php 想打印当天的全名
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want to print full Name of the day
提问by Nisha
I want to print name of the day giving numbers from 1 to 7.
我想打印日期从 1 到 7 的数字。
Ex:
前任：
<?php
echo date("1");
?>
and want to get output as Monday. But it prints 1.
并希望获得输出为Monday. 但它打印1.
Can any one help me?
谁能帮我？
回答by
echo date ('l');
amazing what you find in the manual, its l not 1
您在手册中发现的内容令人惊讶，它不是 1
l (lowercase 'L') A full textual representation of the day of the week Sunday through Saturday
l（小写“L”）星期日到星期六星期几的全文表示
update
更新
sneakiness to not have to create your own array of day names
偷偷摸摸地不必创建自己的日期名称数组
$day='1';
echo date("l", mktime(0,0,0,8,$day,2011));// it will work for day 1-7
回答by PHP Ferrari
Use like:
像这样使用：
$mydate = '2016-01-01';
echo date('l, F jS, Y', strtotime($mydate));
# Friday, January 1st, 2016
note the letter l(lower case of L)
注意字母l（L 的小写）
回答by Mihai Iorga
you cannot do that with date()
你不能这样做 date()
date()= Returns a string formatted according to the given format string using the given integer timestamp or the current time if no timestamp is given.
date()= 返回根据给定格式字符串格式化的字符串，使用给定的整数时间戳或当前时间（如果没有给定时间戳）。
You can set an array with your values:
您可以使用您的值设置一个数组：
$dates = array("", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday");
echo $dates[1]; // will output Monday
if you want the current day name
如果你想要当前的日期名称
echo date ('l'); // (lowercase 'L')
In other words read the manual
换句话说，阅读手册
回答by Mehtab
$mydate = '2016-09-25';
date('l', strtotime($mydate));
It is "l" l for lock :P
锁定是“l” l ：P
回答by Ahmed Mohamed
if some one using Carbon class which is inherited from the PHP DateTime class,
如果有人使用继承自 PHP DateTime 类的Carbon类，
Carbon::parse('yourdate')->formatLocalized('%A');
回答by smutty
function dt($val) {
$arrWeek = array("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday");
return $arrWeek[$val];
}
回答by xaph
You can use
您可以使用
echo date('l');
Also if you want to localize it:
另外，如果您想对其进行本地化：
setlocale(LC_TIME, "C");//you need to change C with your locale.
echo strftime("%A");
will solve your problem.
将解决您的问题。
