JavaScript和removeChild( )魔术

Question

为什么这是可行的？如果不指定哪一项，它如何知道要删除哪一项？JavaScript魔法？

let addButtton = document.getElementById('addButton');
let body = document.querySelector("body");


addButton.onclick = function addToList() {
    let taskName = document.getElementById("AddToList").value;
    let toDoList = document.createElement("input");
    toDoList.setAttribute("type", "checkbox");
    toDoList.id = "task";
    body.appendChild(toDoList);
    let deleteButton = document.createElement("button");
    deleteButton.innerHTML = "Delete";
    let label = document.createElement("label");
    label.setAttribute("for", ("task"));
    label.innerHTML = taskName;
    body.appendChild(label);
    body.appendChild(deleteButton);

    deleteButton.onclick = function deleteTask() {
        body.removeChild(label);
        body.removeChild(toDoList);
        body.removeChild(buttonDel);
        body.removeChild(breakLn);
    }
}

Answer 1

当您在label变量中创建项let label = document.createElement("label");时，它会将引用保留在其中。当您将body.removeChild(label);作为参数传递时，将其从DOM中完全删除。