Reputation: 17
I'm trying to make simple search bar using spring and MySQL query. I used the below code to get list of customers but I'm getting error
org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected token: %.
Code I used:
public List<CustDTO> getCust(String name) throws Exception {
List<CustDTO> customerList=null;
String queryString ="SELECT c FROM Cust c WHERE c.name LIKE %?1%";
Query query=entityManager.createQuery(queryString);
query.setParameter(1, name);
List<Cust> result = query.getResultList();
customerList=new ArrayList<CustDTO>();
for (Cust customerEntity : result) {
CustDTO customer=new CustDTO();
customer.setName(customerEntity.getName());
customer.setCity(customerEntity.getCity());
customerList.add(customer);
}
System.out.println(customerList);
return customerList;
}
}
Upvotes: 1
Views: 3645
Reputation: 2422
Yeah, Hibernate won't let you do that. If you want your like query to work, use:
String queryString ="SELECT c FROM Cust c WHERE c.name LIKE ?1";
Query query=entityManager.createQuery(queryString);
query.setParameter(1, "%"+name+"%”);
Or you can even use a native query.
Upvotes: 2
Reputation: 2496
public List<CustDTO> getCust(String theSearchName) throws Exception {
Query theQuery = null;
theQuery = createQuery("from CustDTO where lower(name) like :theName", CustDTO.class);
query.setParameter("theName", "%" + theSearchName.toLowerCase() + "%");
List<CustDTO> customers = theQuery.getResultList();
return customers;
lower(name) - name is the field from your CustDTO entity.
You can also add checks if your input is empty or not
if (theSearchName != null && theSearchName.trim().length() > 0) {
// do search with like
}else {
// theSearchName is empty ... so just get all customers
theQuery =currentSession.createQuery("from CustDTO", CustDTO.class);
}
Upvotes: 0