[PHP]Zajem podatkov iz baze

Spostovani!

Zakaj mi spodnja skripta ne vraca podatkov, stevilo zadetkov je ok (8), vendar dobim same prazne nize:

<?php
        session_start();

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

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

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

                $query_user=mysql_query("SELECT id_category
                                         FROM ps_category
                                         WHERE id_parent =1 AND active=1");
                $rows=mysql_num_rows($query_user);
                if($rows>0)
                {
                        $output=mysql_fetch_assoc($query_user);
                        $query_categories=mysql_query("SELECT name
                                                       FROM ps_category_lang
                                                       WHERE id_category IN ($output) AND id_lang=1");
                        $rows_categories=mysql_num_rows($query_categories);
                        if($rows_categories>0)
                        {
                                $categories=mysql_fetch_assoc($query_categories);
                                $userdata=array("name" => $output['name']);

                                print(json_encode($userdata));
                                mysql_close();
                        }       // if
                }
                else
                {
                        #echo "User not found.<br/>";
                }       // if
        }       // if
?>

V php sem zacetnik, vendar mislim, da je problem v drugem SQL stavku, kjer uporabim SELECT * FROM X WHERE Y IN (). Znotraj teh oklepajev je treba lociti vrednosti z vejico, kako to storim v php kodi?

Lp,
M.

Fantje, prosim se za malo pomoci! Spodnja skripta mi nikakor noce delovati!!!! Celo noc nisem spal in se zaje.... z to skripto, a kdo vidi kako sintakticno napako:

<?php
	session_start();

	if($_SERVER["REQUEST_METHOD"]=="POST")
	{
		$hostname_localhost ="localhost";
		$database_localhost ="localhost";
		$username_localhost ="test";
		$password_localhost ="test";
 
		$localhost=mysql_connect($hostname_localhost,
					 $username_localhost,
					 $password_localhost) or die(mysql_error());

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

		$query_categories_id_numbers=mysql_query("SELECT id_category
					 		  FROM ps_category
		  					  WHERE id_parent=1 AND active=1");
		$number_of_categories=mysql_num_rows($query_categories_id_numbers);
		if($number_of_categories>0)
		{
			$categories_id_numbers="";
			
			while($category_id_number=mysql_fetch_assoc($query_categories_id_numbers))
			{
				$categories_id_numbers.=$category_id_number.",";
			}	// while
			$categories_id_numbers=substr($categories_id_numbers,
						      0,
						      strlen($categories_id_numbers)-2);

			$query_categories_names=mysql_query("SELECT name
						       	     FROM ps_category_lang
						             WHERE id_category IN ($categories_id_numbers) AND id_lang=1");
			$number_of_categories_names=mysql_num_rows($query_categories_names);
			if($number_of_categories_names>0)
			{
				$categories=mysql_fetch_assoc($query_categories_names);
				$userdata=array("name" => $categories['name']);

	                        print(json_encode($userdata));
			}	// if
		}
		else
		{
		}	// if
	}	// if
?>

Lp,
M.

Ravno to je, error.log ne javi nobene napake, error reporting je vklopljen!

KernelPanic ne rabiš gledati error.log ampak ko nastaviš error reporting se ti error izpiše, ko greš na spletno stran (ta skripta). In tisti error prilepi.

var_dump("SELECT name
FROM ps_category_lang
WHERE id_category IN ($categories_id_numbers) AND id_lang=1");

In rezultat je ... array?

Arto je 10. maj 2012 ob 11:57 izjavil:

$categories_id_numbers.=$category_id_number.",";

$category_id_number je array, ti ga pa kar prilepiš k stringu.

LOL mas prav zgleda, kako pa naj to potem naredim, da bo ok?

Lej v bazi imam dve tabeli (zaenkrat - baza je bazirana na prestashop template-u), ki se nanasata na kategorijo izdelkov. Kategorija, ki ima za parent_id vrednost 1, je vrhnja kategorija (glavna). V php skripti najprej poberem id-je le teh kategorij (z parent_id=1) in nato v drugi tabeli preko dobljenih id-jev (iz prvega mysql stavka) poiscem imena kategorij. Ta imena (in nato se slike, ampak to je drug problem, ker moram najprej nastudirati sistem, kako se slike kategorij shranijo v prestashop template-u) bi rad prenesel v android aplikacijo, da jih prikazem v ImageView-u, da lahko nato pregledujes izdelke v izbrani kategoriji. Torej, v drugem mysql stavku hocem pobrati imena kategorij, jih spraviti v array, ga zakodirati v json in poslati android napravi.

