Question

这是一个基于android-php-mysql连接的简单项目，使用xamp server.Here iam尝试更新查询，并在服务器数据库中成功更新，并获得php响应授予，但如果我的程序中的条件是不工作，我该怎么办？请帮帮我..这是我的代码!! 提前谢谢你

btt.setOnClickListener(new View.OnClickListener() {

    @Override
    public void onClick(View v) {


    try{
            httpclient=new DefaultHttpClient();
            httppost= new HttpPost("http://192.168.0.45/doctorphp/status.php"); // make sure the url is correct.
            //add your data
            nameValuePairs = new ArrayList<NameValuePair>(1);
            // Always use the same variable name for posting i.e the android side variable name and php side variable name should be similar,
            nameValuePairs.add(new BasicNameValuePair("date",Datetxt.getText().toString().trim()));  // $Edittext_value = $_POST['Edittext_value'];

            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            ResponseHandler<String> responseHandler = new BasicResponseHandler();
            final String response = httpclient.execute(httppost, responseHandler);
            System.out.println("Response : " + response);
      runOnUiThread(new Runnable() {
            public void run() {
                    tvView.setText("Response from PHP : " + response);


            }    });

            if(response.equalsIgnoreCase("Granted")){
                runOnUiThread(new Runnable() {
                    public void run() {
                        Toast.makeText(UserPage.this,"Granted successfully", Toast.LENGTH_SHORT).show();

                    }
                });   
                Intent intent=new Intent(UserPage.this, token.class);
                startActivity(intent);
            }
            else{
                  showAlert();    
            }
        }  
        catch(Exception e){

            System.out.println("Exception : " + e.getMessage());
            Toast.makeText(getApplicationContext(),"Error:"+e.toString(), Toast.LENGTH_LONG).show();
        }   



    }
});

Answer 1

试试这个..希望它有效

if(response.toString().trim().equalsIgnoreCase("Granted")){
    runOnUiThread(new Runnable() {
        public void run() {
            Toast.makeText(UserPage.this,"Granted successfully", Toast.LENGTH_SHORT).show();

        }
    });   
    Intent intent=new Intent(UserPage.this, token.class);
    startActivity(intent);
}

Answer 2

请初始化 public static String response =“”;

response = httpclient.execute（httppost，responseHandler）;

如果程序内的条件不起作用

2 个答案: