XMPP IIllegalArgumentException在表＃setAnswer（QUOT;用户名和QUOT;，真正的），而使用UserSearchManager [英] XMPP IIllegalArgumentException at Form#setAnswer("Username",true) while using UserSearchManager

问题描述

我的工作使用ejabberd 2.1.11的Andr​​oid聊天应用程序。搜索一个特定的用户是否存在我使用UserSearchManager

I am working on android chat application using ejabberd 2.1.11. For searching whether a particular user exists or not i'm using UserSearchManager

public boolean isUserExists(IrishContact ic) {
        try {
            UserSearchManager search = new UserSearchManager(connection);
            Collection<String> services = search.getSearchServices();
            if (services.isEmpty()) {
                Log.v("IrishuserSearch ", "no service found");
            }

            Log.v("service name: ", connection.getServiceName());

            Form searchForm = search.getSearchForm("vjud."
                    + connection.getServiceName());

            Form answerForm = searchForm.createAnswerForm();
            answerForm.setAnswer("Username", true);
            answerForm.setAnswer("search", ic.getPhoneNumber());
            org.jivesoftware.smackx.ReportedData data = search
                    .getSearchResults(answerForm,
                            "search." + connection.getServiceName());

            if (data.getRows() != null) {
                Iterator<Row> it = data.getRows();
                if (it.hasNext()) {
                    return true;
                } else
                    return false;
            } else
                return false;

        } catch (XMPPException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return false;

    }

在上这回答

我使用

表searchForm = search.getSearchForm（vjud。
                    + connection.getServiceName（））;

Form searchForm = search.getSearchForm("vjud." + connection.getServiceName());

而不是

表searchForm =
  sea​​rch.getSearchForm（搜索。+ connection.getServiceName（））;

Form searchForm = search.getSearchForm("search."+connection.getServiceName());

后半部分是给503服务未发现异常，所以我改变了我的code和它得到了解决，但现在的问题是我得到在抛出：IllegalArgumentException

the later part was giving 503 service not found exception so i changed my code and it got solved, but now the problem is I am getting IllegalArgumentException at

answerForm.setAnswer（用户名，真正的）;

answerForm.setAnswer("Username", true);

我使用asmack-Android的8-0.8.10。

I am using asmack-android-8-0.8.10.

STACK_TRACE=java.lang.IllegalArgumentException: Field not found for the specified variable name.
at org.jivesoftware.smackx.Form.setAnswer(Form.java:258)
at com.irishtalk.utilities.IrishUserSearch.isUserExists(IrishUserSearch.java:42)

at com.irishtalk.utilities.IrishContactsHelper.addContactToDefaultRoster(IrishContactsHelper.java:51)
at com.irishtalk.utilities.IrishContactsHelper.getRoster(IrishContactsHelper.java:32)
at com.irishtalk.service.IXmppAidlStub$1.run(IXmppAidlStub.java:221)

谁能帮我为什么发生这种情况？谢谢

can anyone help me out why this is happening? Thanks

推荐答案

Ejabberd不支持字段值用户名。它是在Openfire的使用。请使用字段谷用户，并传递参数在ejabberd字符串。谢谢。如果你的错误就解决了，请投票了。

Ejabberd does not support the field value "Username". It is used in openfire. Please use the field vale "user" and pass arguments as string in ejabberd. Thanks. Please Vote up if your error is solved.

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