问题
So I'm trying to use the model.upsert()
of sequelize
and all i receive is inserts , no matter what i change in the query.
I have a Transaction model that has some fields, with the default generated id.
reading the sequelize's upsert
documentation i noticed this:
An update will be executed if a row which matches the supplied values on either the primary key or a unique key is found. Note that the unique index must be defined in your sequelize model and not just in the table.
So i was guessing i have to define the id
of the Transaction in the model definition, and so i did with no luck as it still only creates new entries..
TransactionModel = {
id: {
type: Sequelize.INTEGER,
allowNull: false,
primaryKey: true,
autoIncrement: true
},
{.......}
}
What am i doing wrong, what did i miss?
Any explanation and solution will be highly appreciated, thanks in advance!
EDIT:
This is the upsert code:
createOrUpdateTransaction: {
type: Transaction,
args: {
payerAccountNumber: {type: new GraphQLNonNull(GraphQLInt)},
recipientAccountNumber: {type: new GraphQLNonNull(GraphQLInt)},
amount: {type: new GraphQLNonNull(GraphQLFloat)},
currency: {type: new GraphQLNonNull(GraphQLString)},
paymentMethod: {type: new GraphQLNonNull(GraphQLString)},
cardNumber: {type: GraphQLFloat},
cardName: {type: GraphQLString},
cardNetwork: {type: GraphQLString},
cashMachineId: {type: GraphQLFloat},
receiptNumber: {type: new GraphQLNonNull(GraphQLFloat)},
invoiceNumber: {type: new GraphQLNonNull(GraphQLFloat)},
receiptCopy: {type: new GraphQLNonNull(GraphQLString)},
description: {type: GraphQLString},
bankDescription: {type: GraphQLString},
bankReference: {type: new GraphQLNonNull(GraphQLString)},
bankSubCurrencyAccount: {type: new GraphQLNonNull(GraphQLString)},
tags: {type: new GraphQLList(GraphQLString)},
notes: {type: GraphQLString}
},
resolve: (root, args) => {
return db.models.transaction.upsert({
time: new Date().toString(),
payerAccountNumber: args.payerAccountNumber,
recipientAccountNumber: args.recipientAccountNumber,
amount: args.amount,
currency: args.currency,
paymentMethod: args.paymentMethod,
cardNumber: args.cardNumber,
cardName: args.cardName,
cardNetwork: args.cardNetwork,
cashMachineId: args.cashMachineId,
receiptNumber: args.receiptNumber,
invoiceNumber: args.invoiceNumber,
receiptCopy: args.receiptCopy,
description: args.description,
bankDescription: args.bankDescription,
bankReference: args.bankReference,
bankSubCurrencyAccount: args.bankSubCurrencyAccount,
tags: args.tags,
notes: args.notes,
bankAccountAccountNumber: args.payerAccountNumber
})
}
}
As this is part of a Mutation
in GraphQL
.
It might be worth noting that this was addTransaction
before and all i changed was to db.models.transaction.upsert()
from db.models.transaction.create()
回答1:
In your upsert() example you aren't providing the id of the entry into the upsert method. This means sequelize can't match the id to a row (because the id is undefined) and therefore it inserts a new row.
Even if you use a different primary key it must always be a property for it to match since sequelize uses the primary key to search for an existing row.
createOrUpdateTransaction: {
type: Transaction,
args: {
// Omitted code...
},
resolve: (root, args) => {
return db.models.transaction.upsert({
// The id property must be defined in the args object for
// it to match to an existing row. If args.id is undefined
// it will insert a new row.
id: args.id,
time: new Date().toString(),
payerAccountNumber: args.payerAccountNumber,
recipientAccountNumber: args.recipientAccountNumber,
amount: args.amount,
currency: args.currency,
paymentMethod: args.paymentMethod,
cardNumber: args.cardNumber,
cardName: args.cardName,
cardNetwork: args.cardNetwork,
// Omitted fields ...
})
}
}
来源:https://stackoverflow.com/questions/39399018/sequelize-upsert-never-updates-and-only-inserts