How to force refetch the entire node in Relay? - relayjs

Question

Suppose that we have this mutation:
const [addCryptoAddress] =
useMutation(graphql`
mutation useCrypto_AddCryptoWalletAddressMutation(
$input: AddCryptoAddressInput!
) {
addCryptoAddress(input: $input) {
userAccount {
cryptoCurrencyAddresses {
count
edges {
node {
id
address
}
}
}
}
errors {
message
}
}
}
`);
If successful, a new CryptoCurrencyAddresses will be available under UserAccount.
Now suppose that somewhere else in the code we have a lazyLoadQuery that fetches these addresses, e.g.
const { visitor } = useLazyLoadQuery<ManualPayoutMachineContextQuery>(
graphql`
query ManualPayoutMachineContextQuery {
visitor {
userAccount {
cryptoCurrencyAddresses {
edges {
node {
id
address
canDelete
currency
isDefault
label
}
}
}
}
}
}
`,
{}
);
However, note that this query references additional fields that are not mentioned in the mutation. The result is that all unmentioned fields are undefined immediately after the mutation, i.e.
visitor.userAccount?.cryptoCurrencyAddresses.edges
.map(({ node }) => {
return {
address: node.address,
canDelete: node.canDelete,
currency: node.currency,
id: node.id,
isDefault: node.isDefault,
label: node.label,
};
});
Produces:
[
{
address: '0xc0ffee254729296a45a3885639AC7E10F9d54979',
canDelete: undefined,
currency: undefined,
id: 'WyJDcnlwdG9DdXJyZW5jeUFkZHJlc3MiLDIwMjE3NF0=',
isDefault: undefined,
label: undefined,
}
]
Apart from listing every overlapping field in the mutation, is there a way to force all queries that depend on this data to reload when they detect new data?

Answer 1

You can set fetchPolicy as store-and-network, like this:
const { visitor } = useLazyLoadQuery<ManualPayoutMachineContextQuery>(
graphql`
query ManualPayoutMachineContextQuery {
visitor {
userAccount {
cryptoCurrencyAddresses {
edges {
node {
id
address
canDelete
currency
isDefault
label
}
}
}
}
}
}
`,
{},
{
fetchPolicy: 'store-and-network',
}
);
The store-and-network will use the data of the store and always do a network request without making regardless.