15
请告诉我如何获得一年中的最后几周？PHP - 获取上周的年份数
A
回答
14
由你想要的一年，欧洲/柏林你的时区替换2010：
<?php
date_default_timezone_set('Europe/Berlin');
echo gmdate("W", strtotime("31 December 2010"));
?>
你会得到的值01，52或53之一。
只是为了好玩（Demo）：
<?php
date_default_timezone_set('Europe/Berlin');
echo "ISO-8601 week number of year, weeks starting on Monday, of 31 December:\n\n";
for ($year = 1900; $year < 2100; $year++) {
$date = strtotime("31 December $year");
echo "$year => ", gmdate("W", $date), "\n";
}
?>
输出：
ISO-8601 week number of year, weeks starting on Monday, of 31 December:
1900 => 01
1901 => 01
1902 => 01
1903 => 53
1904 => 52
1905 => 52
1906 => 52
1907 => 01
1908 => 53
1909 => 52
1910 => 52
1911 => 52
1912 => 01
1913 => 01
1914 => 53
1915 => 52
1916 => 52
1917 => 52
1918 => 01
1919 => 01
1920 => 53
1921 => 52
1922 => 52
1923 => 52
1924 => 01
1925 => 53
1926 => 52
1927 => 52
1928 => 52
1929 => 01
1930 => 01
1931 => 53
1932 => 52
1933 => 52
1934 => 52
1935 => 01
1936 => 53
1937 => 52
1938 => 52
1939 => 52
1940 => 01
1941 => 01
1942 => 53
1943 => 52
1944 => 52
1945 => 52
1946 => 01
1947 => 01
1948 => 53
1949 => 52
1950 => 52
1951 => 52
1952 => 01
1953 => 53
1954 => 52
1955 => 52
1956 => 52
1957 => 01
1958 => 01
1959 => 53
1960 => 52
1961 => 52
1962 => 52
1963 => 01
1964 => 53
1965 => 52
1966 => 52
1967 => 52
1968 => 01
1969 => 01
1970 => 53
1971 => 52
1972 => 52
1973 => 52
1974 => 01
1975 => 01
1976 => 53
1977 => 52
1978 => 52
1979 => 52
1980 => 01
1981 => 53
1982 => 52
1983 => 52
1984 => 52
1985 => 01
1986 => 01
1987 => 53
1988 => 52
1989 => 52
1990 => 52
1991 => 01
1992 => 53
1993 => 52
1994 => 52
1995 => 52
1996 => 01
1997 => 01
1998 => 53
1999 => 52
2000 => 52
2001 => 52
2002 => 01
2003 => 01
2004 => 53
2005 => 52
2006 => 52
2007 => 52
2008 => 01
2009 => 53
2010 => 52
2011 => 52
2012 => 52
2013 => 01
2014 => 01
2015 => 53
2016 => 52
2017 => 52
2018 => 52
2019 => 01
2020 => 53
2021 => 52
2022 => 52
2023 => 52
2024 => 01
2025 => 01
2026 => 53
2027 => 52
2028 => 52
2029 => 52
2030 => 01
2031 => 01
2032 => 53
2033 => 52
2034 => 52
2035 => 52
2036 => 01
2037 => 53
2038 => 01
2039 => 01
2040 => 01
2041 => 01
2042 => 01
2043 => 01
2044 => 01
2045 => 01
2046 => 01
2047 => 01
2048 => 01
2049 => 01
2050 => 01
2051 => 01
2052 => 01
2053 => 01
2054 => 01
2055 => 01
2056 => 01
2057 => 01
2058 => 01
2059 => 01
2060 => 01
2061 => 01
2062 => 01
2063 => 01
2064 => 01
2065 => 01
2066 => 01
2067 => 01
2068 => 01
2069 => 01
2070 => 01
2071 => 01
2072 => 01
2073 => 01
2074 => 01
2075 => 01
2076 => 01
2077 => 01
2078 => 01
2079 => 01
2080 => 01
2081 => 01
2082 => 01
2083 => 01
2084 => 01
2085 => 01
2086 => 01
2087 => 01
2088 => 01
2089 => 01
2090 => 01
2091 => 01
2092 => 01
2093 => 01
2094 => 01
2095 => 01
2096 => 01
2097 => 01
2098 => 01
2099 => 01
1
如果你问如何让留在今年的周数，这将做到这一点：
<?php
$year = date('Y');
$week_count = date('W', strtotime($year . '-12-31'));
if ($week_count == '01')
{
$week_count = date('W', strtotime($year . '-12-24'));
}
echo ($week_count - date('W'));
echo ' weeks left in ' . date('Y') . '!';
?>
编辑：添加逻辑以补偿日期（'W'）返回的'01';
+0
这可能不会产生预期的结果，因为如问题评论中所述，上周可能是'01'。在这种情况下，你的代码会在今年剩下几周。您应该以编程方式将边缘案例'01'（通过在倒数第二周加1）。 – 2010-07-23 15:41:41
+0
@paul我刚刚编辑了代码，很确定这会正确地补偿返回的'01'...在第二个想法中，当本周返回01时，它仍然会被扣掉...... sooo close;） – joshtronic 2010-07-23 16:12:41
+0
无论如何，这是沿着这些线的东西。 :) – 2010-07-23 21:54:24
36
function getIsoWeeksInYear($year) {
$date = new DateTime;
$date->setISODate($year, 53);
return ($date->format("W") === "53" ? 53 : 52);
}
的o日期格式给出了ISO-8601年份数字。我们可以使用它，并且“无效”日期自动滚动以生成有效日期（2011-02-31给出2011-03-03），以确定给定年份是否有53周。如果没有，那么它必须有52个。
+3
不错，简单 – Gordon 2012-01-26 13:26:38
20
-1
echo getLastWeekOfYear(2015);
function getLastWeekOfYear($year) {
$date = new DateTime();
return date('W', strtotime(date('Y-m-d', strtotime($date->setISODate($year, 1, "1")->format('Y-m-d') . "-1day"))));
}
什么exac这是你的意思吗？请更具体一些。 – 2010-07-23 15:05:49
@Andy E：不一定......最后一周的数字可能是53.第01周的ISO-8601定义是第一个星期四的星期。 – Powerlord 2010-07-23 15:14:29
我在之前的评论中有一个例子，但我认为它是错误的，所以我删除了它。长话短说：由于52周是364天，一年是365-366天，每隔几年你最终会增加一周。 – Powerlord 2010-07-23 15:20:13