Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given

[_chester]

It's one of those questions again.

I'm getting Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given on line 12, the one with the while() statement:

$sqlTranzactii="select id_tranzactie, DATE_FORMAT(data_tranzactie,'%d-%m-%y') as data_tranzactie, nume_cumparator, adresa_cumparator from tranzactii where comanda_onorata=0";
$resursaTranzactii=mysql_query($sqlTranzactii);
while($rowTranzactie=mysql_fetch_array($resursaTranzactii))
form action="prelucrare_comenzi.php" method="post">
		Data comenzii:
		 <b><?php echo $rowTranzactie['data_tranzactie']?></b>
		 <div style="width:500px; border:1px solid #ffffff; background-color:#c0c0c0; padding:5px">
		 	<b><?php echo $rowTranzactie['nume_cumparator']?></b><br>
		 	<?php echo $rowTranzactie['adresa_cumparator']?>

I tried googling the issue, but I haven't found anything exactly like my problem. I'm sure it's something trivial that I'm missing, though. Any ideas?

[NogDog]

It means that your query failed, so mysql_query() returned a Boolean false instead of a query result resource. You need to test the value of $resursaTranzactii after you call mysql_query(), and if it is false, do some debugging (such as using mysql_error() to get the last error message from MySQL and echoing/error_log()-ing $sqlTranzactii to see what the query actually looked like.

[_chester]

Could you please explain where in the code should I add the mysql_error? I've never used it before.

[NogDog]

$sqlTranzactii="select id_tranzactie, DATE_FORMAT(data_tranzactie,'%d-%m-%y') as data_tranzactie, nume_cumparator, adresa_cumparator from tranzactii where comanda_onorata=0";
$resursaTranzactii=mysql_query($sqlTranzactii);
if($resursaTranzactii == false) {
    error_log(mysql_error()."\n".$sqlTranzactii);
    die("Sorry, there was an unexpected database error.");
}

If you see the error message, check your PHP error log for details. Or just change the error_log() to an echo if you want to see the error output.

[_chester]

I'm not getting anything, the page looks just like before. Page source doesn't show anything either.

[_chester]

Even the line indicated by the warning is the same, line 12, even though line 12 is now error_log(mysql_error()."\n".$sqlTranzactii);

[_chester]

UPDATE:

<?php
include("autorizare.php");
include("admin_top.php");
mysql_connect("localhost", "root", "");
mysql_select_db("anticariat");
?>
<h1>Comenzi</h1>
<b>Comenzi inca neonorate</b><br>
<?php
/* afisam lista comenzilor neonorate*/
$sqlTranzactii="select id_tranzactie, data_tranzactie, nume_cumparator, adresa_cumparator from tranzactii where comanda_onorata=0";
$resursaTranzactii=mysql_query($sqlTranzactii) or die(mysql_error());
print mysql_error();
while($rowTranzactie=mysql_fetch_array($resursaTranzactii))
print_r($sqlTranzactii);
print_r($resursaTranzactii);
{
	?>
	<form action="prelucrare_comenzi.php" method="post">
		Data comenzii:
		 <b><?php echo $rowTranzactie['data_tranzactie']?></b>
		 <div style="width:500px; border:1px solid #ffffff; background-color:#c0c0c0; padding:5px">
		 	<b><?php echo $rowTranzactie['nume_cumparator']?></b><br>
		 	<?php echo $rowTranzactie['adresa_cumparator']?>
		 	<table border="1" cellpadding="4" cellspacing="0">
		 		<tr>
		 			<td align="center"><b>Carte</b></td>
		 			<td align="center"><b>Nr. Buc</b></td>
		 			<td align="center"><b>Pret</b></td>
		 			<td align="center"><b>Total</b></td>
		 		</tr>
		 		<?php
		 		$sqlCarti="select titlu, nume_autor, pret, nr_buc from vanzari, carti, autori where carti.id_carte=vanzari.id_carte and carti.id_autor=autori.id_autor and id_tranzactie=".$rowTranzactie['id_tranzactie'];
				$resursaCarti=mysql_query($sqlCarti);
				while($rowCarte=mysql_fetch_array($resursaCarti))
				{
					print '<tr><td>'.$rowCarte['titlu'].' de '.$rowCarte['nume_autor'].'</td> <td align="right">'.$rowCarte['nr_buc'].'</td> <td align="right">'.$rowCarte['pret'].' </td>';
					$total=$rowCarte['pret'] * $rowCarte['nr_buc'];
					print '<td align="right">'.$total.'</td></tr>';
					$totalGeneral=$totalGeneral+$total;
				}
				?>
				<tr>
					<td colspan="3" align="right">Total comanda:</td><td><?php echo $totalGeneral?> lei</td>
				</tr>
		 	</table>
		 	<input type="hidden" name="id_tranzactie" value="<?php echo $rowTranzactie['id_tranzactie']?>">
		 	<input type="submit" name="comanda_onorata" value="Comanda onorata!">
		 	<input type="submit" name="anuleaza_comanda" value="Anuleaza comanda">
		 </div>
	</form>
	<?php
}
?>
</body>
</html>

I worked on it some more and there's a wee bit of progress. At least the table is showing now. I also added print_r() to $sqlTranzactii and $resursaTranzactii. The result:

I don't know what to make of what print_r() is displaying. Is there no data being extracted from the database?

[vivek151189]

just echo your query

$sqlTranzactii="select id_tranzactie, data_tranzactie, nume_cumparator, adresa_cumparator from tranzactii where comanda_onorata=0";

and fire it in your phpmyadmin and see what result you are getting

aftr seeing the result you will understand the issue and able to solv you problem by yourself.

[_chester]

Solved it. Apparently there was a wrong column name in the query.