排序與串

Question

的境界文檔中的長度域，它被寫入

result = result.sort("name", Sort.DESCENDING); 

簡單的排序，但我想首先

通常在MySQL，我們有

order by CHAR_LENGTH(Field) 

最短名稱排序

在這方面有沒有像這樣的方法？

Answer 1

沒有，但你可以很容易地重新定義的制定者，因爲0.88.0，你做到以下幾點：

public class Word extends RealmObject { 
    private String name; 

    @Index 
    @Required 
    private long nameLength; 

    public String getName() { 
     return name; 
    } 

    public void setName(String name) { 
     this.name = name; 
     this.nameLength = name == null ? 0 : name.length(); 
    } 
} 

而且你可以指定將創建此預先存在的元素

public class Migration implements RealmMigration { 
    @Override 
    public void migrate(final DynamicRealm realm, long oldVersion, long newVersion) { 
     RealmSchema schema = realm.getSchema(); 
     if(oldVersion == 0) { 
      RealmObjectSchema wordSchema = schema.get("Word"); 
      wordSchema.addField("nameLength", long.class, FieldAttribute.INDEXED, FieldAttribute.REQUIRED) 
         .transform(new RealmObjectSchema.Function() { 
          @Override 
          public void apply(DynamicRealmObject obj) { 
           obj.set("nameLength", obj.getString("name").length()); 
          } 
         }); 
      oldVersion++; 
     } 
    } 

    @Override 
    public void equals(Object object) { 
     return object != null && object instanceof Migration; 
    } 

    @Override 
    public int hashCode() { 
     return Migration.class.hashCode(); 
    } 
} 

RealmConfiguration realmConfiguration = new RealmConfiguration.Builder(context) 
              .schemaVersion(1) 
              .migration(new Migration()) 
              .build(); 

遷移

然後，你可以很容易地做到

realm.where(Word.class).findAllSorted(WordFields.NAME_LENGTH, Sort.ASCENDING, 
             WordFields.NAME, Sort.ASCENDING);