(opendj-ldap-sdk-2.6.0) bind method parameter - password char [ ]

Question

I'm using opendj-ldap-sdk-2.6.0 jar library to search LDAP entry. I am following the guide. (https://backstage.forgerock.com/docs/opendj/2.6/dev-guide/#chap-using-the-sdk)

source code :

import org.forgerock.opendj.ldap.Connection;
import org.forgerock.opendj.ldap.LDAPConnectionFactory;
import org.forgerock.opendj.ldap.SearchScope;
import org.forgerock.opendj.ldap.responses.SearchResultEntry;
import org.forgerock.opendj.ldap.responses.SearchResultReference;
import org.forgerock.opendj.ldif.ConnectionEntryReader;
import org.forgerock.opendj.ldif.LDIFEntryWriter;

public class Test {

public static void main(String[] args) {

    final LDIFEntryWriter writer = new LDIFEntryWriter(System.out);
    Connection connection = null; 

    try { 
        final LDAPConnectionFactory factory = new LDAPConnectionFactory("localhost",389);

        connection = factory.getConnection();
        connection.bind("cn = Directory Mangager", password );
        // password is just an example of the password. 

        final ConnectionEntryReader reader = connection.search("dc=example,dc=com", SearchScope.WHOLE_SUBTREE,"(uid=bjensen)","*");
        while (reader.hasNext()) {
            if(reader.isEntry()) {
                final SearchResultEntry entry = reader.readEntry();
                writer.writeComment("Search result entry:" + entry.getName().toString());
                writer.writeEntry(entry);
            } else {
                final SearchResultReference ref = reader.readReference();
                writer.writeComment("Search result reference:" + ref.getURIs().toString());
            }
        }
        writer.flush();
    } catch (final Exception e) { 
        System.err.println(e.getMessage());
    } finally { 
        if (connection !=null) { 
            connection.close(); 
        }
    }
}

connection.bind("cn = Directory Mangager", password );

I'm getting a red line at this line under password because the parameter has to be 'char []'.
I captured Bind method in the below.

If my password is 1234, how can I change that into char [] type?

You're missing a call from factory to obtain a connection. – Ludovic Poitou Dec 21 '18 at 16:58 — Ludovic Poitou, Dec 21 '18 at 16:58

score 1 · Accepted Answer · answered Dec 21 '18 at 17:00

1

You're missing a call from factory to obtain a connection.

  connection = factory.getConnection();
  connection.bind("cn = Directory Mangager", password );

answered Dec 21 '18 at 17:00

Ludovic Poitou

4,788
2
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Even with that, it doesn't solve my question. I keep getting the red line under password, saying its char value or something. I put ' around instead of " (String). still doesn't work. – Jin Lee Dec 26 '18 at 00:02
1

In your code extract, the variable password is not defined. It should be a `char[]` and the value should be a valid UTF-8 set of characters. – Ludovic Poitou Jan 02 '19 at 08:43
thank you. password wasn't a variable. according to the manual, I'm supposed to put my password there. I put 'password[ there as an example of password. I saw the description char[ ] part, but couldn't understand :(. If my password is 1234 , how can I make that into char [ ] type? – Jin Lee Jan 02 '19 at 08:50
Hey thank you for your tips. With your help, I figured it out. Cheers~ – Jin Lee Jan 03 '19 at 01:09

Jin Lee · Answer 2 · 2019-01-03T01:14:16.900

0

I figured it out.

connection.bind("cn=Directory Manager", "yourpassword".toCharArray() );

You can use toCharArray()

Also, as Ludovic Poitou mentioned above, you need to use
connection = factory.getConnection(); with the bind method. The guide says if you are not using anonymous search, use the bind method, but you gotta use them both. (I misunderstood the guide)

edited Jan 03 '19 at 01:14

answered Jan 03 '19 at 01:08

Jin Lee

3,194
12
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86

Well, this is properly documented in the right section: https://backstage.forgerock.com/docs/opendj/2.6/dev-guide/#chap-authenticating – Ludovic Poitou Jan 03 '19 at 08:36

(opendj-ldap-sdk-2.6.0) bind method parameter - password char [ ]

2 Answers2