javascript - 如果 active = true 则在 javascript 中将其涂成红色

Question

我需要检查 active = true 是否将文本涂成红色，但我做不到。

var products = [
	{name: 'Коля', active: true},
	{name: 'Вася', active: true},
	{name: 'Петя', active: false},
];

var ul = document.createElement("ul");

for( var i = 0; i < products.length; i++ ){
    
    var li = document.createElement("li");
    li.innerHTML =  products[i].name + " " + products[i].active;
    if( products[i].active == true ){
    	products[i].name.style.color = "red";
    } 
    ul.appendChild(li);
    document.body.appendChild(ul);
}

Answer 1

使用li代替products元素

li.style.color = "red";

var products = [{
    name: 'Коля',
    active: true
  },
  {
    name: 'Вася',
    active: true
  },
  {
    name: 'Петя',
    active: false
  },
];

var ul = document.createElement("ul");

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

  var li = document.createElement("li");
  li.innerHTML = products[i].name + " " + products[i].active;
  if (products[i].active == true) {
    //products[i].name.style.color = "red";
    li.style.color = "red";
  }
  ul.appendChild(li);
  document.body.appendChild(ul);
}