Firebase query not working as expected React native
I have a query based on multiple WHERE clauses and it returns true if found and deletes record and false if otherwise. The problem is it adds the new record in the database but automatically deletes it afterwards. When commenting out the remove part it works as expected adding the new record only if not present.
This is my code:
var found = false;
firebase.database().ref(`/yums/`)
.orderByChild('recipeId').equalTo(uid)
.on('value', snapshot => {
const likes = snapshot.val();
for (var like in likes) {
if (likes[like].userId == currentUser.uid) {
found = true;
firebase.database().ref(`/yums/`)
.child(like).remove();
}
}
})
if (found == false ) {
firebase.database().ref(`/yums/`)
.push({
recipeId: uid,
userId: currentUser.uid,
timestamp: Date.now(),
})
.then(() => {
dispatch({ type: YUMMED_SUCCESS });
});
}
firebase react-native firebase-realtime-database
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I have a query based on multiple WHERE clauses and it returns true if found and deletes record and false if otherwise. The problem is it adds the new record in the database but automatically deletes it afterwards. When commenting out the remove part it works as expected adding the new record only if not present.
This is my code:
var found = false;
firebase.database().ref(`/yums/`)
.orderByChild('recipeId').equalTo(uid)
.on('value', snapshot => {
const likes = snapshot.val();
for (var like in likes) {
if (likes[like].userId == currentUser.uid) {
found = true;
firebase.database().ref(`/yums/`)
.child(like).remove();
}
}
})
if (found == false ) {
firebase.database().ref(`/yums/`)
.push({
recipeId: uid,
userId: currentUser.uid,
timestamp: Date.now(),
})
.then(() => {
dispatch({ type: YUMMED_SUCCESS });
});
}
firebase react-native firebase-realtime-database
add a comment |
I have a query based on multiple WHERE clauses and it returns true if found and deletes record and false if otherwise. The problem is it adds the new record in the database but automatically deletes it afterwards. When commenting out the remove part it works as expected adding the new record only if not present.
This is my code:
var found = false;
firebase.database().ref(`/yums/`)
.orderByChild('recipeId').equalTo(uid)
.on('value', snapshot => {
const likes = snapshot.val();
for (var like in likes) {
if (likes[like].userId == currentUser.uid) {
found = true;
firebase.database().ref(`/yums/`)
.child(like).remove();
}
}
})
if (found == false ) {
firebase.database().ref(`/yums/`)
.push({
recipeId: uid,
userId: currentUser.uid,
timestamp: Date.now(),
})
.then(() => {
dispatch({ type: YUMMED_SUCCESS });
});
}
firebase react-native firebase-realtime-database
I have a query based on multiple WHERE clauses and it returns true if found and deletes record and false if otherwise. The problem is it adds the new record in the database but automatically deletes it afterwards. When commenting out the remove part it works as expected adding the new record only if not present.
This is my code:
var found = false;
firebase.database().ref(`/yums/`)
.orderByChild('recipeId').equalTo(uid)
.on('value', snapshot => {
const likes = snapshot.val();
for (var like in likes) {
if (likes[like].userId == currentUser.uid) {
found = true;
firebase.database().ref(`/yums/`)
.child(like).remove();
}
}
})
if (found == false ) {
firebase.database().ref(`/yums/`)
.push({
recipeId: uid,
userId: currentUser.uid,
timestamp: Date.now(),
})
.then(() => {
dispatch({ type: YUMMED_SUCCESS });
});
}
firebase react-native firebase-realtime-database
firebase react-native firebase-realtime-database
edited Jan 1 at 20:00
Frank van Puffelen
238k29382408
238k29382408
asked Jan 1 at 15:39
georgcongeorgcon
399
399
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add a comment |
1 Answer
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votes
had to replace .on with .once and put the if found statement in a .then() block
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
had to replace .on with .once and put the if found statement in a .then() block
add a comment |
had to replace .on with .once and put the if found statement in a .then() block
add a comment |
had to replace .on with .once and put the if found statement in a .then() block
had to replace .on with .once and put the if found statement in a .then() block
answered Jan 1 at 16:37
georgcongeorgcon
399
399
add a comment |
add a comment |
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