我能够连接到我的 MySQL 数据库，但问题是。我的 android 没有使用此代码将编辑文本值传递给 PHP。
public class MySQL extends Activity {
EditText inputName;
private static String url_create_product = "http://atlantis-us.com/phpFile.php?";
JSONParser jsonParser = new JSONParser();
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_my_sql);
inputName = (EditText) findViewById(R.id.txt1);
((Button) findViewById(R.id.btnsend)).setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
// TODO Auto-generated method stub
new CreateNewProduct().execute();
}
});
}
class CreateNewProduct extends AsyncTask<String, String, String> {
@Override
protected String doInBackground(String... arg0) {
// TODO Auto-generated method stub
String name = inputName.getText().toString();
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("name", name));
JSONObject json = jsonParser.makeHttpRequest(url_create_product,
"POST", params);
return null;
}
}
}
this is my PHP codes
<?php
$DB_HostName = "localhost";
$DB_Name = "_atlantisdb";
$DB_User = "atlantis_frux";
$DB_Pass = "frux2012";
$DB_Table = "testDB";
if (isset ($_GET["name"]))
$name = $_GET["name"];
else
$name = "Ghalia";
$con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die(mysql_error());
mysql_select_db($DB_Name,$con) or die(mysql_error());
$sql = "insert into $DB_Table (name) values('$name');";
$res = mysql_query($sql,$con) or die(mysql_error());
mysql_close($con);
if ($res) {
echo "success";
}else{
echo "faild";
}// end else
?>
任何人都可以追踪我的代码的磨损情况吗？提前致谢。