英文:
Convert hashmap to json while filtering some keys?
问题
我试图将哈希映射转换为 JSON，同时过滤一些键。我已经尝试了 Jackson，但结果始终包含所有键。
Map<String, String> hashtable = new HashMap<>();
hashtable.put("KEY1", "VALUE1");
hashtable.put("KEY2", "VALUE2");
hashtable.put("KEY3", "VALUE3");
ObjectMapper objectMapper = new ObjectMapper();
FilterProvider filter = new SimpleFilterProvider().addFilter("filter", SimpleBeanPropertyFilter.filterOutAllExcept("KEY1"));
objectMapper.setFilterProvider(filter);
System.err.println(objectMapper.writeValueAsString(hashtable));
我的键是动态的，有没有更有效的方法来做到这一点？
Jackson 过滤的方法对我来说很完美，因为我需要为同一个 HashMap 生成多个具有不同过滤器的 JSON。是否可能使用 Jackson 来实现这一点？
英文:
I am trying to convert hashmap to json while filtering some keys. I have tried with Jackson but the result always have all the keys in it.
Map<String, String> hashtable = new HashMap();
hashtable.put("KEY1", "VALUE1");
hashtable.put("KEY2", "VALUE2");
hashtable.put("KEY3", "VALUE3");
ObjectMapper objectMapper = new ObjectMapper();
FilterProvider filter = new SimpleFilterProvider().addFilter("filter", SimpleBeanPropertyFilter.filterOutAllExcept("KEY1"));
objectMapper.setFilterProvider(filter);
System.err.println(objectMapper.writeValueAsString(hashtable));
My keys are dynamic, Is there more efficent way to do this?
Jackson method of filtering is perfect to me, since i need to generate multiple json's with different filter for the same HashMap. Is is possible to make this work with Jackson?
答案1
得分: 1
你可以设置一个包括一个 SimpleBeanPropertyFilter 的 FilterProvider。
注意： 在这个示例中，validKeys 可以作为参数传递给 createFilterProvider 并存储在 FilterProvider 上，而不是在 KeyFilter 驱动程序上。
英文:
You can set up a FilterProvider that includes a SimpleBeanPropertyFilter.
Note: In this example, the validKeys could be passed as an argument to createFilterProvider and stored on the FilterProvider, rather than on the KeyFilter driver.
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.introspect.Annotated;
import com.fasterxml.jackson.databind.introspect.JacksonAnnotationIntrospector;
import com.fasterxml.jackson.databind.ser.BeanPropertyWriter;
import com.fasterxml.jackson.databind.ser.FilterProvider;
import com.fasterxml.jackson.databind.ser.PropertyWriter;
import com.fasterxml.jackson.databind.ser.impl.SimpleBeanPropertyFilter;
import com.fasterxml.jackson.databind.ser.impl.SimpleFilterProvider;
public class KeyFilter {
private static ObjectMapper mapper;
private static Set<String> validKeys;
static {
mapper = new ObjectMapper();
mapper.setFilterProvider(createFilterProvider());
validKeys = new HashSet<>();
validKeys.add("KEY1");
}
public static void main(String[] args) throws JsonProcessingException {
Map<String, String> hashtable = new HashMap<>();
hashtable.put("KEY1", "VALUE1");
hashtable.put("KEY2", "VALUE2");
hashtable.put("KEY3", "VALUE3");
System.out.println(mapper.writeValueAsString(hashtable)); // {"KEY1":"VALUE1"}
}
// Derived from: https://stackoverflow.com/a/25232485/1762224
private static FilterProvider createFilterProvider() {
SimpleFilterProvider filterProvider = new SimpleFilterProvider();
final String filterName = "valid-key-filter";
mapper.setAnnotationIntrospector(new JacksonAnnotationIntrospector() {
@Override
public Object findFilterId(Annotated a) {
if (Map.class.isAssignableFrom(a.getRawType())) {
return filterName;
}
return super.findFilterId(a);
}
});
filterProvider.addFilter(filterName, new SimpleBeanPropertyFilter() {
@Override
protected boolean include(BeanPropertyWriter writer) {
return true;
}
@Override
protected boolean include(PropertyWriter writer) {
return validKeys.contains(writer.getName());
}
});
return filterProvider;
}
}
答案2
得分: 1
Here is the translated content:
这里 是您问题的答案。您应该创建一个没有任何字段的 mixinSource 类并将其添加到 objectMapper。然后，您可以动态过滤您的映射。
@JsonFilter("filter")
public class DynamicMixIn {
}
Map<String, String> hashtable = new HashMap();
hashtable.put("KEY1", "VALUE1");
hashtable.put("KEY2", "VALUE2");
hashtable.put("KEY3", "VALUE3");
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.addMixIn(Object.class, DynamicMixIn.class);
FilterProvider filter = new SimpleFilterProvider().addFilter("filter", SimpleBeanPropertyFilter.serializeAllExcept("KEY1"));
objectMapper.setFilterProvider(filter);
String filtered = objectMapper.writeValueAsString(hashtable); // {"KEY2":"VALUE2","KEY3":"VALUE3"}
英文:
Here is the answer of your question. You should create mixinSource Class without any fields and add it to the objectMapper. Then you can filter out your map dynamically
@JsonFilter("filter")
public class DynamicMixIn {
}
Map<String, String> hashtable = new HashMap();
hashtable.put("KEY1", "VALUE1");
hashtable.put("KEY2", "VALUE2");
hashtable.put("KEY3", "VALUE3");
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.addMixIn(Object.class, DynamicMixIn.class);
FilterProvider filter = new SimpleFilterProvider().addFilter("filter", SimpleBeanPropertyFilter.serializeAllExcept("KEY1"));
objectMapper.setFilterProvider(filter);
String filtered = objectMapper.writeValueAsString(hashtable); // {"KEY2":"VALUE2","KEY3":"VALUE3"}
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