Question

所以，我试图通过Android中的Post向PHP发送消息这是Android Java功能：

 //enviando para o backend
private void SendtoPHP(String reg) throws IOException {


    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://bubbledev.com.br/gcm/getdevice.php");

    try{
        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
        nameValuePairs.add(new BasicNameValuePair("regid", "" + reg ));
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost);


    }
    catch (ClientProtocolException e) {
        // TODO Auto-generated catch block
    } catch (IOException e) {
        // TODO Auto-generated catch block
    }

}

PHP：

<?php       
include 'conecta.php';
$device_token  = urldecode($_POST['regid']);    
$sql = "INSERT INTO corposa.deviceandroid"
              . " (id,device_token)"
              . " VALUES (NULL,'$device_token')";
mysqli_query($con,$sql);

＆GT;

PHP很好，我已经用另一个Post测试了它并且工作了，但是当Android函数尝试$ device_token时，不要重复任何值，并且SQL保存了一个＆＃34;＆＃34;在桌子上的id

Answer 1

除了ResponseHandler之外，我使用的代码类似于你的代码。它对我有用。

HttpClient Client = new DefaultHttpClient();

List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("id", sID));
nameValuePairs.add(new BasicNameValuePair("etc", sETC));

try {           

    String SetServerString = "";

    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://your-url.com/script.php");
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    ResponseHandler<String> responseHandler = new BasicResponseHandler();

    SetServerString = httpclient.execute(httppost, responseHandler);                

}  catch(Exception ex) {
    // failed
}

Answer 2

我确定有更好的方法可以做到这一点，但这就是我通常处理来自Android的帖子请求的方式：

<?php
{
    $input = json_encode(file_get_contents("php://input"));

    $deviceToken = $input->regId;
    mysqli_query($connection, "INSERT INTO corposa.deviceandroid (`id`, `device_token`) VALUES(NULL, '" . mysqli_real_escape_string($deviceToken) . "')";
}

此外，由于您使用的是POST而不是GET，因此传递给服务器的数据不会被编码，因此此处不需要urldecode。

Answer 3

代码似乎没问题，只有＆＃34;奇怪＆＃34;事情是：

List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);

因为您只在列表中添加一个BasicNameValuePair。也许这就是问题......？

Answer 4

我在当前项目中使用此代码，希望它也适合您

  Map<String, String> kvPairs = new HashMap<String, String>();
    kvPairs.put("regid", reg);
    HttpClient httpclient = this.getNewHttpClient();
    HttpPost httppost = new HttpPost("http://bubbledev.com.br/gcm/getdevice.php");
    if (kvPairs != null && kvPairs.isEmpty() == false) {
        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(kvPairs.size());
        String k, v;
        Iterator<String> itKeys = kvPairs.keySet().iterator();
        while (itKeys.hasNext()) {
            k = itKeys.next();
            v = kvPairs.get(k);
            nameValuePairs.add(new BasicNameValuePair(k, v));
        }
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs, "UTF-8"));
    }
    HttpResponse response;
    response = httpclient.execute(httppost);
    String responseString = EntityUtils.toString(response.getEntity());

在你的班级中也需要这个方法

public static HttpClient getNewHttpClient() {
    try {
        KeyStore trustStore = KeyStore.getInstance(KeyStore.getDefaultType());
        trustStore.load(null, null);

        SSLSocketFactory sf = new RecorridoSSL(trustStore);
        sf.setHostnameVerifier(SSLSocketFactory.ALLOW_ALL_HOSTNAME_VERIFIER);

        HttpParams params = new BasicHttpParams();
        HttpProtocolParams.setVersion(params, HttpVersion.HTTP_1_1);
        HttpProtocolParams.setContentCharset(params, HTTP.UTF_8);

        SchemeRegistry registry = new SchemeRegistry();
        registry.register(new Scheme("http", PlainSocketFactory.getSocketFactory(), 80));
        registry.register(new Scheme("https", sf, 443));

        ClientConnectionManager ccm = new ThreadSafeClientConnManager(params, registry);

        return new DefaultHttpClient(ccm, params);
    } catch (Exception e) {
        return new DefaultHttpClient();
    }
}

希望我没有忘记任何事情，其他方式告诉我

Answer 5

Guys错误发生在我的PHP中

$device_token  = urldecode($_POST['regid']);

这个urldecode弄乱了我的代码哈哈

感谢所有帮助

通过Android发送数据到PHP

5 个答案: