sql query in php

zashto pri...

<?
mysql_connect ('localhost','user','pass');
mysql_select_db ('database');
echo mysql_query ("SELECT `date_time` FROM `news` WHERE 1 ORDER BY `date_time` DESC LIMIT 0, 5");
?>

mi vyrshta 'Resource id #2' pri polojenic che imam celi 3 zapisa v bazata koito mi otgovariat na kriteria....

prosto iskam poslednite 5 zapisa v poleto 'date_time'....

Правилното е:
$getVar = mysql_query ("SELECT `date_time` FROM `news` WHERE 1 ORDER BY `date_time` DESC LIMIT 0, 5");
while (list ($data_time) = mysql_fetch_array($getVar) {
echo $date_time;
};

[линк]

След като ти отговориха вече в два клуба

mysql_query() връща тип resource
Или някой пак го е домързяло да чете

На: bg.php.net/manual/en/function.mysql-query.php

пише:
Description
resource mysql_query ( string query [, resource link_identifier [, int result_mode]])
...
for SELECT
...
mysql_query() returns a resource identifier or FALSE

А на: bg.php.net/manual/en/ref.mysql.php
xample 1. MySQL extension overview example

<?php
/* Connecting, selecting database */
$link = mysql_connect("mysql_host", "mysql_user", "mysql_password")
or die("Could not connect");
print "Connected successfully";
mysql_select_db("my_database") or die("Could not select database");

/* Performing SQL query */
$query = "SELECT * FROM my_table";
$result = mysql_query($query) or die("Query failed");

/* Printing results in HTML */
print "<table>\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
print "\t<tr>\n";
foreach ($line as $col_value) {
print "\t\t<td>$col_value</td>\n";
}
print "\t</tr>\n";
}
print "</table>\n";

/* Free resultset */
mysql_free_result($result);

/* Closing connection */
mysql_close($link);
?>

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