Spring MongoTemplate插入成功或失败

Question

如何检查使用MongoTemplate插入到集合中的值是否成功保存？

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

MongoTemplate中的Insert、save或任何方法不返回true或false。

Answer 1

它将在下面的方法流中抛出异常：

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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以下是insertDBObject的源代码

    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");
        }
    });
}

它会在你写失败的时候抛出异常，它们是运行时异常。运行时执行由MongoExceptionTranslator .So翻译，任何没有抛出任何异常的插入都是成功的。