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我正在尝试编写一个代码,在 mysql 表中注册用户的访问数据。该代码在一周左右的时间内运行良好(打印 2012-07-12 即),但随后它开始在日期列中生成 0000-00-00。

$date_now = date("m.d.y"); // gets date

if (empty($username)) {} else (mysql_query ("INSERT INTO new_visit_data    (user_id, date, exercise, lesson, username ) VALUES ('$user_id','$date_now', '' , 'yes', '$username' ) ON DUPLICATE KEY UPDATE lesson='yes'; "));

我做错了什么?谢谢!

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3 回答 3

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如果日期确实是“日期”类型的列,它应该是

$date_now = date("Y-m-d");
于 2012-12-17T09:43:51.000 回答
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如果列exercise具有数据类型DATETIME,则使用:

INSERT INTO new_visit_data    (user_id, date, exercise, lesson, username ) VALUES ('$user_id', NOW(), '' , 'yes', '$username' ) ON DUPLICATE KEY UPDATE lesson='yes'; 
于 2012-12-17T09:46:51.410 回答
3

这是日期格式

    $today = date("F j, Y, g:i a");                 // March 10, 2001, 5:16 pm
    $today = date("m.d.y");                         // 03.10.01
    $today = date("j, n, Y");                       // 10, 3, 2001
    $today = date("Ymd");                           // 20010310
    $today = date('h-i-s, j-m-y, it is w Day');     // 05-16-18, 10-03-01, 1631 1618 6Satpm01
    $today = date('\i\t \i\s \t\h\e jS \d\a\y.');   // it is the 10th day.
    $today = date("D M j G:i:s T Y");               // Sat Mar 10 17:16:18 MST 2001
    $today = date('H:m:s \m \i\s\ \m\o\n\t\h');     // 17:03:18 m is month
    $today = date("H:i:s");                         // 17:16:18
    $today = date("Y-m-d H:i:s");                   // 2001-03-10 17:16:18 (the MySQL DATETIME format)

日期手册

于 2012-12-17T09:56:12.863 回答