Result is empty for query select * from person where address like '%\\\\%';, but MySQL can get a line of result.
CREATE TABLE `person` (
`id` int(11) NULL,
`name` text NULL,
`age` int(11) NULL,
`class` int(11) NULL,
`address` text NULL
) ENGINE=OLAP
UNIQUE KEY(`id`)
COMMENT 'OLAP'
DISTRIBUTED BY HASH(`id`) BUCKETS 1
PROPERTIES (
"replication_allocation" = "tag.location.default: 1",
"in_memory" = "false",
"storage_format" = "V2",
"disable_auto_compaction" = "false"
);
insert into person values (10001,'test1',30,2,'test\\\\,xxx');
Adding logs:
select * from person where address like '%\\\\%';
I0323 10:26:15.907760 2387043 like.cpp:558] arg str: %\\%, size: 4, pattern LIKE_ENDS_WITH_RE: (?:%+)(((\\%)|(\\_)|([^%_]))+), size: 30
I0323 10:26:15.907789 2387043 like.cpp:562] match 0: \\%, size: 3
I0323 10:26:15.907801 2387043 like.cpp:562] match 1: \%, size: 2
I0323 10:26:15.907811 2387043 like.cpp:562] match 2: \%, size: 2
I0323 10:26:15.907821 2387043 like.cpp:562] match 3: , size: 0
I0323 10:26:15.907830 2387043 like.cpp:562] match 4: \, size: 1
I0323 10:26:15.907842 2387043 like.cpp:615] search_string : \\%
I0323 10:26:15.907855 2387043 like.cpp:619] search_string escape removed: \%
It matchs against the LIKE_ENDS_WITH_RE which is wrong, the meaning of the sql should be: match strings that have one backslash in any place.
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