如何按文件ID删除行

Question

I am able to add data no problem, but i cannot seem to figure out how to use remove properly...

I want the ability to remove records in the DB by each documents unique ID that gets autogenerated. ex: /lab-results/292AU3VhfMXOzieHtmiD

What am i doing wrong here? I keep getting "db.ref is not a function" when i try to issue the remove.

Please see code for my attempts at getting this to work.

HTML CODE

<script src="https://www.gstatic.com/firebasejs/6.1.1/firebase-app.js"></script> 
<!-- The core Firebase JS SDK is always required and must be listed first -->
<script src="https://www.gstatic.com/firebasejs/6.1.1/firebase-firestore.js"></script>
<script src="https://www.gstatic.com/firebasejs/6.1.1/firebase-auth.js"></script>
<!-- TODO: Add SDKs for Firebase products that you want to use
     https://firebase.google.com/docs/web/setup#config-web-app -->


<script>
        // Your web app's Firebase configuration
        var firebaseConfig = {
            apiKey: "",
            authDomain: ".firebaseapp.com",
            databaseURL: "https://.firebaseio.com",
            projectId: "",
            storageBucket: ".appspot.com",
            messagingSenderId: "",
            appId: ""
        };
        // Initialize Firebase
        firebase.initializeApp(firebaseConfig);
        const db = firebase.firestore();
        const auth = firebase.auth();

</script>

JS FILE CODE


//Add to firebase db THIS WORKS FINE

var el = document.getElementById('test');
if(el){
labTestForm.addEventListener('submit', (e) =>{
    e.preventDefault();
    db.collection('tests').add({
        labTestID: labTestForm.labTestID.value,
        labTestCode: labTestForm.labTestCode.value,
        labTestName: labTestForm.labTestName.value,

    })
    //Reset for fields after submit
    labTestForm.reset();


})

}

///////////////////
//TEST JS - THROWS ERROR IN CONSOLE "db.ref is not a function"


var labResultsFB = db.ref("lab-results/0nH95TRY5xwnPJovwEFd")
labResultsFB.remove();

console.log('remove fired for:' + key)


/////////

Answer 1

var labResultsFB = db.doc("lab-results/0nH95TRY5xwnPJovwEFd")
labResultsFB.delete();