php - 没有错误: Data is not inserting in PHP MySql from Android

Question

数据没有从android插入MySql，代码中没有错误，我已经使用AsyncThread - doInBackground和onPostExecute定义了它们。它显示了我在Json对象 try catch 方法“抱歉再试一次”中定义的内容如果未插入数据则处于 else 条件。

InputStream is = null;
String res = null;
String line = null;
int code;

onClick代码

            AsyncTaskRunner runner = new AsyncTaskRunner();
            runner.execute();

            question = addque.getText().toString();
            choice1 = addc1.getText().toString();
            choice2 = addc2.getText().toString();
            choice3 = addc3.getText().toString();
            answer = addanswer.getText().toString();
            Explanation = addexplan.getText().toString();

AsyncTaskRunner 类

 private class AsyncTaskRunner extends AsyncTask<String, String, String> {
    @Override
    protected String doInBackground(String... arg0) {
        // TODO Auto-generated method stub

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

        nameValuePairs.add(new BasicNameValuePair("question", question));
        nameValuePairs.add(new BasicNameValuePair("choice1", choice1));
        nameValuePairs.add(new BasicNameValuePair("choice2", choice2));
        nameValuePairs.add(new BasicNameValuePair("choice3", choice3));
        nameValuePairs.add(new BasicNameValuePair("answer", answer));
        nameValuePairs.add(new BasicNameValuePair("Explanation", 
 Explanation));

        try {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new 
HttpPost("http://192.168.43.4/insert/insert.php");
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();
            Log.e("pass 1", "connection success ");
        } catch (Exception e) {
            Log.e("Fail 1", e.toString());

        }

        try {
            BufferedReader reader = new BufferedReader(
                    new InputStreamReader(is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            res = sb.toString();
            Log.e("pass 2", "connection success ");
        } catch (Exception e) {
            Log.e("Fail 2", e.toString());
        }

        return null;
    }

    @Override
    protected void onPostExecute(String result) {
        // TODO Auto-generated method stub
        super.onPostExecute(result);

        try
        {
                JSONObject json_data;
                json_data = new JSONObject(res);
                code = json_data.getInt("code");

                if(code==1)
                {
            Toast.makeText(getActivity(), "Inserted Successfully",
                Toast.LENGTH_SHORT).show();
                }
                else
                {
             Toast.makeText(getActivity(), "Sorry, Try Again",
                Toast.LENGTH_LONG).show();
                }
        }
        catch(Exception e)
        {
                Log.e("Fail 3", e.toString());
        }
        }

    }

PHP 代码

<?php
$host='localhost';
$uname='root';
$pwd='';
$db="quizDB";

$con = mysql_connect($host,$uname,$pwd) or die("connection failed");
mysql_select_db($db,$con) or die("db selection failed");

$question=$_REQUEST['question'];
$choice1=$_REQUEST['choice1'];
$choice2=$_REQUEST['choice2'];
$choice3=$_REQUEST['choice3'];
$answer=$_REQUEST['answer'];
$Explanation=$_REQUEST['Explanation'];

$flag['code']=0;

 if($r=mysql_query("insert into quizquestion values('$question','$choice1', 
 $choice2, '$choice3', '$answer', '$Explanation') ",$con))
{
    $flag['code']=1;
    echo"hi";
}

print(json_encode($flag));
mysql_close($con);
?>

如果我错了，请纠正我。

Answer 1

首先获取要插入的所有字符串，然后调用AsyncTask。可能是这样解决的。

  question = addque.getText().toString();
  choice1 = addc1.getText().toString();
  choice2 = addc2.getText().toString();
  choice3 = addc3.getText().toString();
  answer = addanswer.getText().toString();
  Explanation = addexplan.getText().toString();

  AsyncTaskRunner runner = new AsyncTaskRunner();
  runner.execute();

并更改这 2 行。将 UTF-8 传递给 UrlEncodeFormEntity

  UrlEncodedFormEntity formEntity = new UrlEncodedFormEntity(nameValuePairs, "UTF-8");
  httppost.setEntity(formEntity);

已编辑

我认为您忘记将连接添加到 $choice2 上，并像这样更改查询。首先传递 fieldName，然后传递 Value。

$sql = "INSERT INTO MyGuests (fieldName1, fieldName2, fieldName3,fieldName4,fieldName5,fieldName6)
VALUES ('$question','$choice1','$choice2', '$choice3', '$answer', '$Explanation')";

if (mysqli_query($conn, $sql)) {
   echo "New record created successfully";
   $flag['code']=1;
   echo"hi";
} else {
   echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}