Question

我想创建请求，我想将参数发送到服务器，例如两个字符串（hello，hello）我搜索了这个，我写了一些代码，但我有调试错误（我的问题是一个URL）我想在此网址中发送2个字符串http：// * / / * * < /strong>.aspx?deviceId={id}&list={list} deviceID = {id）是一个第一个字符串，例如divice id，list = {list}是第二个字符串，只是你好

HttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new HttpPost( "http://****/***/******.aspx?deviceId={id}&list={list}"); List<NameValuePair> nameValuePair = new ArrayList<NameValuePair>(2); nameValuePair.add(new BasicNameValuePair("id", android_id)); nameValuePair.add(new BasicNameValuePair("list","Hello")); try { httpPost.setEntity(new UrlEncodedFormEntity(nameValuePair)); } catch (UnsupportedEncodingException e) { e.printStackTrace(); } try { HttpResponse response = httpClient.execute(httpPost); Log.d("Http Response:", response.toString()); } catch (ClientProtocolException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } Toast.makeText(getApplicationContext(), android_id, Toast.LENGTH_LONG) .show(); }

}

Answer 1

我认为这是一个GET方法，你试图发布它，我是否正确？

如果是GET方法，你必须在URL本身内传递你的请求，就像这样

您之前的代码：

 HttpPost httpPost = new HttpPost(
        "http://****/***/******.aspx?deviceId={id}&list={list}");

现在改为，

 HttpGet httpPost = new HttpGet(
        "http://****/***/******.aspx?deviceId="+android_id+"&list="+"Hello");

Answer 2

由于您使用的是名称值对，因此您无需在URL中对参数进行硬编码。我认为这应该没有问题

HttpClient httpClient = new DefaultHttpClient();
    HttpPost httpPost = new HttpPost(
            "http://****/***/******.aspx");

    List<NameValuePair> nameValuePair = new ArrayList<NameValuePair>(2);
    nameValuePair.add(new BasicNameValuePair("id", android_id));
    nameValuePair.add(new BasicNameValuePair("list","Hello"));

    try {
        httpPost.setEntity(new UrlEncodedFormEntity(nameValuePair));
    } catch (UnsupportedEncodingException e) {

        e.printStackTrace();
    }

    try {
        HttpResponse response = httpClient.execute(httpPost);

        Log.d("Http Response:", response.toString());
    } catch (ClientProtocolException e) {

        e.printStackTrace();
    } catch (IOException e) {

        e.printStackTrace();

    }
    Toast.makeText(getApplicationContext(), android_id, Toast.LENGTH_LONG)
            .show();
}
}

Android发送参数服务器.create请求

2 个答案: