <objectYouWantToModify> <attributeYouWantToModify> "quotedStringValue". If it is not an attribute but an edge, the value has to be another <uid>.
The full mutation would be for example
{
set {
<0x2> <name> "modified-name" .
}
}
The . terminates the sequence and there is an optional fourth parameter you can use to also assign a label.
Check https://www.w3.org/TR/n-quads/ for further details.
answered Nov 11 at 8:24
TommyF
1,0651726
answered Nov 11 at 8:24
TommyF
1,0651726
answered Nov 11 at 8:24
TommyF
1,0651726
answered Nov 11 at 8:24
TommyF
1,0651726
1,0651726
add a comment |
add a comment |
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