Getting .key from Firebase using VueFire - firebase

Question

I am trying to get a single key from firebase nodes and I can't get any from the code I have right now. Here is my code:
let app = Firebase.initializeApp(config)
let db = app.database()
let bdRef = db.ref()
export default {
name: 'hello',
firebase: {
businesses: bdRef.orderByChild('.key').equalTo('306')
}
}
I get this error when doing this:
validation.js?5c80:234 Uncaught Error: Query.orderByChild failed: First argument was an invalid path = ".key". Paths must be non-empty strings and can't contain ".", "#", "$", "[", or "]"`
When I do this with my code:
businesses: bdRef.orderByChild('title').equalTo('Feather Animation Wood Carving Supplies')
It comes with this array:
0:Object
.key:"3021"
address:"Hello Avenue"
city:""
description:"Wood carving tools and supplies. Please contact us by phone or internet."
email:"hi#gmail.com"
employees:"1"
How do I get the .key property?

Answer 1

Did you try use this command:
businesses['.key']

Answer 2

Its very simple answer that you have json with ".key" key and "3021" as its value. But in ".key" you have included "." which refers that you are giving empty path or maybe invalid path.
So if you name it as "key" or any name as "keyid" would be cool, unless you include as your error says can't contain ".", "#", "$", "[", or "]"
Hope this explanation helped!

Answer 3

As stated, you cannot query a property with a dot in its name. From the documentation, you need to use the built in orderByKey() filter instead:
export default {
name: 'hello',
firebase: {
businesses: bdRef.equalTo('306').orderByKey()
}
}