[MySql]Sintaksa poizvedbe

Spostovani!

Zgubljam zivce z sledecim enostavnim stavkom:

SELECT id_customer, lastname, firstname FROM 'ps_customer' WHERE 'email=user@gmail.com' AND 'passwd=geslo'

Namrec, na server strani imam skripto za logiranje uporabnika v bazo. Uporabnisko ime in geslo pa dobim iz Android aplikacije iz telefona. Debugiral sem in php skripta vrne namesto resource boolean vrednost (false), kar pomeni, da se stavek ni izvedel. Identicen stavek sem probal pognati direktno na mysql serverju in mi javi sledeco napako:

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''ps_customer' WHERE 'email=user@gmail.com' AND 'geslo' at line 1

Kaj je narobe z sintakso? Vem, da je ena enostavna stvar, ampak enostavno ne najdem.

Lp,
M.

darkolord je 20. apr 2012 ob 23:16 izjavil:

SELECT id_customer, lastname, firstname FROM 'ps_customer' WHERE email='user@gmail.com' AND passwd='geslo'

Ista napaka v /var/log/apache2/error.log. Tukaj je skripta:

<?php
        session_start();

        if($_SERVER["REQUEST_METHOD"]=="POST")
        {
                $hostname_localhost ="localhost";
                $database_localhost ="test";
                $username_localhost ="test";
                $password_localhost ="test";

                $localhost=mysql_connect($hostname_localhost,
                                         $username_localhost,
                                         $password_localhost) or die(mysql_error());

                echo "Connected to Server.<br/>";

                mysql_select_db($database_localhost) or die(mysql_error());

                echo "Connected to Database.<br/>";

                $username=mysql_real_escape_string($_POST['username']);
                $password=mysql_real_escape_string($_POST['password']);

                $query_user=mysql_query("SELECT id_customer, lastname, firstname
                                         FROM 'ps_customer'
                                         WHERE email='$username' AND passwd='$password'");
                $number_of_users=mysql_num_rows($query_user);
                if($number_of_users==1)
                {
                        echo "User found.<br/>";
                        $output=mysql_fetch_assoc($query_user);
                        print(json_encode($output));
                        mysql_close();
                }
                else
                {
                        echo "User not found.<br/>";
                }       // if
        }       // if
?>

Tukaj je Android koda:

    private void loginToDatabase(final String strUsername,
                                 final String strPassword)
    {
        HttpClient client=new DefaultHttpClient();
        HttpPost post=new HttpPost("http://moja-domena.com/php/login.php");
        HttpResponse response=null;
        HttpEntity entity=null;
        InputStream is=null;
        String result=null;

        try
        {
            List<NameValuePair> loginInfo=new ArrayList<NameValuePair>(2);

            loginInfo.add(new BasicNameValuePair("username",
                                                 strUsername));
            loginInfo.add
                    (new BasicNameValuePair("password",
                                                 strPassword));

            post.setEntity(new UrlEncodedFormEntity(loginInfo));

            response=client.execute(post);
            entity=response.getEntity();
            is=entity.getContent();

            try
            {
                BufferedReader reader=new BufferedReader(new InputStreamReader(is,
                                                                               ("iso-8859-1")),
                                                                               8);
                StringBuilder sb=new StringBuilder();
                String line=null;
                while((line = reader.readLine())!=null)
                {
                    sb.append(line).append("\n");
                }
                is.close();

                result=sb.toString();
                int i=0;    // test breakpoint code
            }
            catch(Exception e)
            {
            }
            finally
            {

            }   // try-catch-finally
        }
        catch(Exception e)
        {
        }
        finally
        {
        }   // try-catch-finally
    }   // loginToDatabase

Spremenljivka result vsebuje poleg ostalih echo-jev iz php skripte tudi "User not found", kar pomeni, da uporabnika preko tega sql stavka ne najde. Zakaj?

darkolord je 20. apr 2012 ob 23:43 izjavil:

Aha, pa ime tabele je brez narekovajev.

DELAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA STARI HVALAAAAAAAAAAAAAAA, TI SI MI PA ZE TOKRAT POMAGU, KDAJ TE LOHKA PELEM NA PIR, CE SI PA BABA TE BOM PA KR KUSNU (BREZ VEDNOSTI ZENE)!!!!

Spura je 21. apr 2012 ob 14:06 izjavil:

Prazni catch in finally stavki?

Ravno sedaj saniram tvojo pripombo. Drugace pa hvala za kritiko.