Neo4jRepository是繼承了Spring data的CrudRepository的，所以是在CrudRepository的基礎上針對Neo4j做了擴展的。
初步使用的實驗是，你能想到的CRUD的方法，基本不用你自己寫Cypher（也就是Neo4j的SQL）。
比如要模糊查詢:
MATCH (a:Artifact) WHERE a.artifactId CONTAINS 'red' return a
那麼，對應地你只要在你自己定義的Repository中繼承Neo4jRepository，就可以像堆積木一樣寫出你的get請求對應的方法名“findAllByArtifactIdContains”，那麼這個方法名就對應了上面的查詢語句，且不用我們自己顯式用@Query在接口方法上寫出查詢語句
//@Query("MATCH (a:Artifact) WHERE a.artifactId CONTAINS $artifactId RETURN a")
List<Artifact> getAllByArtifactIdContains(String artifactId);
findAll在Neo4jRepository裏面有多種重載
如果你在一個圖數據庫裏面使用不帶參數的findAll()方法，那麼就會報下面的錯誤。大概就是因爲你Node之間是彼此關聯的，findAll()方法執行的Cypher如下：
MATCH (n:`Organization`) WITH n RETURN n,[ [ (n)<-[r_h1:`HAS_ORG`]-(a1:`Artifact`) | [ r_h1, a1 ] ] ]
這個Cypher在數據庫裏面會查到Organization和Artifact兩個節點和他們彼此間的關係。如下，
那麼，這邊不加上控制，就會通過a去找到g，在通過g去找到a，直接導致stackOverFlowError內存溢出了。
Swagger上面call這個findAll方法，直接返回這樣一個結果
表示，雖然Cypher正確執行也拿到了結果，大師數據過大，Swagger無力將其轉換成JSON，只能給你看個原數據。
顯示的數據如下，基本就是重複數據無線循環：
can't parse JSON. Raw result:
[{"id":5,"name":"redhat","url":"redhat-7.org","artifacts":[{"id":20,"groupId":"neo4j","artifactId":"neo4j","version":"1.2.3","availability":true,
"scope":null,"artifacts":[],"dependencies":[],"parent":null,"licenses":[],"organization":
{"id":5,"name":"redhat","url":"redhat-7.org","artifacts":
[{"id":20,"groupId":"neo4j","artifactId":"neo4j","version":"1.2.3","availability":true,
"scope":null,"artifacts":[],"dependencies":[],"parent":null,"licenses":[],"organization":
{"id":5,"name":"redhat","url":"redhat-7.org","artifacts":
[{"id":20,"groupId":"neo4j","artifactId":"neo4j","version":"1.2.3","availability":true,
"scope":null,"artifacts":[],"dependencies":[],"parent":null,"licenses":[],"organization":
{"id":5,"name":"redhat","url":"redhat-7.org","artifacts":
[{"id":20,"groupId":"neo4j","artifactId":"neo4j","version":"1.2.3","availability":true,
"scope":null,"artifacts":[],"dependencies":[],"parent":null,"licenses":[],"organization":
{"id":5,"name":"redhat","url":"redhat-7.org","artifacts":
...
解決方法：
其實就給這個findAll查詢加個深度就可以了
artifactRepository.findAll(0);