javascript - ConstraintError: Key already exists in the object store - Stack Overflow

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I am working with React 16.3.2, Redux 4 and Dexie 2.0.3.

when I am going to store data second time it throws this error message.

Error: ConstraintError: Key already exists in the object store.

   return dispatch => {
        db.table
        .add(data)
        .then (function(id){
            console.log(id)
        })
        .catch (function (error) {
            console.log("Error: " + error);
        });
    }

My Db schema:

   const db = new Dexie('ReactReduxDexieJsCRUD');
  db.version(1).stores({table:'++id,name,age,bloodGroup,donateBefore,weight' });

The first time it stores date well but after it gives the error.

I am working with React 16.3.2, Redux 4 and Dexie 2.0.3.

when I am going to store data second time it throws this error message.

Error: ConstraintError: Key already exists in the object store.

   return dispatch => {
        db.table
        .add(data)
        .then (function(id){
            console.log(id)
        })
        .catch (function (error) {
            console.log("Error: " + error);
        });
    }

My Db schema:

   const db = new Dexie('ReactReduxDexieJsCRUD');
  db.version(1).stores({table:'++id,name,age,bloodGroup,donateBefore,weight' });

The first time it stores date well but after it gives the error.

• javascript
• redux
• indexeddb
• dexie

Share Follow edited May 18, 2018 at 11:12 MD Ashik asked May 17, 2018 at 18:25 MD AshikMD Ashik 9,8451010 gold badges5555 silver badges6161 bronze badges 8

• 2 Try put(data) instead of add(data). – Oblosys Commented May 17, 2018 at 18:59
• I didn't worked with Dexie, but similar problem on MySql can be solved with "ON DUPLICATE KEY UPDATE". If you want on second call to add new data as new record, you need to set table key to be autoincrement, or to specify new key manually as second parameter to add call. If you want to add data or update data if exist, try [dexie/docs/Table/Table.put()](put) instead of add. – stolex Commented May 17, 2018 at 18:59
• Lovely ment #Oblosys & #stolex but it creates another issue it doesn't change the ID and it works like #update in database – MD Ashik Commented May 17, 2018 at 19:04
• works as expected, but you dont want an update, so what do you actually want to happen when a duplicate key is used? – user9487972 Commented May 17, 2018 at 22:51
• @smith I don't want to use the duplicate key. and why it gives me error duplicate key . i am submitted my from second time with new value so it should be store new data with new ID ? – MD Ashik Commented May 18, 2018 at 9:05

 |  Show 3 more ments

1 Answer 1

Sorted by: Reset to default 8

How does your schema look like? (the part db.version(x).stores({...}) ?

The most mon is to have inbound primary key, example:

db.version(1).stores({
  table: 'id, idx1, idx2...'
});

Here id is the primary key.

• db.table.add({id: 1, foo: 'bar'}) will add object with id 1.
• db.table.add({id: 1, foo: 'bar2'}) 2nd time will fail because id 1 exists.
• db.table.put({id: 1, foo: 'bar2'}) will update object with id 1.

So what do you really want to do? You say you want to add new object with new key. If so, I suppose the error is that you give the same key second time.

You can also let the id be generated by the db

db.version(2).stores({
  table: '++id, idx1, idx2...'
});

Then you don't need to supply id in calls to add():

• db.table.add({foo: 'bar'}) will add object with id 1.
• db.table.add({foo: 'barX'}) 2nd time will add new obj with id 2
• ...

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