Question

我正在创建待办事项列表应用程序。我创建了 newNode 并将其附加到父节点。我为 newNode 提供了一个删除按钮，其中包含一个从DOM中删除 newNode 的事件侦听器。有时当我在应用程序运行时点击删除按钮时，待办事项项目节点将被删除（按照需要），其他时候什么也不会发生，它会给出 NodeFoundError：节点是找不到。即使我执行console.log（newNode），它也会打印出节点信息，但它会产生NodeFoundError？任何想法为什么？错误通常在我删除1来执行项目后发生，当我尝试删除其他项目时，会出错。

for(var i = 0; i < serverArray.length ; i++){

    //Parent node I append newNode to           
    var node = document.getElementById("toDoList");            

    var todo2 = new todo(serverArray[i].name, serverArray[i].deadline, serverArray[i].rating, serverArray[i].id, serverArray[i].done);
    myArray.push(todo2);

    //create a new node with values of current to do item being added
    var newNode = document.createElement("p");
    newNode.setAttribute("id", todo2.id);
    newNode.setAttribute("class", "todosection");

    var descNode = document.createElement("p");
    descNode.setAttribute("id","descr");
    descNode.innerHTML = todo2.name;

    var deadlineNode = document.createElement("p");
    deadlineNode.setAttribute("id","dead");
    deadlineNode.innerHTML = todo2.deadline;

    var priorityNode = document.createElement("p");
    priorityNode.setAttribute("id","prio");
    priorityNode.innerHTML = todo2.rating;

    //My newNode gets three nodes added to it as paragraphs for styling purposes
    newNode.appendChild(descNode);
    newNode.appendChild(deadlineNode);
    newNode.appendChild(priorityNode);


    //Create **DELETE** button with appopriate type and attributes
    var removeToDo = document.createElement('input');
    removeToDo.setAttribute('type', 'button');
    removeToDo.setAttribute("value", "Delete");
    removeToDo.setAttribute("id", id);
    removeToDo.addEventListener('click', function(event) {
        newNode.parentNode.removeChild(newNode);

        for(var i = myArray.length - 1; i >= 0; i--) {
            if(myArray[i].id == removeToDo.id) {
                myArray.splice(i, 1);
            }
        }

    }, false);//Add an event listener which when clicked deletes to do node 
    newNode.appendChild(removeToDo);//appends delete button to newNode  

    node.appendChild(newNode);
}

找不到节点错误javascript

0 个答案: