A form passes variable "$chiave" to result.php
RESULT.PHP CODE:
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<?
include("top_foot.inc.php");
include("config.inc.php");
top();
$db = mysql_connect($db_host, $db_user, $db_password);
if ($db == FALSE)
die ("Error, Verify settings in config.inc.php");
mysql_select_db($db_name, $db)
or die ("Can't select database. Verify settings in config.inc.php");
$keys = explode (" ", $_POST ["chiave"]);
$query = "";
foreach ($Keys as $single_word)
{$single_word = trim($single_word);
if ($single_word != "")
$query .= "titolo LIKE '%$single_word%' OR testo LIKE '%$single_word%' OR autore LIKE
'%$single_word%' OR ";
}
$query .= "'0'";
$query = "SELECT id, titolo, data FROM news WHERE " . $query;
$result = mysql_query($query, $db);
while ($row = mysql_fetch_array($result))
{ echo "<a href=\"view.php?id=$row[id]\">" . date("j/n/y", $row[data]) . " -
$row[titolo]</a><br>"; }
foot();
?>
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Everything goes right till
foreach ($Keys as $single_word)
that RETURNS:
Warning: Invalid argument supplied for foreach() in c:\programmi\apache
group\apache\test\prova\result.php on line
Do U know why?
Thanks in advance
Marco
