如何跳过 document.querySelectorAll 值为 null 或不存在

Posted 2023-02-19

【中文标题】如何跳过 document.querySelectorAll 值为 null 或不存在

【英文标题】：how to skip document.querySelectorAll value is null or does not exist

【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？

【参考方案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)