计数SUM查询生成的记录


Count records made by SUM query

我有一个由查询填充的HTML表:

         $sql = "SELECT user_id , pag_title , SUM(pag_views) FROM views WHERE user_id = '". $id ."' GROUP BY pag_title";

查询的输出可以是(p.e.):

title 1 35
title 2 25。

我如何计算35 + 25 + ...?我想要一笔总额。

使用:

$sql = "SELECT sum(view) from(SELECT user_id , pag_title , SUM(pag_views) as view FROM views WHERE user_id = '". $id ."' GROUP BY pag_title) as a"; 

这就足够了。它选择特定用户的所有视图的总和:

$sql = "SELECT SUM(pag_views) FROM views WHERE user_id = '". $id ."'";

更新

假设您将PHP与mySQL一起使用,则应该可以使用类似的方法:

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT SUM(pag_views) FROM views WHERE user_id = '". $id ."'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "views: " . $row[0]. "<br>";
    }
} else {
    echo "0 results";
}
$conn->close();
?> 

有一个PHP变量(例如称为$sum),并在迭代行的同时将和放在那里怎么样?

或者,您可以将WITH ROLLUP放在GROUP BY的末尾,这样您的结果集中就有了一行所需的和。