2023-05-08 09:19 回答问题
可以参考下面的代码,但是需要自己确定迭代次数 t=table(1..7 as nodeID,`a1`a2`a3`a4`a5`a6`a7 as nodeName,0 1 1 2 2 3 3 as parentID)def getchildnode(t1,t){ return select * from t1 union select t.nodeID,t.nodeName,t1.pathid+"->"+string(t.nodeID) as pathid,t1.pathname+"-&g
2023-05-04 12:28 回答问题
问题复现不了,可能是浮点数精度的问题,请问可以提供一下您的脚本吗?
2023-05-04 11:09 回答问题
将price列乘3 t1=table(`A`A`B`B as symbol, 2021.04.15 2021.04.16 2021.04.15 2021.04.16 as date, 12 13 21 22 as price)col="price"t1[col]=t1[col]*3 //方法1t1[col]=expr(sqlCol(col),*,3) //方法2sqlUpdate("t1",sqlColAlias(expr(sqlCol(col),*,3),col)).eval() //方法3 元编
2023-05-04 11:06 回答问题
x = 1 2 1 2 1 2 1 2 1 2 1 2;y = 0 0 1 1 1 2 2 1 1 3 3 2;t = table(x,y);select *, iif(isDuplicated(segment(y,false),LAST),NULL,sum(x)) as z from t context by segment(y,false)
2023-05-04 11:04 回答问题
x = 1 2 1 2 1 2 1 2 1 2 ; y = 0 0 1 1 1 2 2 1 1 3 ; x-move(y,1)