3
我有一個連接數據庫和Android應用程序的問題。我正在嘗試執行this教程。一切似乎都很好,但我沒有取得任何成功而不是錯誤。連接mysql數據庫和Android應用程序
有一個按鈕偵聽器,它可以在點擊時向PHP文件發送帖子並獲得結果。下面是它的代碼: -
ok.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
postParameters.add(new BasicNameValuePair("username", un.getText().toString()));
postParameters.add(new BasicNameValuePair("password", pw.getText().toString()));
//String valid = "1";
String response = null;
try {
response = CustomHttpClient.executeHttpPost("http://10.0.2.2/check.php", postParameters);
String res=response.toString();
Log.d("res:", res);
// res = res.trim();
res= res.replaceAll("\\s+","");
//error.setText(res);
if(res.equals("1"))
error.setText("Correct Username or Password");
else
error.setText("Sorry!! Incorrect Username or Password");
} catch (Exception e) {
un.setText(e.toString());
}
}
});
這裏是HTTP POST方法: -
public static String executeHttpPost(String url, ArrayList<NameValuePair> postParameters) throws Exception {
BufferedReader in = null;
try {
HttpClient client = getHttpClient();
HttpPost request = new HttpPost(url);
UrlEncodedFormEntity formEntity = new UrlEncodedFormEntity(postParameters);
request.setEntity(formEntity);
HttpResponse response = client.execute(request);
in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
StringBuffer sb = new StringBuffer("");
String line = "";
String NL = System.getProperty("line.separator");
while ((line = in.readLine()) != null) {
sb.append(line + NL);
}
in.close();
String result = sb.toString();
Log.d("postMethodReturn", result);
return result;
} finally {
if (in != null) {
try {
in.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
的PHP代碼如下: -
<?php
$un=$_POST['username'];
$pw=$_POST['password'];
//connect to the db
$user = "xyz";
$pswd = "xyz";
$db = "mydb";
$host = "localhost";
$conn = mysql_connect($host, $user, $pswd);
mysql_select_db($db);
//run the query to search for the username and password the match
$query = "SELECT * FROM mytable WHERE user = '$un' AND pass = '$pw'";
$result = mysql_query($query) or die("Unable to verify user because : " . mysql_error());
//this is where the actual verification happens
if(mysql_num_rows($result) --> 0)
echo 1; // for correct login response
else
echo 0; // for incorrect login response
?>
是否有任何錯誤該程序?我嘗試在活動代碼中記錄res(http響應)的中間值並導致execute post方法,但沒有記錄任何內容。嘗試將「localhost」更改爲「127.0.0.1」,並將其更改爲公開可用的虛擬主機,但所有數據庫環境均未成功。所有這些都在模擬器和公共主機上,也使用真實設備進行嘗試。從瀏覽器檢查時,服務器似乎正在運行。數據庫與值存在。所有運行的服務(apache,mysql)。
主要問題是沒有錯誤!任何建議發生了什麼問題? 找不到有相同問題的人。
你能告訴打印結果嗎? –
它總是給出輸出爲0(不正確的登錄響應),不幸的是沒有要求的日誌。 – gkris
這告訴prblm是PHP功能不在android端...所以先檢查你的php連接 –