我正在重构我的代码,只是碰到一个障碍。下面代码中的第一个查询获取最新提交的选项卡的tab_id。第二个查询获取每个选项卡的详细信息。 在我以前的做法中,我嵌入了php和html,它确实是一团糟。现在我想将2个查询合并为1并/或将其加载到数组中。多个mysql查询到一个php二维数组
随意问及/或屠夫
function get_newest_tabs()
{
$db_open;
$sql = "SELECT tab_id, song_id, user_id FROM tabs ORDER BY time_added DESC ". "LIMIT 15";
$result = mysql_query($sql) or die("ERROR - newest tabs function: ".mysql_error());
if (mysql_num_rows($result) > 0)
{
for($i = 0; $i < mysql_num_rows($result); $i++)
{
$tab_id = mysql_result($result, $i, "tab_id");
$db_open;
$sql =
"SELECT tabs.tab_id, tabs.tab_version, tabs.number_of_hits, artist.artist_name, users.user_alias, songs.song_name, tabs.time_added
FROM tabs, users, artist, songs
WHERE tabs.tab_id ='".$tab_id."' AND tabs.user_id = users.user_id AND tabs.song_id = songs.song_id AND songs.artist_id = artist.artist_id";
$result2 = mysql_query($sql) or die("ERROR - i3p mysql - 4: ".mysql_error());
if(mysql_num_rows($result2) == 1)
{
$song_name = mysql_result($result2, 0, "songs.song_name");
$artist_name = mysql_result($result2, 0, "artist.artist_name");
$user_alias = mysql_result($result2, 0, "users.user_alias");
$tab_version = mysql_result($result2, 0, "tabs.tab_version");
$number_of_hits = mysql_result($result2, 0, "tabs.number_of_hits");
$time_added = mysql_result($result2, 0, "tabs.time_added");
}
}
}
}
此链接可能有用。解释为什么使用mysql_num_rows可能不是最好的。 http://stackoverflow.com/questions/1165298/is-mysqlnumrows-efficient-and-or-standard-practice – Smandoli 2009-07-23 15:07:00