java neo4j檢查是否存在關係

Question

如果這是重複的，則排除，儘管目前爲止我沒有找到答案。

我有一個應用程序通過cypher語句針對REST-API創建節點和關係。我創建關係與下面的代碼：

public URI createRelationship(GraphNodeTypes sourceType, URI sourceNode, 
         GraphNodeTypes targetType, URI targetNode, 
    GraphRelationshipTypes relationshipType, String[] jsonAttributes) { 
URI relationShipLocation = null; 

String cypherArt = getNodeIdFromLocation(sourceNode)+"-[:"+relationshipType+"]->"+getNodeIdFromLocation(targetNode); 

logger.info("creating relationship ({}:{}) -[:{}]-> ({}:{})", 
           sourceType, 
           getNodeIdFromLocation(sourceNode), 
           relationshipType, 
           targetType, 
           getNodeIdFromLocation(targetNode)); 

try { 
    URI finalUrl = new URI(sourceNode.toString() + "/relationships"); 
    String cypherStatement = generateJsonRelationship(targetNode, 
                 relationshipType, 
                 jsonAttributes); 

    logger.trace("sending CREATE RELATIONSHIP cypher as {} to endpoint {}", cypherStatement, finalUrl); 
    WebResource resource = Client.create().resource(finalUrl); 

    ClientResponse response = resource 
      .accept(MediaType.APPLICATION_JSON) 
      .type(MediaType.APPLICATION_JSON) 
      .entity(cypherStatement) 
      .post(ClientResponse.class); 

    String responseEntity = response.getEntity(String.class).toString(); 
    int responseStatus = response.getStatus(); 

    logger.trace("POST to {} returned status code {}, returned data: {}", 
      finalUrl, responseStatus, 
      responseEntity); 

    // first check if the http code was ok 
    HttpStatusCodes httpStatusCodes = HttpStatusCodes.getHttpStatusCode(responseStatus); 
    if (!httpStatusCodes.isOk()){ 
     if (httpStatusCodes == HttpStatusCodes.FORBIDDEN){ 
      logger.error(HttpErrorMessages.getHttpErrorText(httpStatusCodes.getErrorCode())); 
     } else { 
      logger.error("Error {} sending data to {}: {} ", response.getStatus(), finalUrl, HttpErrorMessages.getHttpErrorText(httpStatusCodes.getErrorCode())); 
     } 
    } else { 
     JSONParser reponseParser = new JSONParser(); 
     Object responseObj = reponseParser.parse(responseEntity); 
     JSONObject jsonResponseObj = responseObj instanceof JSONObject ?(JSONObject) responseObj : null; 
     if(jsonResponseObj == null) 
      throw new ParseException(0, "returned json object is null"); 

     //logger.trace("returned response object is {}", jsonResponseObj.toString()); 
     try { 
      relationShipLocation = new URI((String)((JSONObject)((JSONArray)((JSONObject)((JSONArray)((JSONObject)((JSONArray)jsonResponseObj.get("results")).get(0)).get("data")).get(0)).get("rest")).get(0)).get("self")); 
     } catch (Exception e) { 
      logger.warn("CREATE RELATIONSHIP statement did not return a self object, returning null -- error was {}", e.getMessage()); 
      relationShipLocation = null; 
     } 
    } 
} catch (Exception e) { 
    logger.error("could not create relationship "); 
} 
return relationShipLocation; 
} 

private static String generateJsonRelationship(URI endNode, 
    GraphRelationshipTypes relationshipType, String[] jsonAttributes) { 
StringBuilder sb = new StringBuilder(); 
sb.append("{ \"to\" : \""); 
sb.append(endNode.toString()); 
sb.append("\", "); 

sb.append("\"type\" : \""); 
sb.append(relationshipType.toString()); 
if (jsonAttributes == null || jsonAttributes.length < 1){ 
    sb.append("\""); 
} else { 
    sb.append("\", \"data\" : "); 
    for (int i = 0; i < jsonAttributes.length; i++) { 
     sb.append(jsonAttributes[i]); 
     if (i < jsonAttributes.length - 1){ 
      // Miss off the final comma 
      sb.append(", "); 
     } 
    } 
} 

sb.append(" }"); 
return sb.toString(); 
} 

我的問題是，我想檢查是否給定類型的給定關係的兩個節點之間已經存在PRIOR創建它。

有人可以告訴我，如何查詢關係？

隨着節點我不喜歡這樣的搭配：

MATCH cypher {"statements": [ {"statement": "MATCH (p:SOCIALNETWORK {sn_id: 'TW'}) RETURN p", "resultDataContents":["REST"]} ] } 

對端點

http://localhost:7474/db/data/transaction/<NUMBER> 

我將如何構建，檢查是否有關係的聲明，節點6和5或任何之間說？

由於提前，

克里斯

Answer 1

你可能要考慮通過暗號這樣做，並使用MERGE/ON CREATE/ON MATCH關鍵字。

例如，你可以做這樣的事情：

create (a:Person {name: "Bob"})-[:knows]->(b:Person {name: "Susan"}); 

MATCH (a:Person {name: "Bob"}), (b:Person {name: "Susan"}) 
MERGE (a)-[r:knows]->(b) 
ON CREATE SET r.alreadyExisted=false 
ON MATCH SET r.alreadyExisted=true 
RETURN r.alreadyExisted; 

這MATCH/MERGE查詢，我在這裏提供將返回true或false，這取決於關係是否已經存在與否。

另外，FWIW它看起來像你用來通過StringBuilder累積JSON的代碼很可能會很笨重和容易出錯。有很多優秀的庫，如Google GSON可以爲你做JSON，所以你可以創建JSON對象，數組，基元等等 - 然後讓庫擔心將它正確序列化爲一個字符串。這往往會使你的代碼變得更清潔，更容易維護，當你把你的JSON格式化（我們都這麼做）的東西搞砸時，比你累積字符串更容易找到。

Answer 2

在Java

Relationship getRelationshipBetween(Node n1, Node n2) { // RelationshipType type, Direction direction 
    for (Relationship rel : n1.getRelationships()) { // n1.getRelationships(type,direction) 
     if (rel.getOtherNode(n1).equals(n2)) return rel; 
    } 
    return null; 
}