Hibernate IN clause from a JSONArray

Question

I am trying get a list with Hibernate. I get a JSONArray with the IN values. JSONArray rolesEnSession is, for example, ["ROLE1", "ROLE2"]

public List<MenuLeft> getMenu(JSONArray rolesEnSession){


        String listaRoles = "";

        for (int i = 0; i < rolesEnSession.length(); i++) {
            try {
                listaRoles =  listaRoles + "'"+rolesEnSession.getString(i)+",";
            } catch (JSONException ex) {
                Logger.getLogger(LinkDAOImpl.class.getName()).log(Level.SEVERE, null, ex);
            }
        }
        return getCurrentSession()
                .createQuery("from MenuLeft "
                        + "WHERE menuRol in (:rolesEnSession)")
                .setParameter("rolesEnSession", listaRoles)
                .list();
    }

But I get a blank value. The column menuRol in database has values ROLE1 and ROLE2

score 0 · Answer 1 · answered Jan 16 '15 at 20:15

0

Consider listaRoles after two iterations with your current code. It will have value:

'ROLE_1,'ROLE_2,

Instead of trying to build a well-formed string of parameter values, just change listaRoles to be an actual List and pass that in as a parameter. Let Hibernate handle the conversion to a valid SQL.

answered Jan 16 '15 at 20:15

darrengorman

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How can I do it? Only `List listaRoles = (List) rolesEnSession;` ?? – Giancarlo Ventura Granados Jan 16 '15 at 20:30
@GiancarloVenturaGranados see [this](http://stackoverflow.com/a/13172820/69352) answer – darrengorman Jan 16 '15 at 20:42
I try it but I get `java.util.ArrayList cannot be cast to java.lang.String` – Giancarlo Ventura Granados Jan 16 '15 at 21:20
@GiancarloVenturaGranados can you show me your code? – darrengorman Jan 16 '15 at 22:08
The code above it is the code. What code do you want to show you? – Giancarlo Ventura Granados Jan 17 '15 at 00:37
You tried "it". Presumably you now have code that is attempting to cast a list to a string? I'd like to see that code – darrengorman Jan 17 '15 at 14:37
I try everyone. I realize the query without parameters and it results. :/ – Giancarlo Ventura Granados Jan 21 '15 at 00:29

Hibernate IN clause from a JSONArray

1 Answers1