afisare dupa cautare

Quote

$sql='SELECT * FROM adrese WHERE "'.$_post['categorie'].'" = "'.$_post['partial'].'" ';

$sql="SELECT * FROM adrese WHERE ".$_post['categorie']." = ".$_post['partial'];

echo $sql

Edited by worm, 05 November 2008 - 22:14.

worm, on Nov 5 2008, 23:10, said:

eu as pune asa:

$sql="SELECT * FROM adrese WHERE ".$_post['categorie']." = ".$_post['partial'];

eventual pune un

echo $sql

sa vezi exact cum arata query-ul.
my 0.02$

GooDy, on Nov 5 2008, 23:05, said:

aha...deci se pare ca de la niste ghilimele buclucase am dat de furca la lume ... neatentia .

acum in sfarsit imi da o eroare    ( am inceput sa ma bucur ca imi da erori ...e grav ) ..

Edited by OriginalCopy, 05 November 2008 - 22:15.

<?php
require_once('config.php');
if(isset($_POST['submit']))
		 {
		
	   $sql='SELECT * FROM `adrese` WHERE "'.$_POST['categorie'].'"  =  "'.$_POST['partial'].'"  ';
	   //$sql='SELECT * FROM `adrese` WHERE `'.$_POST['categorie'].'` = "'.$_POST['partial'].'"';
	  $res = mysql_query($sql);
	var_dump(mysql_errno());
	   if(mysql_errno()) {
		   die('Cererea SQL:<pre>"'.htmlspecialchars($sql).'"</pre> a returnat eroarea:<pre>'.mysql_error().'</pre>');
	   }
	   var_dump($res);
	   $r=mysql_fetch_array($res);
 //var_dump($r);

echo mysql_error();
	
	 {
}

} else {
?>
<form action="cauta.php" method="post" meta http-equiv=Refresh content=1;url=cauta.php>
<table>
	
	<tr>
		<td>
			Cautare dupa: <input type="text" name="categorie" />
</td>	
	</tr>
	<tr>
		<td>
			Incepe cu: <input type="text" name="partial" /><br /> </td></tr>
  <tr><td>
			<br />
			<input type="submit" name="submit" value="Trimite" />
		</td>
	</tr>		
</table>
</form>	

<?php
}
   mysql_close();
?>

int(0) resource(4) of type (mysql result)

	  

........................
	   }
	   var_dump($res);
	   $r=mysql_fetch_array($res);
 //var_dump($r);


echo $r['site'] , ''.$r['mail'].''.$r['parola'].'<br>';

// AICI ESTE TREABA CU AFISAREA 

?>