Spring MongoTemplate插入成功或失敗

Question

如何檢查使用MongoTemplate插入到集合中的值是否已成功保存？

MongoTemplate template; 
User user=new User(); 
user.name="Mark"; 
user.email="mark@mark.com" 
template.insert(user); 

插入，保存或MongoTemplate中的任何方法不會返回true或false。

Answer 1

它會拋出此異常是方法的流程：

template.insert(object) 
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templateinsert(Object objectToSave, String collectionName) 
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protected <T> void doInsert(String collectionName, T objectToSave, MongoWriter<T> writer) 
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protected Object insertDBObject(final String collectionName, final DBObject dbDoc, final Class<?> entityClass) 
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這裏是insertDBObject

protected Object insertDBObject(final String collectionName, final DBObject dbDoc, final Class<?> entityClass) { 
    if(LOGGER.isDebugEnabled()) { 
     LOGGER.debug("Inserting DBObject containing fields: {} in collection: {}", dbDoc.keySet(), collectionName); 
    } 

    return this.execute(collectionName, new CollectionCallback() { 
     public Object doInCollection(DBCollection collection) throws MongoException, DataAccessException { 
      MongoAction mongoAction = new MongoAction(MongoTemplate.this.writeConcern, MongoActionOperation.INSERT, collectionName, entityClass, dbDoc, (DBObject)null); 
      WriteConcern writeConcernToUse = MongoTemplate.this.prepareWriteConcern(mongoAction); 
      WriteResult writeResult = writeConcernToUse == null?collection.insert(new DBObject[]{dbDoc}):collection.insert(dbDoc, writeConcernToUse); 
      MongoTemplate.this.handleAnyWriteResultErrors(writeResult, dbDoc, MongoActionOperation.INSERT); 
      return dbDoc.get("_id"); 
     } 
    }); 
} 

的源代碼，這將拋出當你寫失敗例外，他們是運行時異常。運行時問題由MongoExceptionTranslator 轉換。因此，任何不會拋出任何異常的插入都會成功。