Lp,
M.

Matr, zdele mi pa javlja sledece:

"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 ') AND id_lang=1' at line 3

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in /var/www/moja-domena.com/php/getCategories.php on line 39

"

Tole je pa malo dodelana skripta sedaj:

<?php
	session_start();

	if($_SERVER["REQUEST_METHOD"]=="POST")
	{
		$hostname_localhost ="localhost";
		$database_localhost ="localhost";
		$username_localhost ="test";
		$password_localhost ="test";
 
		$localhost=mysql_connect($hostname_localhost,
					 $username_localhost,
					 $password_localhost) or die(mysql_error());

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

		$query_categories_id_numbers=mysql_query("SELECT id_category
					 		  FROM ps_category
		  					  WHERE id_parent=1 AND active=1");
		echo mysql_error();	// debug code

		$number_of_categories=mysql_num_rows($query_categories_id_numbers);
		if($number_of_categories>0)
		{
			$categories_id_numbers="";
			
			while($category_id_number=mysql_fetch_assoc($query_categories_id_numbers))
			{
				$categories_id_numbers.=$category_id_number['id_category'].",";
			}	// while
			substr($categories_id_numbers,0,-1);
			
			$query_categories_names=mysql_query("SELECT name
						       	     FROM ps_category_lang
						             WHERE id_category IN ($categories_id_numbers) AND id_lang=1");

			echo mysql_error();	// debug code

			$number_of_categories_names=mysql_num_rows($query_categories_names);
			if($number_of_categories_names>0)
			{
				$categories=mysql_fetch_assoc($query_categories_names);
				$userdata=array("name" => $categories['name']);

	                        print(json_encode($userdata));
			}	// if
		}
		else
		{
		}	// if
	}	// if
?>

Raje print_r, če je array

_Dormage_ je 10. maj 2012 ob 13:08 izjavil:

Raje print_r, če je array

Aja sm tud jes en bumbar. Ja veš kaj nared, novo spremenljivko za SQL, pa ga še sprinti preden ga pokličeš. Tkole:

...
$sql = "SELECT name FROM ps_category_lang WHERE id_category IN ($categories_id_numbers) AND id_lang=1";
echo $sql;
$query_categories_names=mysql_query($sql);
echo mysql_error(); // debug code
...

Takoj bom probal, sedaj sem dobil paket v androida (gledam v debuger watch):

"16,74,75,87,159,199,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 ') AND id_lang=1' at line 3

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in /var/www/moja-domena.com/php/getCategories.php on line 43

"

Cifre zdej koncno dobim, samo vejica je na kocnu stringa, moram se je resit! Takoj javim ce bo delalo!

Zdele je pa neki zacel migat, ja, hvala, samo se par stvari preverim!

if($number_of_categories_names>0)
            {
              $userdata = array();
                while($categories=mysql_fetch_assoc($query_categories_names))
                {
                    $userdata[] = array("name" => $categories['name']);
                }   // while
 
                            print(json_encode($userdata));
            }   // if

 $backupname = date("d-m-y");
$žeObstaja = false;
		$query_backups = $this->db->get('backups');
		foreach ($query_backups->result() as $row){
		if($row->date == $backupname){ // preveri če obstaja že zapis backupa z tem datumom, in začne naprej zapisovat z časom komplet
			$žeObstaja = true
break;
		}
}

$backupname = $žeObstaja ? date("d-m-y-G-i-s") : date("d-m-y");

Lion: upam da je dovolj grdo, da bo popravil na kaj lepšega ;).

Sicer pa bi v bistvu zadostovalo ...

foreach ($query_backups->result() as $row) {
  if($row->date == $backupname) {
    $backupname = date("d-m-y-G-i-s");
    break;
  }
}

(Upam da tisti break deluje tako kot si predstavljam - torej, prekine foreach zanko.)

Sičnikov zagotovo ne dopušča! Kam pa pridemo s sičniki v kodi!?! :P

Ma ok, neumno vprasanje, vem, to sem samo tako vprasal, kot zanimivost ...

cobrica je 11. maj 2012 ob 09:39 izjavil:

@ZihaDolhar hvala zadeva deluje iz nule!

Mate mogoče idejo na kakšen način uvozit preko phpja dumpano bazo nazaj? Sepravi da bi restoral bazo?

Ja, z Import gumbom.