PHP date to string method: first find the difference between the two dates; then use JS to get the current date, and convert the normal date format into a timestamp. The code is [$year=(( int)substr("2008-12-04",0,4));】.
php method to convert date to string:
1. Find the difference between two dates
For example, the date difference between 2007-3-5 ~ 2007-3-6
echo abs(strtotime("2007-3-5") - strtotime("2007-3-6"))/60/60/24; echo "天
";
2. JS to get the current date
var myDate = new Date(); myDate.getYear(); //获取当前年份(2位) myDate.getFullYear(); //获取完整的年份(4位,1970-????) myDate.getMonth(); //获取当前月份(0-11,0代表1月) myDate.getDate(); //获取当前日(1-31) myDate.getDay(); //获取当前星期X(0-6,0代表星期天) myDate.getTime(); //获取当前时间(从1970.1.1开始的毫秒数) myDate.getHours(); //获取当前小时数(0-23) myDate.getMinutes(); //获取当前分钟数(0-59) myDate.getSeconds(); //获取当前秒数(0-59) myDate.getMilliseconds(); //获取当前毫秒数(0-999) myDate.toLocaleDateString(); //获取当前日期 var mytime=myDate.toLocaleTimeString(); //获取当前时间 myDate.toLocaleString( ); //获取日期与时间
3. PHP date and timestamp conversion to each other
There are two types of PHP time: one istimestamp type(1228348800), the other isNormal date format(2008-12-4)
So there are two forms of saving to the database. I save the timestamp type as a string, which is more convenient.
Normal date type is saved asDATE
type.
Please pay attention to these two. I usually use two types, and save the time type as DATE. I still keep Saved with timestamp, the data could not be written into the table. After debugging for a long time, I found out the error. The type was inconsistent and the data was not written into the library.
Convert 1228348800 to 2008-12-4 format code as follows:
$date3=date('Y-m-d H:i:s',"1228348800");
This is OK. If you still want to get hours, minutes and seconds, just change 'Y-m-d' , but please note that the PHP time still seems to have an error of 8 hours. If you add it, it will be OK.
The timestamp can be converted to a normal date. On the contrary, how can the normal date format be converted to a timestamp? , please look at the following code:
$year=((int)substr("2008-12-04",0,4));//取得年份 $month=((int)substr("2008-12-04",5,2));//取得月份 $day=((int)substr("2008-12-04",8,2));//取得几号 echo mktime(0,0,0,$month,$day,$year);
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