大写 SQL 语句在雪花存储过程中不起作用

Question

在下面的示例中，working_one存储过程工作，而broken_one则不工作。两者之间唯一的区别是SQL语句的字母大小写。

create table tmp (
raw_json variant
);
-- 2019-01-01 = 1546347600000
-- 2018-01-01 = 1514811600000
insert into tmp select parse_json('{ "timestamp":1514811600000}');
create or replace procedure working_one(TIME_VALUE varchar)
returns varchar
language javascript
as
$$
var stmtString = "delete from tmp where to_timestamp(raw_json:timestamp::string) < to_timestamp(:1);"
var stmt = snowflake.createStatement({sqlText: stmtString, binds: [TIME_VALUE]})
var rs = stmt.execute()
rs.next()
return rs.getColumnValue(1)
$$;
create or replace procedure broken_one(TIME_VALUE varchar)
returns varchar
language javascript
as
$$
var stmtString = "DELETE FROM TMP WHERE TO_TIMESTAMP(RAW_JSON:TIMESTAMP::STRING) < TO_TIMESTAMP(:1);"
var stmt = snowflake.createStatement({sqlText: stmtString, binds: [TIME_VALUE]})
var rs = stmt.execute()
rs.next()
return rs.getColumnValue(1)
$$;
call broken_one('1546347600000');
call working_one('1546347600000');

Answer 1

我认为问题不在于SQL的大小写敏感性，甚至不在于它是一个存储过程。问题是JSON中的属性区分大小写。试试这个，告诉我这对你是否更有效。

create or replace procedure fixed_one(TIME_VALUE varchar)
returns varchar
language javascript
as
$$
var stmtString = "DELETE FROM TMP WHERE TO_TIMESTAMP(RAW_JSON:timestamp::STRING) < TO_TIMESTAMP(:1);"
var stmt = snowflake.createStatement({sqlText: stmtString, binds: [TIME_VALUE]})
var rs = stmt.execute()
rs.next()
return rs.getColumnValue(1)
$$;