当我使用ksql提示符创建表时,它可以工作。当试图对ksql二进制文件执行相同的操作时,相同的命令不起作用。sudo echo "CREATE TABLE movie_table AS SELECT title, count(*) AS appearance FROM test WHERE year > 1988TABLE movie_table AS SELECT title, count(*)
在流-流连接查询结果上,我面临着一种意想不到的行为。KSQL版本: 5.1.3#1.参考文献ksql> select a.id as `id`, a.date as `date`, a.count as `count` from streamA a inner join streamBb within 1 day on a.id = b.id;
# 00000001 | 2020-06-2
我对KSQL表的理解是,它们显示的是数据的"as “视图,而不是所有数据。因此,如果我有一个简单的聚合查询,并且我从我的表中选择,我应该在此时看到数据。a place45 | CIRCUS | Joffrey Three | a place
创建表MY_TABLESELECT \* FROM MY\_TABLE; APPLICATION NUM\_APPLICANTS 2
我通过聚合创建了一个表,如下所示: CREATE TABLE withdrawal_less_than_5min AS FROM TB3_WITHDRAW_RECORD_EXCLUDE_INTERNAL_USERS
GROUP BY executedate,status; 当我执行DESCRIBE EXTENDEDKey field : KSQL_INTERNAL_COL_0|+|KSQL