php - 如何用PHP回显此MySQL命令

以下代码提供了以下信息：

$getexpenses = "SELECT sum(expense_amount) FROM projects_expense WHERE project_id=$project_id";
    $showexpenses = @mysqli_query ($dbc, $getexpenses); // Run the query.

    echo ' . $showexpenses . ';

给我一个结果

' . $showexpenses . '

而不是实际的总和。。。
我肯定我错过了一些简单的事情。。。谢谢！

最佳答案

我添加了对mysqli_fetch_assoc($showexpenses)的调用以使其完全正常工作。
我还清理了$project_id以避免注入，并使用别名通过关联数组访问sum。

$getexpenses = "SELECT SUM(`expense_amount`) as `sum_amount` FROM `projects_expense` WHERE `project_id`= ".intval($project_id);
$showexpenses = mysqli_query ($dbc, $getexpenses);
$sums = mysqli_fetch_assoc($showexpenses);
$sum = $sums['sum_amount'];

echo $sum;

我们也可以这样使用mysqli_fetch_row：

$getexpenses = "SELECT SUM(`expense_amount`) FROM `projects_expense` WHERE `project_id`= ".intval($project_id);
$showexpenses = mysqli_query ($dbc, $getexpenses);
$sums = mysqli_fetch_row($showexpenses);
$sum = $sums[0];

echo $sum;

不再需要别名，因为我们知道第一列（0）是匹配的。
注意：如果您想使用GROUP BY project_id

关于php - 如何用PHP回显此MySQL命令，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/5809255/