Question

所以这是我的MySQL表的结构，我想用它来完成：

 ID type     category_id amount
 12 Expense     3        963.39
 13 Expense     5        1200.50
 14 Expense     3        444.12
 15 Expense     5        1137.56
  ..............................

期望的输出：

 1407,41 (for category_id = 3)
 2338,06 (for category_id = 5)
 ....... (and for other category_id)

我现在得到的是什么：

 1407,41 (only for category_id = 3)

我的查询不会添加或显示其他category_id的总和。

这是我正在尝试的查询：

$query = "SELECT SUM(amount) AS TotalAmount FROM spendee WHERE type = 'Expense' 
group by category_id having count(*) >1 ";
    $expense_query = mysqli_query($connection, $query);
    $expense_count = mysqli_fetch_array($expense_query);
    echo $expense_count[0];

过去几天一直坚持这一点。很感谢任何形式的帮助。谢谢！

Answer 1

您只需拨打mysqli_fetch_array()一次。您需要在循环中调用它以获取所有总计。您还应该在SELECT列表中包含类别ID。

$query = "SELECT category_id, SUM(amount) AS TotalAmount FROM spendee WHERE type = 'Expense' 
            group by category_id having count(*) >1 ";
$expense_query = mysqli_query($connection, $query);
while ($row = mysqli_fetch_assoc($expense_query)) {
    echo "{$row['TotalAmount']} (for category_id = {$row['category_id']}<br>\n";
}

Answer 2

此处的查询有效。只是你只选择$ expense_count变量的第一个结果。 $ expense_count [1]将返回列出的第二个类别，$ expense_count [2将返回第三个类别，等等... 试试function createNetworkResource (resourceName, inputDirectory) { request.post({ url: browser.params.baseRestUrl + 'resources/connections', auth: browser.params.auth, json: true, body: { displayName: "RESOURCE_TEST", name: "RESOURCE_TEST", type: "NETWORK" } }, function(error, response, user){ if (error) return console.error('error(%s):', resourceName, error) console.log("Status code of createNetworkResource(%s): %s", resourceName, response.statusCode) } ); };

度过愉快的一天。

由其他列分组时列的总和

2 个答案: