Parameter in like clause JPQL

spring data jpa like wildcard
criteriabuilder substring example
jpa group_concat
jpa substring
querydsl like example
jpa like escape
hibernate like query example
jpa query with optional parameters

I am trying to write a JPQL query with a like clause:

LIKE '%:code%'

I would like to have code=4 and find

455
554
646
...

I cannot pass :code = '%value%'

namedQuery.setParameter("%" + this.value + "%");

because in another place I need :value not wrapped by the % chars. Any help?

If you do

LIKE :code

and then do

namedQuery.setParameter("code", "%" + this.value + "%");

Then value remains free from the '%' sign. If you need to use it somewhere else in the same query simply use another parameter name other than 'code' .

JPA + Hibernate - JPQL LIKE Expression Examples, Following examples shows how to use JPQL LIKE keyword to apply wildcard pattern createEntityManagerFactory("example-unit"); public static void //not like and using parameter findEmployeeByName2(); //escaping  The SELECT clause denotes the query result. More than one value may be returned from the SELECT clause of a query. The SELECT clause may contain one or more of the following elements: a single range variable or identification variable that ranges over an entity abstract schema type, a single-valued path expression,

I don't use named parameters for all queries. For example it is unusual to use named parameters in JpaRepository.

To workaround I use JPQL CONCAT function (this code emulate start with):

@Repository
public interface BranchRepository extends JpaRepository<Branch, String> {
    private static final String QUERY = "select b from Branch b"
       + " left join b.filial f"
       + " where f.id = ?1 and b.id like CONCAT(?2, '%')";
    @Query(QUERY)
    List<Branch> findByFilialAndBranchLike(String filialId, String branchCode);
}

I found this technique in excellent docs: http://openjpa.apache.org/builds/1.0.1/apache-openjpa-1.0.1/docs/manual/jpa_overview_query.html

How to Use LIKE % Expression in JPA SQL Query, In the example above I am creating a Native JPA SQL Query which uses advanced LIKE expression together with Named Parameters. In SQL  JPQL LIKE Expression Examples [Updated: May 16, 2018, Created: May 16, 2018] Previous Page Next Page

You could use the JPA LOCATE function.

LOCATE(searchString, candidateString [, startIndex]): Returns the first index of searchString in candidateString. Positions are 1-based. If the string is not found, returns 0.

FYI: The documentation on my top google hit had the parameters reversed.

SELECT 
  e
FROM 
  entity e
WHERE
  (0 < LOCATE(:searchStr, e.property))

LIKE Queries in Spring JPA Repositories, Learn how to create LIKE queries in Spring JPA Repositories. Then, we'll cover the @Query annotation with named and ordered parameters. 2. Setup. For our example, we'll be querying a movie table. Let's define our  Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Learn more Named Query with like in where clause

I don't know if I am late or out of scope but in my opinion I could do it like:

String orgName = "anyParamValue";

Query q = em.createQuery("Select O from Organization O where O.orgName LIKE '%:orgName%'");

q.setParameter("orgName", orgName);

Strings in JPA JPQL/Criteria Queries (LIKE, LOWER, UPPER , Explains string expressions in JPQL queries, including LIKE, LOCATE, LOWER, The [NOT] LIKE operator checks if a specified string matches a specified pattern. and the right operand is always the pattern (usually a parameter or literal). Thanks, the problem of deplouing was that jpql is case sensitive, now it is deploying, but 'LIKE' does not work as I want. It finds just records that are equal to input parameter completely(not containing it). – Viktoriia Jan 30 '14 at 15:42.

There is nice like() method in JPA criteria API. Try to use that, hope it will help.

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery criteriaQuery = cb.createQuery(Employees.class);
Root<Employees> rootOfQuery = criteriaQuery.from(Employees.class);
criteriaQuery.select(rootOfQuery).where(cb.like(rootOfQuery.get("firstName"), "H%"));

Parameters (and setParameter) in the JPA Query Language (JPQL), The WHERE clause reduces the query results to Country objects whose name field value is equal to :name , which is a parameter that serves as a placeholder  Adding IN clause List to a JPA Query. is there any good way of building query like that? java jpa jpql. JPQL IN clause with other parameter, using ordinal

10.2. JPQL Language Reference, JPQL FROM Clause and Navigational Declarations JPQL Named Parameters String literals in queries, like Java String literals, use unicode character  A list is not a valid parameter for a native SQL query, as it cannot be bound in JDBC. You need to have a parameter for each argument in the list. where u.user_id in (?id1, ?id2) This is supported through JPQL, but not SQL, so you could use JPQL instead of a native query.

JPA Tutorial - JPA Query Like Example, JPQL supports the SQL LIKE operator to provide a limited form of string pattern matching. Each LIKE expression consists of a string expression to search for, and​  T-SQL and the WHERE LIKE %Parameter% clause. Ask Question Asked 7 years, 3 months ago. Active 1 year, 7 months ago. Viewed 118k times 75. 6. I was trying to write a

JPA - JPQL, Having SQL like syntax is an advantage because SQL is a simple structured query For example, a JPQL query can retrieve an entity object rather than field result set from It also passes the query parameters rather than embedding literals  The :keyword is a named parameter which will be replaced with a value passed in as a method argument annotated with a @Param annotation,; The named parameter :keyword is surrounded with % which means that the last_name can start with any String, and it can also end with any String but it must contain the value with which the named parameter :keyword will be replaced.

Comments
  • @Manuele Piastra: Is the answer below not what you were looking for?
  • For the record, this does not leave you open to JPQL injection attacks because this.value is automatically properly escaped for you.
  • This "%" + this.value + "%" is what is escaped.
  • How do I make this case-insensitive?
  • Note: CONCAT(?2, '%') will add '%' to the end of the parameter, use CONCAT('%', ?2, '%') to add it to the beginning and end of parameter.
  • for me the best solution - no concatenation, no SQL injection.
  • Downvote: not working, this changes the parameter name to code%