EDIT:见底部上的较小的STAT /模式集“值不是记录或记录ID”时传递批次变量设置为“设定=”在更新
I”重现该问题在OrientDB时间序列用例的实验中,我试图链接两棵树的每个级别。所以,我没有这个SQL批处理:
let $w1 = select expand(W1[2406]) from #12:2;
let $w2 = select expand(W1[2407]) from #12:2;
let $d1 = select expand(D1[4]) from $w1;
let $d2 = select expand(D1[0]) from $w2;
let $h1 = select expand(H1[23]) from $d1;
let $h2 = select expand(H1[0]) from $d2;
let $m1 = select expand(M1[59]) from $h1;
let $m2 = select expand(M1[0]) from $h2;
update $w1 set next = $w2;
update $w2 set previous = $w1;
update $d1 set next = $d2;
update $d2 set previous = $d1;
update $h1 set next = $h2;
update $h2 set previous = $h1;
update $m1 set next = $m2;
update $m2 set previous = $m1;
但我得到的第一个更新此错误:
The field 'W1.next' has been declared as LINK but the value is not a record or a record-id
...我不明白,因为当我尝试:
return [$w1, $w2, $d1, $d2, $h1, $h2, $m1, $m2];
我得到了预期的记录...
所以有两个问题:
- 为什么会出现此错误?
- 有没有更好的方法来做到这一点,因为我还是一个初学者?
(注:我想留在SQL/SQL批处理)
类是这样的:
Symbol{
W1 : LINKMAP W1
}
W1 {
next : LINK W1
previous : LINK W1
D1 : LINKMAP D1
}
D1 {
next : LINK D2
previous : LINK D2
H1 : LINKMAP H1
}
H1 [...]
M1 [...]
编辑:如何重现上一个较小的数据集的问题:
创建模式:
create class A extends V;
create class B extends V;
create property A.B LINKMAP B;
create property B.B LINK B;
注:AB是状阵列乙元件的,BB是1to1链路
虚设插入值:
insert into B CONTENT {}
insert into B CONTENT {}
现在,得到两个假人的RID插入到甲
select from B
现在插入如下的两个项目(#13:0
和#13:1
在我的情况)
insert into A(B) values ({'0' : #13:0, '1' : #13:1})
insert into A(B) values ({'0' : #13:0, '1' : #13:1})
最后,尝试此批:
let $b1 = select expand(B[0]) from A;
let $b2 = select expand(B[1]) from A;
update $b1 SET B = $b2;
update $b2 SET B = $b1;
看到错误
字段“B.B”已被宣布为链接,但该值不是一个记录或记录ID
,但如果你这样做:
let $b1 = select expand(B[0]) from A;
let $b2 = select expand(B[1]) from A;
return $b1;
和
let $b1 = select expand(B[0]) from A;
let $b2 = select expand(B[1]) from A;
return $b2;
和
let $b1 = select expand(B[0]) from A;
let $b2 = select expand(B[1]) from A;
return {'b1': $b1, 'b2' : $b2};
你可以看到它们不是收藏品,但是真实记录
EDIT2由Isavio提供的解决方案的工作,但我想知道为什么它在以前的结果,因为他们似乎并不收集?
let $b1 = select expand(B[0]) from A;
let $b2 = select expand(B[1]) from A;
update $b1 SET B = $b2[0];
update $b2 SET B = $b1[0];
嗨,你有一个简单的数据库测试分享? – LucaS
呃...是的,我想我可以给你一些东西 –