PHP-选择时间与未来日期相同的数据

oliverbj 发表于 Dev

Oliverbj

我的数据库中users_added有一个名为的表，该表中有一个名为的字段expire，用于存储unix时间戳。

我的问题是，我该如何循环x天数，并查看用户还剩下多少天（直到达到到期日期）

例子：

    1 day remaining:
    10 users

    2 days remaining: 
    0 users

    3 days remaining 
    50 users

等，等等。

图形示例：

在此处输入图片说明

这是我到目前为止所拥有的：

    // Start date
    $date = date("Y-m-d",$time);
    // End date
    $end_date = date('Y-m-d', strtotime($date . " +10 days"));



    while (strtotime($date) <= strtotime($end_date)){


        $stmt = $dbh->prepare("SELECT * FROM users_added WHERE user_by=:user AND expire<:time");
        $stmt->bindParam(":user",$userdata['username']);
        $stmt->bindParam(":time",$end_date);
        $stmt->execute();
        $expireData = $stmt->fetchAll();
        $remaining = $date-$end_date;
        echo "$remaining";
        echo "day(s) remaining:";

        $date = date ("Y-m-d", strtotime("+1 day", strtotime($date)));
    }

我被困住了，无法真正走得更远。我不知道该如何实现。

miken32

这为我工作：

mysql> CREATE TABLE users (expiry INT UNSIGNED);
Query OK, 0 rows affected (0.12 sec)

mysql> INSERT INTO users (expiry) VALUES
        (UNIX_TIMESTAMP() + 1000),
        (UNIX_TIMESTAMP() + 12348),
        (UNIX_TIMESTAMP() + 89284),
        (UNIX_TIMESTAMP() + 99438),
        (UNIX_TIMESTAMP() + 333552),
        (UNIX_TIMESTAMP() + 883718),
        (UNIX_TIMESTAMP() + 994872);
Query OK, 7 rows affected (0.05 sec)
Records: 7  Duplicates: 0  Warnings: 0

mysql> SELECT
        ROUND((expiry - UNIX_TIMESTAMP()) / 86400) AS days,
        COUNT(*) AS cnt
       FROM users
       WHERE expiry > UNIX_TIMESTAMP()
       GROUP BY days;
+------+-----+
| days | cnt |
+------+-----+
|    0 |   2 |
|    1 |   2 |
|    4 |   1 |
|   10 |   1 |
|   12 |   1 |
+------+-----+
5 rows in set (0.00 sec)

mysql>