How to query '%' character using LIKE operator in Cassandra 3.7? - cassandra

Question

I use Cassandra 3.7 and have a text column with SASI index.
Let's assume that I want to find column values that contain '%' character somewhere in the middle.
The problem is that '%' is a command char for LIKE clauses.
How to escape '%' char in a query like LIKE '%%%'?
Here is a test script:
DROP keyspace if exists kmv;
CREATE keyspace if not exists kmv WITH REPLICATION = { 'class' : 'SimpleStrategy', 'replication_factor':'1'} ;
USE kmv;
CREATE TABLE if not exists kmv (id int, c1 text, c2 text, PRIMARY KEY(id, c1));
CREATE CUSTOM INDEX ON kmv.kmv ( c2 ) USING 'org.apache.cassandra.index.sasi.SASIIndex' WITH OPTIONS = {
'analyzed' : 'true',
'analyzer_class' : 'org.apache.cassandra.index.sasi.analyzer.NonTokenizingAnalyzer',
'case_sensitive' : 'false',
'mode' : 'CONTAINS'
};
INSERT into kmv (id, c1, c2) values (1, 'f22', 'qwe%asd');
SELECT c2 from kmv.kmv where c2 like '%$$%$$%';
The select query returns nothing.

Answer 1

I think you can use the $$ syntax to achieve this. Your where clause would be:
LIKE '%$$%$$%'
Source: https://docs.datastax.com/en/cql/3.3/cql/cql_reference/escape_char_r.html

Answer 2

The percent sign is treated as special characters only at the beginning and the end. So LIKE '%%%' works fine for your case.
cqlsh:kmv> SELECT c2 from kmv.kmv where c2 like '%%%';
c2
----------
qwe%asd
Looking at the source, however, I don't think there is a way to escape the percent sign if it's the first or the last character, which means you can't do like queries to find values that start with %.