试图从一类更新/更新插入到另一个

Question

可以用来创建方案

CREATE CLASS v1 EXTENDS v 
CREATE CLASS v2 EXTENDS v 
CREATE PROPERTY v1.p1 STRING 
CREATE PROPERTY v1.p2 STRING 
CREATE PROPERTY v1.p3 STRING 
CREATE PROPERTY v2.p1 STRING (MANDATORY TRUE) 
CREATE PROPERTY v2.p2 STRING (MANDATORY TRUE) 
CREATE PROPERTY v2.p3 STRING 
CREATE INDEX v2.p1 UNIQUE 
INSERT INTO v1 (p1,p2,p3) VALUES ('a','b','c') 
INSERT INTO v1 (p1,p2,p3) VALUES ('d','e','f') 
INSERT INTO v1 (p1,p2,p3) VALUES ('h','i','j') 
INSERT INTO v2 (p1,p2,p3) VALUES ('a','y','z') 

和这些是以下中选择的查询的结果从

选择下列V1

{ 
    "result": [ 
     { 
      "@type": "d", 
      "@rid": "#33:0", 
      "@version": 1, 
      "@class": "v1", 
      "p1": "a", 
      "p2": "b", 
      "p3": "c" 
     }, 
     { 
      "@type": "d", 
      "@rid": "#34:0", 
      "@version": 1, 
      "@class": "v1", 
      "p1": "d", 
      "p2": "e", 
      "p3": "f" 
     }, 
     { 
      "@type": "d", 
      "@rid": "#35:0", 
      "@version": 1, 
      "@class": "v1", 
      "p1": "h", 
      "p2": "i", 
      "p3": "j" 
     } 
    ], 
    "notification": "Query executed in 0.011 sec. Returned 3 record(s)" 
} 

选择V2

{ 
    "result": [ 
     { 
      "@type": "d", 
      "@rid": "#41:0", 
      "@version": 1, 
      "@class": "v2", 
      "p1": "a", 
      "p2": "y", 
      "p3": "z" 
     } 
    ], 
    "notification": "Query executed in 0.013 sec. Returned 1 record(s)" 
} 

我已经尝试了多个sql语句，尝试将v1和v2的字段合并到近两天，但无法找到答案。你可以告诉我要去的地方错了

LET $t = SELECT FROM v1 
UPDATE v2 SET (p1,p2) VALUES ($t.$current.p1, $t.$current.p2) UPSERT WHERE $current.p1 = $t.$current.p1 

期望的结果将是V2有值“A，B，Z”和已添加了额外的两个记录，因为他们在V1

更新当前记录

Answer 1

这是一个JavaScript功能来做到这一点。

初步方案：

JS功能：

var g = orient.getGraph(); 
var v1 = g.command("sql","SELECT FROM v1"); 

for(i=0; i<v1.length; i++){ 
    var check_index = g.command("sql","select from index:v2.p1 where key='"+v1[i].getProperty("p1")+"'"); 
    if(check_index.length>0){ 
    //v1 record already present in v2 index - UPDATE 
    print("#### "+"update "+check_index[0].getProperty("rid").getId()+" set p2='"+v1[i].getProperty("p2")+"'"+" ####"); 
    g.command("sql","update "+check_index[0].getProperty("rid").getId()+" set p2='"+v1[i].getProperty("p2")+"'"); 
    } else { 
    //v1 record not present in v2 index - INSERT 
    print("#### "+"insert into v2 set p1='"+v1[i].getProperty("p1")+"',p2='"+v1[i].getProperty("p2")+"',p3='"+v1[i].getProperty("p3")+"'"+" ####"); 
    g.command("sql","insert into v2 set p1='"+v1[i].getProperty("p1")+"',p2='"+v1[i].getProperty("p2")+"',p3='"+v1[i].getProperty("p3")+"'"); 
    } 
    g.commit(); 
} 

结果：