我有这段代码对我来说很好,并且可以访问数据库。唯一的问题是,我想添加消息,敬酒,或者在重复的主键值时显示错误...根据Json响应添加Toast
这是我的asyncTask的代码(我添加了Toast,但没有工作:()
Button con=(Button)findViewById(R.id.inscription);
con.setOnClickListener(new OnClickListener() {
public void onClick(View v){
new CreateNewUser().execute();
}
});
}
class CreateNewUser extends AsyncTask<String, String, String> {
@Override
protected void onPreExecute() {
super.onPreExecute();
}
protected String doInBackground(String... args) {
@SuppressWarnings("deprecation")
Date d=new Date(an-2800,mn,jn);
Date d1=new Date(ap-2800,mp,jp);
String datenaiss=d.toString();
String deliv=d1.toString();
EditText pseud=(EditText) findViewById(R.id.pseud);
String pseudo = pseud.getText().toString();
EditText name=(EditText) findViewById(R.id.nom);
String nom = name.getText().toString();
EditText prenom=(EditText) findViewById(R.id.pren);
String pren =prenom.getText().toString();
EditText cinn=(EditText) findViewById(R.id.cin);
String cin = cinn.getText().toString();
EditText ag=(EditText) findViewById(R.id.age);
String age = ag.getText().toString();
EditText tele=(EditText) findViewById(R.id.tel);
String tel = tele.getText().toString();
EditText mail=(EditText) findViewById(R.id.email);
String email = mail.getText().toString();
EditText adress=(EditText) findViewById(R.id.adresse);
String adresse = adress.getText().toString();
EditText motdp=(EditText) findViewById(R.id.pwd);
String pwd = motdp.getText().toString();
EditText vill=(EditText) findViewById(R.id.ville);
String ville = vill.getText().toString();
EditText numpermi=(EditText) findViewById(R.id.numperm);
String numperm = numpermi.getText().toString();
String x="http://192.168.1.5/add_user.php";
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("pseudo", pseudo));
params.add(new BasicNameValuePair("mdp", pwd));
params.add(new BasicNameValuePair("datenaiss", datenaiss));
params.add(new BasicNameValuePair("deliv", deliv));
params.add(new BasicNameValuePair("nom", nom));
params.add(new BasicNameValuePair("prenom", pren));
params.add(new BasicNameValuePair("cin", cin));
params.add(new BasicNameValuePair("age", age));
params.add(new BasicNameValuePair("tel", tel));
params.add(new BasicNameValuePair("email", email));
params.add(new BasicNameValuePair("adresse", adresse));
params.add(new BasicNameValuePair("ville", ville));
params.add(new BasicNameValuePair("numperm", numperm));
JSONObject json;
try {
json = jsonParser.makeHttpRequest(x,"POST", params);
Log.d("Create Response", json.toString());
try {
int success = json.getInt(TAG_SUCCESS);
if (success == 1) {
Toast.makeText(MainActivity.this,"Ajouté avec succés", Toast.LENGTH_LONG).show();}
else
Toast.makeText(getBaseContext(),"echec",Toast.LENGTH_LONG).show();}
catch(JSONException e) {
e.printStackTrace();
}
} catch (JSONException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
// check log cat fro response
return null;}
protected void onPostExecute(String file_url) {
// dismiss the dialog once done
}
这里是我的PHP ADD_USER文件的代码: ...........
// mysql inserting a new row
$result = mysql_query("INSERT INTO utilisateur VALUES('$pseudo', '$mdp', '$nom', '$prenom','$cin', '$datenaiss', '$tel', '$email', '$adresse', '$ville', '$numperm', '$deliv')");
// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["success"] = 1;
$response["message"] = "Product successfully created.";
// echoing JSON response
echo json_encode($response);
} else {
// failed to insert row
$response["success"] = 0;
$response["message"] = "Oops! An error occurred.";
// echoing JSON response
echo json_encode($response);
}
}?>
你们能告诉我,我能做些什么?如果有不同的方式来做同样的事情,我很乐意尝试它们:D
的JSON可以为空,所以如果你试图解析它,它会去你的渔获物和那里你可以把烤面包 –
我感谢您的效应初探但是,当用户添加一个已经存在的主键(伪)的值时,我可以添加什么来获得一个arror? – AbdallahJg
所以它不是你的android代码的一部分,它现在在PHP的部分 –