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JSON转成List结构数据

先要引入对应的jar,然后调用net.sf.json库的


    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

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://smilenicky.blog.csdn.net/article/details/100145430

posted @ 2019-08-31 10:59  smileNicky  阅读(13290)  评论(0编辑  收藏  举报