如何接收POST在PHP [英] How to receive POST in php

问题描述

我试图将数据从我下方的Andr​​oid客户端发送到PHP一边是code：

I am trying to send data to php side from my Android client below is the code:

（我已经建造从各种渠道这个code在互联网上，所以我不知道，如果一切顺利这里）

(I have constructed this code from various sources on the internet so I am not sure if all is well here)

private void postJSON(String myurl) throws IOException {
        java.util.Date date= new java.util.Date();
        Timestamp timestamp = (new Timestamp(date.getTime()));
        try {
            JSONObject parameters = new JSONObject();
            parameters.put("timestamp",timestamp);
            parameters.put("jsonArray", new JSONArray(Arrays.asList(makeJSON())));
            parameters.put("type", "Android");
            parameters.put("mail", "[e-mail]");
            URL url = new URL(myurl);
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            //  conn.setReadTimeout(10000 /* milliseconds */);
            //  conn.setConnectTimeout(15000 /* milliseconds */);
            conn.setRequestProperty( "Content-Type", "application/json" );
            conn.setDoOutput(true);
            conn.setRequestMethod("POST");
            OutputStream out = new BufferedOutputStream(conn.getOutputStream());
            BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(out, "UTF-8"));
            writer.write(parameters.toString());
            writer.close();
            out.close();

            int responseCode = conn.getResponseCode();
            System.out.println("\nSending 'POST' request to URL : " + url);
            System.out.println("Response Code : " + responseCode);

            BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            String inputLine;
            StringBuffer response = new StringBuffer();

            while ((inputLine = in.readLine()) != null) {
                response.append(inputLine);
            }
            in.close();
            System.out.println(response.toString());

        }catch (Exception exception) {
            System.out.println("Exception: "+exception);
        }
    } 

我很困惑在这行：

I am confused at this line:

writer.write(parameters.toString());

基本上我只是将一个字符串的PHP端。我如何接受它呢？会是什么POST变量名？

essentially I am only sending one string to the php side. How do I receive it there? What would be the POST variable name?

推荐答案

像CarlosAllende我对Android的没有经验。但希望这可以帮助你。

like CarlosAllende i have no experience in Android. But hope this might help you.

这是你在你的'$ _POST'得到什么？

this is what you get in your '$_POST' ?

 Response Code : 200 01-21 10:46:44.969: I/System.out(21561): <br /> 
 <b>Catchable fatal error</b>: Object of class stdClass could not be converted to string in 
<b>/opt/lampp/htdocs/Pdrive/digital_anniversaries/wordpress-3.9.1/wordpress/auth‌​enticate_user.php</b> on line <b>15</b><br /> 

如果因此它是一个捕获的致命错误，告诉类stdClass的客体不能转换成字符串。所以，如果我是正确的它试图把对象插入到$ _POST数组，它是无法将其转换为字符串。而是你获得你的$ _POST数组错误。

if so it is a Catchable fatal error telling that 'Object of class stdClass could not be converted to string'. so if i am correct it's trying to take object into your $_POST array and it's is unable to convert it to string. and instead your getting the error in your $_POST array.

希望这有助于

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