Question

运行脚本时出现此错误：在布尔值上调用成员函数fetch_assoc（）。这让我发疯了。

我从输入框中获取数据如下：2017-12-14 06:42:10

$sql = "SELECT count(*) as total FROM `purchase` where purchase_datetime BETWEEN str_to_date($date1, 'Y-m-d H:i:s') and str_to_date($date2, 'Y-m-d H:i:s')";

或

$sql = "SELECT count(*) as total FROM `purchase` where purchase_datetime BETWEEN '$date1' and '$date2'";

或
试着这个：

$date1 = date('Y-m-d H:i:s', $date1);
$date2 = date('Y-m-d H:i:s', $date2);

完整代码：

            <?php
            if (isset($_POST['submit']) && isset($_POST['date1']) && isset($_POST['date2'])) {
                $date1 = $_POST['date1'];
                $date2 = $_POST['date2'];

                $date1 = date('Y-m-d H:i:s', $date1);
                $date2 = date('Y-m-d H:i:s', $date2);


                $sql = "SELECT count(*) as total FROM `purchase` where purchase_datetime BETWEEN '$date1' and '$date2'";
                $query_result = $conn->query($sql);
                $rows = $query_result->fetch_assoc();
                echo "Result is: " . $rows['total'] . " purchases";
            }
            ?>

Answer 1

date功能是格式化今天的日期，而不是给定日期。使用DateTime::format函数格式化给定日期。试试这个：

$date1 = $date1->format('Y-m-d H:i:s');
$date2 = $date2->format('Y-m-d H:i:s');
$sql = "SELECT COUNT(*) AS total FROM purchase WHERE purchase_datetime BETWEEN '$date1' AND '$date2';";

或者，您可以使用date_format别名功能，如下所示：

$date1 = date_format($date1, 'Y-m-d H:i:s');
$date2 = date_format($date2, 'Y-m-d H:i:s');
$sql = "SELECT COUNT(*) AS total FROM purchase WHERE purchase_datetime BETWEEN '$date1' AND '$date2';";

Answer 2

确保您的格式为：

$date1=$_POST['date1']//1496620800;
$date2=$_POST['date2']//1496707200;
$date1 = date('Y-m-d', $date1);//2017-06-05
$date2 = date('Y-m-d', $date2);//2017-06-06

purchase_datetime格式：2017-xx-xx

两个日期时间之间的记录数

2 个答案: