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
IT程序员