如何在MySQL数据库中用PHP语言制作预订ID？

Question

我有这样的案例：

JS100829 0001
JS100829 0002

JS=code
10=year
08=month
29=date
0001=the sequence of today first entry
0002=the sequence of today second entry

我需要生成这段代码。有人能帮我吗？谢谢。

Answer 1

例如，您有一个用于这些预订条目的数据库，这是我最好的猜测：

//sequences
$datenow = date("Y-m-d H:i:s");
//i suppose u has load mysql db somewhere before this code
$q = mysql_query("SELECT COUNT(*) FROM yourtablename WHERE bookingdate = '$datenow'");
//how many squences today
$sequencedtoday = mysql_result($q, 0, 0);
//generate code:
$code = 'JS';
$ymd = date('ymd');
$squence = $squencedtoday+1;
$squence = str_pad($squence,4,0,STR_PAD_LEFT);
//return
echo $code.$ymd.' '.$squence;
//return: JS100829 0001

Answer 2

不知道“顺序计数器”是在哪里维护的，这是我最好的尝试(有意的冗长)：

$code = 'JS';
$date = date('ymd');
$seq  = 1;
$id = sprintf('%s%s %04d', $code, $date, $seq);

Answer 3

将$sequence的值保存在数据库或配置文件中。每次生成ID时保持递增。包含列的数据库表：

• date : date(也是主键)
• LatestID: int

然后，像这样做：

$sequence = mysql_fetch_array(mysql_query("select LatestID from seqtable where Date = CURDATE()"));
$sequence = $sequence + 1;

$code = "JS";
$mycode = $code . date("ymd") . str_pad($sequence, 4, '0', STR_PAD_LEFT);

mysql_query("insert into seqtable set LatestID = $sequence, Date = CURDATE ON DUPLICATE KEY UPDATE LatestID = $sequence");