CODEWITHSUNDEEP
iosxmppopenfire
Kapil Choubisa
Kapil Choubisa

Reputation: 5232

XEP 0055 (Jabber search) with iphone SDK

I am creating a XMPP sample on ios. What I am trying to do is search user over xmpp. I checked the opernfire server and it has the search plugin enabled. I gone through XEP:0055 and found the stanza. I also gone through this so question but did

I am done with xmpp connection and login. Now here are the details regarding to my server:

Host Name = 10.x.x.x; Server name = abcd;

The request format I got from the XEP 0055 link is:

<iq type='set' from='[email protected]/home' to='characters.shakespeare.lit' id='search2' xml:lang='en'>
    <query xmlns='jabber:iq:search'>
        <last>Capulet</last>
    </query>
</iq>

I don't know what characters.shakespeare.lit is. I assume this is the server name. So I created following request:

<iq to="abcd" id="search2" xml:lang="en">
    <query xmlns="jabber:iq:search">
       <Username>kapil</Username>
    </query>
</iq>

but I am not getting any response. I tried to replace the to value with following search.abcd and 10.x.x.x but I am not even receiving any response IQ for this rquest.

Any help plz.

Upvotes: 1

Views: 813

Answers (3)

rubStackOverflow
rubStackOverflow

Reputation: 6163

    var iq = $iq({
          type: 'set',
          id: 'search2',
          to: 'vjud.yourserver.org'
    })
.c('query', {xmlns: 'jabber:iq:search'})
.c('x', {xmlns: 'jabber:x:data', type:'submit'})
.c('field', {var: 'first'}).c('value','Rub*').up(); 
conn.sendIQ(iq);

note: to: characters.shakespeare.lit

Not your server but the vcard service search.

Upvotes: 0

Karun
Karun

Reputation: 890

shakespeare.lit is the xmpp host name. 'characters' is the Search service Name.

Upvotes: 1

tigloo
tigloo

Reputation: 654

It is the name of a directory service. If you connect to a server, you can retrieve the list of services and use that server's directory service. Alternatively, you can use any server's directory service (depending on the service's configuration, you do not have to be connected to the server hosting the service).

Upvotes: 0

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