作为避免此问题的一种解决方法?运算符，您可以创建新运算符进行完全相同的操作.

As a workaround to avoid the ? operator, you could create a new operator doing exactly the same.

这是原始运算符的代码:

This is the code of the original operator:

CREATE OPERATOR ?(
  PROCEDURE = jsonb_exists,
  LEFTARG = jsonb,
  RIGHTARG = text,
  RESTRICT = contsel,
  JOIN = contjoinsel);

SELECT '{"a":1, "b":2}'::jsonb ? 'b'; -- true

使用没有任何冲突的其他名称(例如#-#)并创建一个新名称:

Use a different name, without any conflicts, like #-# and create a new one:

CREATE OPERATOR #-#(
  PROCEDURE = jsonb_exists,
  LEFTARG = jsonb,
  RIGHTARG = text,
  RESTRICT = contsel,
  JOIN = contjoinsel);

SELECT '{"a":1, "b":2}'::jsonb #-# 'b'; -- true

在您的代码中使用此新运算符，它应该可以工作.

Use this new operator in your code and it should work.

对于所有使用?的运算符，请检查pgAdmin-> pg_catalog->运算符.的名字.

Check pgAdmin -> pg_catalog -> Operators for all the operators that use a ? in the name.