Java Array.asList 错误
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【中文标题】Java Array.asList 错误【英文标题】:Java Array.asList error 【发布时间】:2018-09-05 09:23:38 【问题描述】:我已经使用 asList() 创建了一个数组列表,但出现了很多错误
@Service
public class ArticleService
List<Article> articleList = new ArrayList<Article>(Arrays.asList(
new Article( id:"1", name: "Article 01", description: "Description 01" ),
new Article( id:"2", name: "Article 02", description: "Description 02" ),
new Article( id:"3", name: "Article 03", description: "Description 03" )
));
public List<Article> getAllArticles()
return articleList;
【问题讨论】:
很多错误意味着什么样的错误? 请发布任何编译器错误,并编写一个最小且完整的示例。 是的……那不是 Java。new Article( id:"1", name: "Article 01", description: "Description 01" )
???也许从学习语言的基本语法开始。
我正在尝试上传图片,但应用程序崩溃了
类似 假设您的文章类如下所示,如果不是,请进行适当的类,
class Article
String id;
String article;
String description;
Article(String id ,String articleName,String description)
this.id=id;
this.articleName = articleName;
this.description = description;
........
//your getter/setters are defined here.
现在你应该在你的主类中使用它们,如下所示,
List<Article> articleList = new ArrayList<Article>(Arrays.asList(
new Article( "1","Article 01", "Description 01" ),
new Article( "2","Article 02","Description 02" ),
new Article("3","Article 03","Description 03" )
));
//this is the correct way of using it.
【讨论】:
【参考方案2】:Arrays.asList 的正确使用方法:
List<Article> articleList = Arrays.asList(new Article(),new Article(),new Article());
此外,您没有正确地将参数传递给承包商,如果您有三个字符串,那么它应该如下所示:
new Article("arg0","arg1","arg1");
【讨论】:
【参考方案3】:// you can execute your business logic like below code
List<Article> articleList=null;
articleList = (List<Article>) new ArrayList();
Article art = new Article();
art.setId(0);
art.setName("Article 01");
art.setDescription("Description 01");
articleList.add(art);
【讨论】:
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