Question

我想将一个值发送到网址的末尾。

前：

如果我有id=1;，我想将此ID发送到我的网址末尾（获取ID）：

www.example.com/get/id

id值不同。（例如：ID = 2，ID = 3，ID = 4 ...）

。

有可能吗？如何在此场景中使用HttpPost？

我正在使用这些功能，但我总是收到此消息：

没有发现参数！

在我的网址中：

www.example.com/get/id：

function get($id=0){

$id = (int)$id;
if(!$id) exit("no parametrs was sended !");

$trac  = $this->m_general->get('tractions' , array('id' => $id ) , true );



if(!$trac ) $resp = "-1";
else
if($trac->expired != 0  ||   $trac->cradit_end_date < date('Y-m-d'))
{
    $resp = 0;
}else
$resp = 1;

echo json_encode(array('response'=>$resp));

}


function set(){

    $data = $this->input->post('data');
    if(!$data) exit("no parametrs was sended !");


    $message = $data;
    $message = substr($message,7,-6);
    list($qr,$date,$time) = explode("&",$message);

    $insert = array(
    'qr'=>$qr ,
    'date'=>$date ,
    'time'=>$time ,
    'main'=>$message
    );

    $this->m_general->add('qr' , $insert );

}

private void postData(String valueIWantToSend) {
    // Create a new HttpClient and Post Header
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost(this.url_server_side);
    HttpResponse response = null;
    try {
        // Add your data
        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
        nameValuePairs.add(new BasicNameValuePair("data", valueIWantToSend));
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

        // Execute HTTP Post Request
        response = httpclient.execute(httppost);
        String res = EntityUtils.toString(response.getEntity());
        Log.e("Response = ", res);

        isok = 1 ;

    } catch (ClientProtocolException e) {
        e.printStackTrace();
        // TODO Auto-generated catch block
        isok = -1 ;
    } catch (IOException e) {
        e.printStackTrace();
        // TODO Auto-generated catch block
        isok = -1 ;
    }
    //Log.e("res", response.toString()) ;
}

Answer 1

如上所述，您的Web应用程序期望GET方法传递变量。在java代码中，您正在发送POST请求。您应该使用GET方法传递数据。

String url = "http://www.example.com/id/YOUR_ID_DATA/data/YOUR_DATA";

HttpClient client = HttpClientBuilder.create().build();
HttpGet request = new HttpGet(url);

// add request header
request.addHeader("User-Agent", USER_AGENT);
HttpResponse response = client.execute(request);

BufferedReader rd = new BufferedReader(
    new InputStreamReader(response.getEntity().getContent()));

StringBuffer result = new StringBuffer();
String line = "";
while ((line = rd.readLine()) != null) {
    result.append(line);
}

或者在POST请求中发送id，数据并从PHP端接受POST响应。

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://www.example.com");
HttpResponse response = null;
try {
    // Add your data
    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("id", YOUR_ID_DATA));
    nameValuePairs.add(new BasicNameValuePair("data", YOUR_DATA));
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

然后在PHP Side

$id = $_POST["id"];
$data = $_POST["data"];

使用HttpPost对象发送值

1 个答案: