java - 无法在 Android 中登录 OpenERP

Question

引用这里的问题，

Android OpenErp Search Request XMLRPC

我正在尝试从我的 Android 项目建立对 OPENERP 的搜索和登录请求。这就是我正在做的事情，

    private URI uri;
    uri = URI.create("http://192.168.100.29:8069");
    client = new XMLRPCClient(uri);

    Array[] arr;
    arr = (Array[]) client.call("search", "DevDB", "admin",
                "password", "product.product", "execute");


   HttpResponse response;
   response = client.execute(postMethod); 

   String res = null;
     if (null != response) {
            HttpEntity resEntity = response.getEntity();
            res = EntityUtils.toString(resEntity);
        }

更新:

 OpenErpConnect op = OpenErpConnect.connect("http://192.168.100.29:8069/", 8069, "DevDB", "admin", "openerp");

在 XMLRPClient 中，

    URL loginUrl = new URL("http://192.168.100.29:8069/");
    XMLRPCClient client = new XMLRPCClient(loginUrl);
    Integer id = (Integer)client.call("login", db, user, pass);
    connection = new OpenErpConnect(server, port, db, user, pass, id);

但我总是得到 html 文件形式的响应。我不知道我哪里错了。有谁可以帮助我吗？

更新:

首先，我真的很抱歉要从这里继续。但因为我最初的问题是关于搜索，所以我只在这里继续。请道歉。

我现在可以使用此代码登录，

       OpenErpConnect localOpenErpConnect = new OpenErpConnect(
                paramString1, paramString2, paramInteger, paramString3,
                paramString4, paramString5,
                (Integer) new XMLRPCClient(new URL(paramString1,
                        paramString2, paramInteger.intValue(),
                        "/xmlrpc/common")).call("login", paramString3,
                        paramString4, paramString5));

但是当我执行这样的搜索请求时，

        Long[] ids = conn.search("product.product", new Object[0]);
        System.out.println(ids); 


        Object[] responseIds = (Object[]) client.call("execute",
                    parameters);

我收到错误，

            Traceback (most recent call last):
                  File "/opt/openerp/v7/server/openerp/service/wsgi_server.py", line 82,      in xmlrpc_return
             result = openerp.netsvc.dispatch_rpc(service, method, params)
            File "/opt/openerp/v7/server/openerp/netsvc.py", line 292, in dispatch_rpc
            result = ExportService.getService(service_name).dispatch(method, params)
            File "/opt/openerp/v7/server/openerp/service/web_services.py", line 611, in dispatch
             (db, uid, passwd ) = params[0:3]
         ValueError: need more than 1 value to unpack

你能帮我吗？

Answer 1

您正在向错误的 URI 发出请求。 URL 应为 http://192.168.100.29:8069/xmlrpc/object 才能通过 openerp 对象执行任何函数。

您也可以引用此链接。 https://doc.openerp.com/6.0/de/developer/6_22_XML-RPC_web_services/#java