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我正在尝试登录网站并取回代码，但我收到“无效密码或登录名”，但我确定它们都可以

这是我的代码的一部分：

        HttpPost httppost = new HttpPost("http://www3.mackenzie.com.br/tia/index2.php");
        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
        nameValuePairs.add(new BasicNameValuePair("username", username));
        nameValuePairs.add(new BasicNameValuePair("password", password));
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

        // Execute HTTP Post Request
        HttpResponse response = httpclient.execute(httppost);
        String str = inputStreamToString(response.getEntity().getContent()).toString();

我正在尝试访问此页面： http ://www3.mackenzie.com.br/tia/index2.php

我不确定是否应该输入“用户名”和“密码”或其他内容？我怎么能识别呢？我的问题可能是什么？

感谢您的帮助 =)

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2 回答 2

1

对 Maxwell Weru 的回答稍作修改，我相信他的网址是错误的。& 添加了一个参数。

HttpPost httppost = new HttpPost("http://www3.mackenzie.com.br/tia/verifica.php");
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("alumat", username));
nameValuePairs.add(new BasicNameValuePair("pass", password));
nameValuePairs.add(new BasicNameValuePair("unidade", "001"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
于 2012-06-19T20:27:02.183 回答
1

根据您提供的网站，我认为这是您应该编写代码的方式

HttpPost httppost = new HttpPost("http://www3.mackenzie.com.br/tia/index2.php");
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("alumat", username));
nameValuePairs.add(new BasicNameValuePair("pass", password));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

特别强调增加的价值对

于 2012-06-19T20:09:08.413 回答