Java基础之JSON转成List结构数据

2022-05-07 17:30:25 浏览数 (1)

先要引入对应的jar,然后调用net.sf.json库的

代码语言:javascript复制
    ObjectMapper mapper = new ObjectMapper();
		JavaType javaType = mapper.getTypeFactory().constructParametricType(List.class,FormModel.class);
		/*List<FormModel> writUnionFormList = 
				JSON.parseArray(params, FormModel.class);*/
		List<FormModel> writUnionFormList=new ArrayList<FormModel>();
		try {
			writUnionFormList = (List<FormModel>)mapper.readValue(params, javaType);
		} catch (JsonParseException e) {
			e.printStackTrace();
		} catch (JsonMappingException e) {
			e.printStackTrace();
		} catch (IOException e) {
			e.printStackTrace();
		}

fastjson的简单用法,fastjson转换相对语法简单点,不过如果出现Bean类过大,或者在ie模式有时候会出现一些报错

代码语言:javascript复制
List<FormModel> writUnionFormList = 
				JSON.parseArray(params, FormModel.class);

附录,如果在ie出现中文乱码问题,可以参考我之前教程:https://cloud.tencent.com/developer/article/1995392

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