Like expression in query

I have to create a filter for table based on the string from text field. From my controller I create the query string like this 'SELECT e FROM seabis$TaskQueueItem e WHERE e.title LIKE :(?!)custom$title ’ and set it to collection DS linked with the table. The custom$title is equal to string from text field. So I made all steps from here but got the exception:
‘ArgumentException: Encountered “e . title LIKE : (”…’

    at org.apache.openjpa.kernel.jpql.JPQL.generateParseException(
    at org.apache.openjpa.kernel.jpql.JPQL.jj_consume_token(
    at org.apache.openjpa.kernel.jpql.JPQL.conditional_primary(
    at org.apache.openjpa.kernel.jpql.JPQL.conditional_factor(
    at org.apache.openjpa.kernel.jpql.JPQL.conditional_term(
    at org.apache.openjpa.kernel.jpql.JPQL.conditional_expression(
    at org.apache.openjpa.kernel.jpql.JPQL.where_clause(
    at org.apache.openjpa.kernel.jpql.JPQL.select_statement(
    at org.apache.openjpa.kernel.jpql.JPQL.parseQuery(
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder$ParsedJPQL.parse(
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder$ParsedJPQL.<init>(
    at org.apache.openjpa.kernel.jpql.JPQLParser.parse(

Am I miss something?

You have mistyped the prefix: use “(?i)” instead of “(?!)”

Oh… i instead of !. Thanks!