如何跳过 document.querySelectorAll 值为 null 或不存在答案

【问题标题】：how to skip document.querySelectorAll value is null or does not exist如何跳过 document.querySelectorAll 值为 null 或不存在
【发布时间】：2022-01-20 19:52:40
【问题描述】：

<script>
(function(){
    var data = {
  "@context": "https://schema.org",
  "@type": "Movie",
  "actor": [],
for (i = 0; i < document.querySelectorAll('span.movie-cast-title').length; i++) 
{
  if (!document.querySelectorAll('span.gcharacter')[i]) {
        data.actor.push({
    "@type": "PerformanceRole",
    "actor": {
      "@type": "Person",
      "name": document.querySelectorAll('span.movie-cast-title')[i].innerText,
      "url": document.querySelectorAll('a.movie-cast-url')[i].href,

      
    },
    "characterName": document.querySelectorAll('span.gcharacter')[i].innerText,
          
        });
  }
  else
  {
  data.actor.push({
    "@type": "PerformanceRole",
    "actor": {
      "@type": "Person",
      "name": document.querySelectorAll('span.movie-cast-title')[i].innerText,
      "url": document.querySelectorAll('a.movie-cast-url')[i].href,  
    },
     });
        
  }  
  };
var script = document.createElement('script');
  script.type = "application/ld+json";
  script.innerHTML = JSON.stringify(data);
  document.getElementsByTagName('head')[0].appendChild(script);
})(document);
</script>

如何跳过“characterName”：document.querySelectorAll('span.gcharacter')[i].innerText，值为空。 for is const characterName is does not exist 仅跳过 characterName var。 if else 语句是正确的还是使用其他 any 语句？

【问题讨论】：

标签： if-statement schema google-tag-manager

【解决方案1】：

要走的路是创建一个对象来存储始终存在的属性。然后，如果 characterName 不为空，则将此属性添加到对象并将其推送到 data.actor 数组：

var actor = {
    "@type": "PerformanceRole",
    "actor": {
      "@type": "Person",
      "name": document.querySelectorAll('span.movie-cast-title')[i].innerText,
      "url": document.querySelectorAll('a.movie-cast-url')[i].href      
    }
}

if (!document.querySelectorAll('span.gcharacter')[i])
    actor["characterName"] = document.querySelectorAll('span.gcharacter')[i].innerText

data.actor.push(actor)

【讨论】：