user
10/22/2017, 8:49 AM.graphcoolrc
is in my home dir, if you've got a terminal try ls .graphcoolrc*
user
10/22/2017, 8:50 AMuser
10/22/2017, 8:52 AMgraphcool.old
is de legacy one , i think with more graphcool.old
you can check this , it should contain a json structure starting with {
instead of c
user
10/22/2017, 8:54 AMgraphcoolrc.old
to .graphcoolrc
Manfred Neustifter
10/22/2017, 8:54 AMuser
10/22/2017, 8:55 AM.graphcoolrc
you see cluster
the unexpected c
Manfred Neustifter
10/22/2017, 8:55 AMuser
10/22/2017, 8:58 AMmv .graphcoolrc graphcool.next
ln -s .graphcoolrc.old .graphcoolrc
Manfred Neustifter
10/22/2017, 9:03 AMManfred Neustifter
10/22/2017, 9:04 AMtfiwm
10/22/2017, 10:43 AMnilan
10/22/2017, 10:55 AMtfiwm
10/22/2017, 11:01 AMManfred Neustifter
10/22/2017, 11:12 AMnilan
10/22/2017, 11:14 AMMaslov
10/22/2017, 2:14 PMMaslov
10/22/2017, 2:18 PMMaslov
10/22/2017, 2:18 PMrein
10/22/2017, 2:42 PMDejan Nesic
10/22/2017, 2:48 PMtype User @model {
id: ID! @isUnique
name: String
email: String! @isUnique
number: String
}
query code
const getUserQuery = gql`
query User($email: String!) {
User(email: $email){
id
name
}
}
`;
checkProfileSubscription: Subscription
myemail:String = ‘exampleEmail@mail.com'
this.checkProfileSubscription = this.apollo.watchQuery({
query: getUserQuery,
variables: {
$email: this.myemail
}
}).subscribe(({data}: any) => {
console.log(data)
});
I get error like
Error: GraphQL error: Variable '$email' expected value of type 'String!' but value is undefined. Reason: Expected non-null value, found null. (line 1, column 12): query User($email: String!)
Vishwaraj Malik
10/22/2017, 2:55 PMnikolasburk
Maslov
10/22/2017, 2:59 PMMaslov
10/22/2017, 3:00 PMMaslov
10/22/2017, 3:02 PMVishwaraj Malik
10/22/2017, 3:06 PMtfiwm
10/22/2017, 3:15 PMuser
10/22/2017, 3:17 PMdynn27
10/22/2017, 3:19 PMdynn27
10/22/2017, 3:19 PM