Suppressing/Catching the ValidationError error from Mongoose - mongodb

Question

I have a static method setup that will verify if an object follows the schema of the model properly, and I've got the validation itself working just fine, but I can't get Mongoose to hide the ValidationError error that displays in the console.
The static method for the validation is:
fieldSchema.statics.testValidate = function( field ) {
return new Promise( ( res, rej ) => {
new this( field ).validate( err => {
if ( err ) return rej( err )
res( true )
} )
})
}
Then when I execute it:
Field.testValidate( fieldData )
.then(data => {
console.log('OK!', data)
})
.catch( err => {
console.error('FAILED:',err)
})
.finally(() => Mongoose.connection.close())
And like I said, I can get it to validate the data, but it will always show this error:
Unhandled rejection ValidationError: Field validation failed
at MongooseError.ValidationError ....
(Note: Im using Bluebird for my JS promises, incase it has anything to do with the then/catch somehow)
The script execution keeps going on with the code, so that means that it doesnt actually kill the script or anything, it just logs the error to the console. But since this function is meant to validate, its not really an "error", so is there a way to hide this from the console? (Just for this function, not for ALL Mongoose errors)
Thanks

Answer 1

validate already returns a promise so you don't need to wrap it in a new promise. The reason you see the error logged to the console is because of the fact that you're wrapping it in a new promise thus preventing the original promise from ever being caught i.e. Unhandled rejection.
fieldSchema.statics.testValidate = function(fieldData) {
return new this(fieldData).validate()
}