它想念你的表结构来知道你的字段名,但是,如果我很好地理解你的问题,你可以在 mysql中使用 count + distinct 。你也可以检查这个答案。
SELECT DISTINCT(from_user) AS user,
COUNT(from_user) AS num
FROM tracks
GROUP BY from_user
ORDER BY num DESC";
对于第二个问题,您可以进行第二次查询,或者进行 join tracking 。
我认为,在您的情况下,您更容易在循环内进行第二次查询以从“用户”结果中获取所有详细信息。
$query1="SELECT DISTINCT(from_user), COUNT(*) AS num
FROM tracks
GROUP BY from_user
ORDER BY COUNT(*) DESC";
$query2="SELECT * FROM tracks";
$result1=mysql_query($query1) or die(mysql_error());
$result2=mysql_query($query2) or die(mysql_error());
$user_array = array();
while ($row = mysql_fetch_array($result1)) {
$user = $row['from_user'];
$num = $row['num'];
$uploads_array = array();
while ($sub_row = mysql_fetch_array($result2)) {
if( $sub_row['from_user'] == $user ) {
//for example only due to the unknown structure of your table
$uploads_array[] = array(
"file_name" => $sub_row['file_name'],
"file_url" => $sub_row['file_url']
);
}
}
$user_array[] = array(
"name" => $user,
"num_entry" => $num,
"actions" => $uploads_array
);
}
// now the table with all data is stuctured and you can parse it
foreach($user_array as $result) {
$upload_html_link_arr = array();
$user = $result['name'];
$num_entry = $result['num_entry'];
$all_actions_from_user_array = $result['actions'];
foreach($all_actions_from_user_array as $upload) {
$upload_html_link_arr[] = sprintf('<a href="%s">%s</a>', $upload["file_url"],$upload["file_name"]);
}
$upload_html_link = implode(', ',$upload_html_link_arr);
$full_row = sprintf("<tr><td>%s</td><td>uploads : %s</td><td>favourites (%d)</td></tr>", $user, $upload_html_link, $num_entry);
// now just echo the full row or store it to a table for the final echo.
echo $full_row;
}
我希望这有帮助,迈克