Revert "Use format instead of type property for Content deserialization"

Using format caused problems because Jackson does not support multiple
type names being mapped to a single class.

This reverts commit 96b0cf42.

src/main/java/de/thm/arsnova/entities/Content.java

 package de.thm.arsnova.entities;
 
-import com.fasterxml.jackson.annotation.JsonSubTypes;
 import com.fasterxml.jackson.annotation.JsonTypeInfo;
 import com.fasterxml.jackson.annotation.JsonView;
 import de.thm.arsnova.entities.serialization.View;
@@ -12,12 +11,8 @@ import java.util.Objects;
 @JsonTypeInfo(
 		use = JsonTypeInfo.Id.NAME,
 		include = JsonTypeInfo.As.EXISTING_PROPERTY,
-		property = "format",
-		visible = true
+		property = "type"
 )
-@JsonSubTypes({
-		@JsonSubTypes.Type(value = ChoiceQuestionContent.class, name = "ChoiceQuestionContent")
-})
 public class Content extends Entity {
 	public enum Format {
 		CHOICE,

src/main/java/de/thm/arsnova/entities/serialization/CouchDbDocumentModule.java

@@ -36,12 +36,8 @@ public class CouchDbDocumentModule extends SimpleModule {
 		context.setMixInAnnotations(Entity.class, CouchDbDocumentMixIn.class);
 		context.setMixInAnnotations(de.thm.arsnova.entities.migration.v2.Entity.class, CouchDbDocumentV2MixIn.class);
 		context.registerSubtypes(
-				new NamedType(Content.class, Content.Format.TEXT.toString()),
-				new NamedType(Content.class, Content.Format.NUMBER.toString()),
-				new NamedType(ChoiceQuestionContent.class, Content.Format.CHOICE.toString()),
-				new NamedType(ChoiceQuestionContent.class, Content.Format.BINARY.toString()),
-				new NamedType(ChoiceQuestionContent.class, Content.Format.SCALE.toString()),
-				new NamedType(ChoiceQuestionContent.class, Content.Format.GRID.toString()),
+				new NamedType(Content.class, Content.class.getSimpleName()),
+				new NamedType(ChoiceQuestionContent.class, ChoiceQuestionContent.class.getSimpleName()),
 				new NamedType(Answer.class, Answer.class.getSimpleName()),
 				new NamedType(ChoiceAnswer.class, ChoiceAnswer.class.getSimpleName()),
 				new NamedType(TextAnswer.class, TextAnswer.class.getSimpleName()));