使用猫鼬从数组中删除单个项目

Question

I am new to web development and am making a todo app. I have the following schema and model:

const tdSchema = new mongoose.Schema({
  category: {
    type: String,
    required: true,
    unique: true
  },
  tds: {
    type: [{
      type: String
    }]
  }
});
const ToDo = mongoose.model("ToDo", tdSchema);

Here is an example of a document:

{ "_id" : ObjectId("5f7e2c3d1a151704382ce109"), 
"todos" : [ "Update Resume", "Apply for jobs" ], 
"category" : "Career", "__v" : 0 }

I would like to delete "Update Resume" from the array "todos" while leaving everything else in the document as is.

How do I do this using Mongoose JS?

Answer 1

First find the document:

const doc = await ToDo.findOne({ _id: "5f7e2c3d1a151704382ce109" });
console.log(doc);  // optional (used to illustrate how code works)

This will log:

{ "_id" : ObjectId("5f7e2c3d1a151704382ce109"), 
"todos" : [ "Update Resume", "Apply for jobs" ], 
"category" : "Career", "__v" : 0 }

From here just access "todos" and remove from it just as you would remove from a typical JavaScript array.

doc.todos = doc.todos.filter(e => e !== "Update Resume");

Then save your document:

const updated = await doc.save();
console.log(updated);  // optional (used to illustrate how code works)

And you'll get this in your log:

{ "_id" : ObjectId("5f7e2c3d1a151704382ce109"), 
"todos" : [ "Apply for jobs" ], 
"category" : "Career", "__v" : 0 }

Answer 2

To build on @Bilal Saleem's answer, if you want to only run one query and ensure it works if multiple people make requests at the same time, you can do this in an update query using the $pull operator.

The $pull operator is used to remove items from a mongodb array, and supports query conditions for the items you wish to pull.

const doc = await ToDo.updateOne(
  { _id: "5f7e2c3d1a151704382ce109" },
  {
    $pull: {
      todos: "Update resume",
    }
  }
);

In the above case, this will $pull from the todos array, and remove any items equal to "Update resume"