I've tried inserting the same entry to mongodb twice on nodejs with mongoskin. Unique index was specified for the collection. However, no error was given on the second insertion. What should I do to detect if there is a duplicate error?
Thanks!
Be sure you're using safe mode. To use safe mode as the default for your connection you can specify it as a parameter when creating the connection. From the mongoskin docs:
var db = mongoskin.db([
'192.168.0.1:27017/?auto_reconnect=true',
'192.168.0.2:27017/?auto_reconnect=true',
'192.168.0.3:27017/?auto_reconnect=true'
], {
database: 'testdb',
safe: true
}, {
connectArbiter: false,
socketOptions: {
timeout: 2000
}
});
Safe mode instructs the client to call getLastError after each operation, which is how it knows whether there was an error. So you usually want to be sure to have safe mode on!