Question

fastjson 过滤多余的字段！

List<实体类> list = oper.ListAll();实体类中有3个字段 有一个字段是不需要的，请问怎么用fastjson过滤？

Answer 1

SimplePropertyPreFilter filter = new SimplePropertyPreFilter(实体类.class, "字段名","字段名");
JSON.toJSONString(list, filter);

Answer 2

@JSONField(serialize = false)

Answer 3

/**
* 过滤
*
* @param keys 需要过滤的属性
* @return
*/
public static SerializeFilter filter(String... keys) {
List<String> arr = (keys != null && keys.length > 0) ? Arrays.asList(keys) : null;
PropertyFilter propertyFilter = new PropertyFilter() {
@Override
public boolean apply(Object o, String s, Object o1) {
if (arr != null && arr.contains(s)) {
return false;
}
return true;
}
};
return propertyFilter;
}