使用 PHP，我如何找到周数：
- 一周的第一天是星期一，
- 第 1 周是今年第一周有星期一，
- 范围是1-53？
我正在尝试实现与 MySQL 代码等效的 PHP：
SELECT WEEK(date('Y-m-d'), 7);
如本网站所示： http: //dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_week
示例（我试图用错误的语言实现，MySQL）：
$dbh = new PDO("mysql:host=localhost;dbname=test", "root", "");
$stmt = $dbh->prepare("SELECT WEEK('2012-12-30', 7) as week");
$stmt->execute();
$result = $stmt->fetch(PDO::FETCH_ASSOC);
echo "2012-12-30 = ".$result['week']."<br />";
$stmt = $dbh->prepare("SELECT WEEK('2012-12-31', 7) as week");
$stmt->execute();
$result = $stmt->fetch(PDO::FETCH_ASSOC);
echo "2012-12-31 = ".$result['week']."<br />";
$stmt = $dbh->prepare("SELECT WEEK('2013-01-01', 7) as week");
$stmt->execute();
$result = $stmt->fetch(PDO::FETCH_ASSOC);
echo "2013-01-01 = ".$result['week']."<br />";
$stmt = $dbh->prepare("SELECT WEEK('2013-01-06', 7) as week");
$stmt->execute();
$result = $stmt->fetch(PDO::FETCH_ASSOC);
echo "2013-01-06 = ".$result['week']."<br />";
$stmt = $dbh->prepare("SELECT WEEK('2013-01-07', 7) as week");
$stmt->execute();
$result = $stmt->fetch(PDO::FETCH_ASSOC);
echo "2013-01-07 = ".$result['week']."<br />";
输出：（我正在寻找的输出）
2012-12-30 = 52
2012-12-31 = 53
2013-01-01 = 53
2013-01-06 = 53
2013-01-017 = 1
我的尝试：（使用正确的语言 PHP，但结果错误）
$date = new DateTime('2012-12-30');
echo "2012-12-30 = ".$date->format('W')."<br />";
$date = new DateTime('2012-12-31');
echo "2012-12-31 = ".$date->format('W')."<br />";
$date = new DateTime('2013-01-01');
echo "2013-01-01 = ".$date->format('W')."<br />";
$date = new DateTime('2013-01-06');
echo "2013-01-06 = ".$date->format('W')."<br />";
$date = new DateTime('2013-01-07');
echo "2013-01-07 = ".$date->format('W')."<br />";
输出（错误结果）
2012-12-30 = 52
2012-12-31 = 01
2013-01-01 = 01
2013-01-06 = 01
2013-01-07 = 02
我觉得我错过了一些相当基本的东西。对此的任何帮助将不胜感激，并在此先感谢您！