如何替换 snowflake.execute 语句中的循环结果？

Posted 2023-03-29

【中文标题】如何替换 snowflake.execute 语句中的循环结果？

【英文标题】：How to replace loop results inside snowflake.execute statement?

【发布时间】：2021-01-15 14:13:00

【问题描述】：

我正在寻找一种方法来迭代具有列 id1、id2 的结果集并传递要在内部查询中使用的结果集列值，如下所示。

结果集的变量

列 1 列2

例如：

select * from months cross join (select column1 AS "id1",column2 AS "id2") Select "GEOGRAPHY" from mytable WHERE id1 = column1 and id2 = column2

但是，当我执行存储过程时，我遇到了以下错误。任何指针都非常感谢。谢谢

错误：

存储过程 TEST_PROC_STMT 中的执行错误：SQL 编译错误：错误第 8 行，位置 81 无效标识符 'COLUMN2' At Snowflake.execute，第 14 行位置 10

完整查询

create or replace procedure TEST_PROC_STMT()
    returns varchar not null
    language javascript
    EXECUTE AS CALLER
as
$$
 
 var distinct_sql_command = "select distinct id1, id2 from mytable ";
    var statement1 = snowflake.createStatement( sqlText: distinct_sql_command );
    var result_set1 = statement1.execute();
    // Loop through the results, processing one row at a time... 
    while (result_set1.next())  
       var column1 = result_set1.getColumnValueAsString(1);
       var column2 = result_set1.getColumnValueAsString(2);
       
snowflake.execute( sqlText: `

INSERT INTO mytable("GEOGRAPHY")(
    Select  * from (
             with months as (
                 select dateadd(month, seq4(), '2020-02-01') "REPORTING MONTH" from table (generator(rowcount => 12))
             ), months_ids as (
                 select * from months cross join (select column1 AS "id1",column2 AS "id2")
             ) ,
                event_months as (
                      Select * from (Select *,ROW_NUMBER() OVER (PARTITION BY id1,id2 ORDER BY "REPORTING MONTH") As rn FROM mytable) Where rn =1
                  ) ,
                  final as (
                      select "REPORTING MONTH",id1,id2
                           , (select array_agg("GEOGRAPHY") within group (order by "REPORTING MONTH" desc)  from mytable where a."REPORTING MONTH">="REPORTING MONTH" and a.id1=id1

                      from months_ids a order by "REPORTING MONTH"
                  )
             Select a."GEOGRAPHY" from final a left join event_months b on  a.id1=b.id1 and a.id2 = b.id2 where a."REPORTING MONTH" > b."REPORTING MONTH"
                 Except
             Select "GEOGRAPHY" from mytable WHERE id1 = column1 and id2 = column2
         ) 
)
`);



return "success";

$$
;

CALL TEST_PROC_STMT();

【参考方案1】：

代码至少有两个问题。首先，“var”定义了一个变量，它是一次性操作。在这种情况下，需要将其移出 while 循环：

// Loop through the results, processing one row at a time... 
while (result_set1.next())  
   var column1 = result_set1.getColumnValueAsString(1);
   var column2 = result_set1.getColumnValueAsString(2);

这是一个简单的解决方法：

// Loop through the results, processing one row at a time... 
var column1;
var column2;
while (result_set1.next())  
   column1 = result_set1.getColumnValueAsString(1);
   column2 = result_set1.getColumnValueAsString(2);

另一个问题是代码将“column2”指定为 SQL 中的文字列名，而不是变量。您可以告诉这一点，因为错误消息将变量名称大写，因此 Snowflake 正在寻找“COLUMN2”但找不到它。还有一个“column1”的用法，它应该是一个替换变量。

您可以通过将其设为替换变量来解决此问题。更改这两行：

select * from months cross join (select column1 AS "id1",column2 AS "id2")

...还有这个...

Select "GEOGRAPHY" from mytable WHERE id1 = column1 and id2 = column2

到这里：

select * from months cross join (select $column1 AS "id1", $column2 AS "id2")

...还有这个...

Select "GEOGRAPHY" from mytable WHERE id1 = column1 and id2 = $column2

请注意，当您使用反引号定义字符串时，$replacement_variable 语法仅适用于 JavaScript。当使用单引号或双引号终止字符串时，它将不起作用。

超过这两个可能会暴露其他人，但可以让它运行。