Delete single item in Firebase with AngularJS

Question

I am struggling a bit understanding how to remove a single item from an object in Firebase with AngularJS.

This is my Firebase (only the interesting bit):

projects
 |
 `--- $id
 |     `-- Name
 |     `-- CreatorID
 |
user_project
 |
 `--- CreatorID
       `-- projectid

Every time a user creates an entry the same is added to the entries with a Creator ID equal to the user ID that I get from Firebase Simple Login. At the same time, my code creates a new item inside user_entries/CreatorID (so that I have a list of entries associated with every user).

Example:

projects
 |
 `--- JlagYdBNX1gyMVCoXBF
 |     `-- Name: "Activity 1"
 |     `-- CreatorID: "simplelogin:29"
 |
user_projects
 |
 `--- simplelogin:29
       `--JleI106xJgZf6_mGHUI: "JlagYdBNX1gyMVCoXBF"

Now my problem is that I can delete an entry, but I don't know how to track back and delete the same from user_entries (so that I am getting many orphans there).

I am using a factory:

app.factory('Project', function ($firebase, FIREBASE_URL) {
  var ref = new Firebase(FIREBASE_URL);
  var projects = $firebase(ref.child('projects')).$asArray();

  var Project = {
    all: projects,
    create: function (project) {
      return projects.$add(project).then(function(projectRef) {
        $firebase(ref.child('user_projects').child(project.creatorUID))
        .$push(projectRef.name());
        return projectRef;
      });
    },
    delete: function (project) {
      var userid = project.creatorUID;
      return projects.$remove(project).then(function(projectRef) {
        //$firebase(ref.child('user_projects').child(userid))
        console.log(project.$id);
      });
    },
  };

  return Project;

});

I know that $remove has a callback returning the $id of the processed item so I tried to change delete to:

delete: function (project) {
  var userid = project.creatorUID;
  return projects.$remove(project).then(function(projectRef) {
    $firebase(ref.child('user_projects').child(userid))
    .$remove(projectRef.$id);
  });
},

but I get

Error: Firebase.child failed: First argument was an invalid path: "undefined". Paths must be non-empty strings and can't contain ".", "#", "$", "[", or "]"

Any hint on how to approach this problem?

I'm assuming you already tried .$remove(project.$id)? Apparently from the docs the object passed to the call back is a regular Firebase object, not an Angular firebase object. I don't know why they decided to do it that way. — Austin Mullins
– Austin Mullins, Commented Apr 3, 2015 at 19:04
And of course, following the docs, you should be able to use .$remove(projectRef.key()). — Austin Mullins
– Austin Mullins, Commented Apr 3, 2015 at 19:14
Yes I tried with project.$id, no luck. But I guess it's because project.$id is the value, not the key in user_projects. — Mirko G.
– Mirko G., Commented Apr 4, 2015 at 7:02

Austin Mullins · Accepted Answer · 2015-04-03 19:21:41Z

1

The AngularFire documentation indicates that the object passed to the promise callback is a regular Firebase object, not an AngularFire reference. This means that you need to use .key() rather than .$id:

delete: function (project) {
  var userid = project.creatorUID;
  return projects.$remove(project).then(function(projectRef) {
    $firebase(ref.child('user_projects').child(userid))
    .$remove(projectRef.ref());
  });
},

answered Apr 3, 2015 at 19:21

Austin Mullins

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1 Comment

Mirko G. Over a year ago

Oh, I understand now... I tried that before and I got this:

Error: Firebase.child failed: First argument was an invalid path: "https://my_firebase.firebaseio.com/projects/-Jm24tOoN23k5PFhoNGB". Paths must be non-empty strings and can't contain ".", "#", "$", "[", or "]"

Mirko G. · Accepted Answer · 2015-04-04 08:32:02Z

It may be not the cleanest solution but I found a way to achieve what I needed saving the unique ID of the user_projects entry id into the project itself:

 create: function (project) {
      return projects.$add(project).then(function(projectRef) {
        var projectId = projectRef.name();
        $firebase(ref.child('user_projects').child(project.creatorUID))
        .$push(projectId).then(function(userProjectRef) {
          var userProjectId = userProjectRef.name();
          $firebase(ref.child('projects').child(projectId).child('LinkID')).$set(userProjectId);
        });
        return projectRef;
      });
    },

So that in my firebase I get:

projects
 |
 `--- JlagYdBNX1gyMVCoXBF
 |     `-- Name: "Activity 1"
 |     `-- CreatorID: "simplelogin:29"
 |     `-- LinkID: "JleI106xJgZf6_mGHUI"
 |
user_projects
 |
 `--- simplelogin:29
       `--JleI106xJgZf6_mGHUI: "JlagYdBNX1gyMVCoXBF"

And then it's much easier to delete the entry:

delete: function (project) {
  var userId = project.creatorUID;
  var userProjectId = project.LinkID;
  return projects.$remove(project).then(function(projectRef) {
 $firebase(ref.child('user_projects').child(userId)).$remove(userProjectId);
  });
},

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