How can you find a literal percent sign (%) in PostgreSQL using a LIKE query?

问题

I have character varying entries in a table where some (not all) values contain percentages, e.g., '12%', '97%', etc. I want to find all the values that contain percentages. In other words, I want to find all values that end with a percent sign ('%').

回答1:

You can try like this:

SELECT * FROM my_table  WHERE my_column LIKE '%\%%' ESCAPE '\';

Format

 <like predicate> ::=
      <match value> [ NOT ] LIKE <pattern>
        [ ESCAPE <escape character> ]

 <match value> ::= <character value expression>

 <pattern> ::= <character value expression>

 <escape character> ::= <character value expression>

回答2:

You have to escape the literal % sign. By default the escape character is the backslash:

SELECT * FROM my_table  WHERE my_column LIKE '%\%';

In this case the first % sign matches any starting sequence in my_column. The remaining \% are interpreted as a literal % character. The combination is therefore: match anything that ends in a % character.

SQLFiddle

回答3:

I ended up using regular expressions:

select * from my_table where my_column ~ '%$';

However, I'd still like to know if it's possible using the LIKE operator/comparison.

来源：https://stackoverflow.com/questions/33007164/how-can-you-find-a-literal-percent-sign-in-postgresql-using-a-like-query