Reputation: 33
Data is not inserting in MySql from android there is no error in the code, i have defined them using AsyncThread - doInBackground and onPostExecute..Its showing the toast what i have defined in Json Object Try Catch Method 'Sorry Try Again' in else condition if data is not inserted.
InputStream is = null;
String res = null;
String line = null;
int code;
onClick Code
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 Class
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 Code
<?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);
?>
Correct me if i am wrong.
Upvotes: 0
Views: 652
Reputation: 1343
first get all string you want to insert and then call AsyncTask. may be this solve .
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();
and also change this 2 line . pass UTF-8 to UrlEncodeFormEntity
UrlEncodedFormEntity formEntity = new UrlEncodedFormEntity(nameValuePairs, "UTF-8");
httppost.setEntity(formEntity);
Edited
I think you forget to add concatination onto $choice2 , and change query like this. pass first fieldName and then Values.
$sql = "INSERT INTO MyGuests (fieldName1, fieldName2, fieldName3,fieldName4,fieldName5,fieldName6)
VALUES ('$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);
}
Upvotes: 1