Firebase query not working as expected React native












0















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 });
});
}









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    0















    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 });
    });
    }









    share|improve this question



























      0












      0








      0








      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 });
      });
      }









      share|improve this question
















      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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      edited Jan 1 at 20:00









      Frank van Puffelen

      238k29382408




      238k29382408










      asked Jan 1 at 15:39









      georgcongeorgcon

      399




      399
























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          had to replace .on with .once and put the if found statement in a .then() block






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            1 Answer
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            1 Answer
            1






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            had to replace .on with .once and put the if found statement in a .then() block






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              had to replace .on with .once and put the if found statement in a .then() block






              share|improve this answer


























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                had to replace .on with .once and put the if found statement in a .then() block






                share|improve this answer













                had to replace .on with .once and put the if found statement in a .then() block







                share|improve this answer












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                share|improve this answer










                answered Jan 1 at 16:37









                georgcongeorgcon

                399




                399
































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