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我將從mysql檢索到的數據設置爲TextView
,現在我需要從「有價值的textview」中插入值。我認爲我需要2個php/api來做到這一點,但我不知道如何在一個活動中使用它。它成功檢索到該值並將其設置爲TextView
,但未能將其發送到mysql。如何將有價值的textviews中的值插入到mysql
這裏是我的腳本:
public class ListJawaban extends Activity {
JSONArray mk = null;
ProgressDialog pDialog;
TextView idm, nama_mk, kls;
Button submit;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.spin_main);
Intent in = getIntent();
String kode = in.getStringExtra("id_stu");
String link_url = "http://xxxxxxxx/ffffff/gggggg/mkQuiz.php?kode="+kode;
JSONParser jParser = new JSONParser();
JSONObject json = jParser.AmbilJson(link_url);
try {
mk = json.getJSONArray("mk");
for(int i = 0; i < mk.length(); i++){
JSONObject ar = mk.getJSONObject(i);
TextView mhs = (TextView) findViewById(R.id.nim_mhs);
TextView mkname = (TextView) findViewById(R.id.nama_mk);
TextView kelas = (TextView) findViewById(R.id.kelas);
String mhs_get = ar.getString("id_mhw");
String mkname_get = ar.getString("mk_mhw");
String kelas_get = ar.getString("kelas");
mhs.setText(mhs_get);
mkname.setText(mkname_get);
kelas.setText(kelas_get);
}
} catch (JSONException e) {
e.printStackTrace();
}
submit = (Button) findViewById(R.id.submit);
submit.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
new DataMasuk().execute();
Intent forum = new Intent(getApplicationContext(), ForumActivity.class);
forum.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
startActivity(forum);
finish();
}
});
}
public class DataMasuk extends AsyncTask<String, String, String> {
String success;
String url_link = "http://xxxxxxxx/ffffff/gggggg/postAns.php";
@Override
protected String doInBackground(String... params) {
String strnama = nama_mk.getText().toString();
String stremail = idm.getText().toString();
String strpassword = kls.getText().toString();
List<NameValuePair> nvp = new ArrayList<NameValuePair>();
nvp.add(new BasicNameValuePair("mk_mhs", strnama));
nvp.add(new BasicNameValuePair("id_mhs", stremail));
nvp.add(new BasicNameValuePair("kelas", strpassword));
JSONParser jParser = new JSONParser();
JSONObject json = jParser.makeHttpRequest(url_link, "POST", nvp);
try {
success = json.getString("success");
} catch (Exception e) {
Toast.makeText(getApplicationContext(), "Error",
Toast.LENGTH_LONG).show();
}
return null;
}
}
}
,這是插入值的PHP
<?php
$id_mhs=$_POST['id_mhs'];
$mk_mhs = $_POST['mk_mhs'];
$kelas=$_POST['kelas'];
include "koneksi.php";
$query = "INSERT INTO tbl_tampung (id_mhs,mk_mhs,kelas)VALUES('$id_mhs','$mk_mhs','$kelas')";
$hasil = mysql_query($query);
if($hasil)
{
$response["success"] = "1";
$response["message"] = "Successfully Added";
echo json_encode($response);
}
else
{$response["success"] = "0";
$response["message"] = "Sorry somethings wrong";
echo json_encode($response);
}
?>
,但它不工作,任何人都可以幫我嗎?它的功課,並應在週四進行:(
刪除意圖論壇=新意圖(getApplicationContext(),ForumActivity.class); 論壇。 addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP); startActivity(論壇); 完成();並開始ForumActivity如果更迭中doinbackground – eurosecom
感謝先生= 1,它的工作原理:) –