Kļūda whilā

[yeahz]

	if (!empty($_GET['atdod'])) {	

	if ($_GET['atdod'] == $_GET['nem']) {
	echo "Izvēlies atšķirīgus parametrus, ko mainīt.";
	footer();
	exit;
	}

	$yo = mysql_query("SELECT * FROM tirgus_maina WHERE maina=$_GET[atdod] AND pret=$_GET[nem]");
	echo "<table width='100%' border='1' cellspacing='0' cellpadding='0'>";
	echo "<td>Lietotājs</td><td>Maina, daudzums</td><td>Pret, daudzums</td><td>Pievienošanas laiks</td>";
	while ($r = mysql_fetch_array($yo)) {
		$he = mysql_query("SELECT username,id FROM users WHERE id=$r[user_id]");
		$user = mysql_fetch_array($he);
		echo "<td>$user[username]</td><td>$r[maina]<br />$r[maina_daudzums]</td><td>$r[pret]<br />$r[pret_daudzums]</td><td>" . date("Y",$r['laiks']) . "</td>";
	}
	echo "</table>";
}

 

Kur ir kļūda?

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ... on line 22

 

22 līnija ir while ($r = mysql_fetch_array($yo)) {

Edited March 1, 2009 by yeahz

[codez]

$_GET[atdod]=>$_GET['atdod']

[yeahz]

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in ... on line 19

 

Labi, pats izlaboju.

Edited March 1, 2009 by yeahz