Question

我正在尝试开发一个Android应用程序，它调用一个php文件来查询和从数据库中提取数据。该URL可以通过Web浏览器在我的手机上访问，但我似乎无法通过下面的java代码调用它。任何人都可以帮助我从我的Java代码中调用我的PHP文件。

            URL url = new URL("http://10.0.3.2/MYCODE/app/login.php");
            String urlParams = "name="+name+"&password="+password;

            HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
            httpURLConnection.setDoOutput(true);
            OutputStream os = httpURLConnection.getOutputStream();
            os.write(urlParams.getBytes());
            os.flush();
            os.close();

            InputStream is = httpURLConnection.getInputStream();
            while((tmp=is.read())!=-1){
                data+= (char)tmp;
            }

            is.close();
            httpURLConnection.disconnect();

我收到以下错误：java.io.FileNotFoundException:

我正在使用POST方法，因为它更安全。

Answer 1

如果是GET方法，那么我们可以使用下面的代码行，调用文件的问题可能是你的.php文件路径不正确

    //Integrating url with values [Starts]
    Map<String, String> request = new HashMap<String, String>();
    request.put("name", name);
    request.put("password", password);
    Uri.Builder uriBuilder = new Uri.Builder();
    uriBuilder.encodedPath("http://10.0.3.2/MYCODE/app/login.php");
    if (mapOfStrings != null) {
        for (Map.Entry<String, String> entry : request.entrySet()) {
            Log.d("buildSanitizedRequest", "key: " + entry.getKey()
                    + " value: " + entry.getValue());
            uriBuilder.appendQueryParameter(entry.getKey(),
                    entry.getValue());
        }
    }
    String uriString;
    try {
        uriString = uriBuilder.build().toString(); // May throw an
                                                   // UnsupportedOperationException
    } catch (Exception e) {
        Log.e("Exception", "Exception" + e);
    }
    //Integrating url with values [Ends]
    HttpURLConnection connection = null;
    try {
        URL url = new URLuriBuilder.build().toString());
        connection = (HttpURLConnection) url.openConnection();
        connection.setRequestProperty("Content-Type", "application/json");
        connection.setRequestProperty("Accept", "application/json");
        connection.setRequestProperty("Accept-Charset", "utf-8,*");
        Log.d("Get-Request", url.toString());
        try {
            BufferedReader bufferedReader = new BufferedReader(
                    new InputStreamReader(connection.getInputStream()));
            StringBuilder stringBuilder = new StringBuilder();
            String line;
            while ((line = bufferedReader.readLine()) != null) {
                stringBuilder.append(line).append("\n");
            }
            bufferedReader.close();
            Log.d("Get-Response", stringBuilder.toString());
            return new JSONObject(stringBuilder.toString());
        } finally {
            connection.disconnect();
        }
    } catch (Exception e) {
        Log.e("ERROR", e.getMessage(), e);
        return null;
    }`

Answer 2

请确保Manifest和WIFI中的以下两行已启用并连接到子网10.0.3.x

var str = getDataValue();
//str value is in this format = "aVal,bVal,cVal,dVal,eVal"

如果获得post方法，那么“？”在你的网址？

Answer 3

您可以使用此方法ByteArrayOutputStream

同样这个我用于我的一个项目cluesshop.com

通过URL调用Java中的PHP文件

3 个答案: