Question

我正在尝试使用pjsua2发送消息。但这是一个例外。请建议我。

这是我的代码

public void sendInstantMessage(String buddy_uri) {

    buddy_uri = "sip:aaaaa@xxx.yyyyyy.zzz";

    BuddyConfig cfg = new BuddyConfig();
    cfg.setUri(buddy_uri);
    cfg.setSubscribe(true);

    MyBuddy im = new MyBuddy(cfg);
    SendInstantMessageParam prm = new SendInstantMessageParam();        
    prm.setContent("Hi This is X, sending message");

    boolean valid = im.isValid();
    Log.e(TAG, "valid ======= "+valid);
    try {
        im.sendInstantMessage(prm);
    } catch (Exception e) {
        Log.e(TAG, "sendInstantMessage ==== "+e);
        e.printStackTrace();
        return;
    }

}

抛出异常，这里是log

05-27 15:42:40.705: E/SipApi(27611): valid ======= false
05-27 15:42:40.706: A/libc(27611): ../src/pjsua-lib/pjsua_pres.c:231: pjsua_buddy_get_info: assertion "pjsua_buddy_is_valid(buddy_id)" failed

Answer 1

最后，我得到了使用pjsip-2.4发送短信的解决方案

这是代码

/**Send message to this number
 * @param String number
 * @param String msgBody*/
public void sendInstantMessage(String number, String msgBody) {
    String sipServer = "aaa.ggg.net";
    String buddy_uri = "<sip:" + number + "@" + sipServer + ">";

    BuddyConfig bCfg = new BuddyConfig();
    bCfg.setUri(buddy_uri);
    bCfg.setSubscribe(false);

    MyBuddy myBuddy = new MyBuddy(bCfg);
    SendInstantMessageParam prm = new SendInstantMessageParam();
    prm.setContent(msgBody);

    try {
        myBuddy.create(account, bCfg);
        myBuddy.sendInstantMessage(prm);
        myBuddy.delete();
    } catch (Exception e) {
        e.printStackTrace();
        return;
    }
}

如何使用pjsua2 android

1 个答案: