Converting string to Date and DateTime

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If I have a PHP string in the format of mm-dd-YYYY (for example, 10-16-2003), how do I properly convert that to a Date and then a DateTime in the format of YYYY-mm-dd? The only reason I ask for both Date and DateTime is because I need one in one spot, and the other in a different spot.

Answers

Use strtotime() on your first date then date('Y-m-d') to convert it back:

$time = strtotime('10/16/2003');

$newformat = date('Y-m-d',$time);

echo $newformat;
// 2003-10-16

Make note that there is a difference between using forward slash / and hyphen - in the strtotime() function. To quote from php.net:

Dates in the m/d/y or d-m-y formats are disambiguated by looking at the separator between the various components: if the separator is a slash (/), then the American m/d/y is assumed; whereas if the separator is a dash (-) or a dot (.), then the European d-m-y format is assumed.

To avoid potential ambiguity, it's best to use ISO 8601 (YYYY-MM-DD) dates or DateTime::createFromFormat() when possible.

Thursday, September 15, 2022

muaaz-ff453966a9ad

Try using something like this:

 if(strtotime('2011-08-19 17:14:40') < strtotime('-30 days')) {
     // this is true
 }

Besides, this string looks like it is stored in SQL as datetime/timestamp field. You can directly select all entries from your database with old dates using:

SELECT ... WHERE `datetime` + INTERVAL 30 DAY < NOW()

Sunday, August 21, 2022

christian_grün

See strtotime() and date()

e.g., 2010-02-03 22:21:26 to 3rd February 2010 at 22:21:

$DateTimeStr = '2010-02-03 22:21:26';
echo date('jS F Y at G:i', strtotime($DateTimeStr));

Friday, August 19, 2022

andreyt

You are using the format pattern string yyyyMMdd. You are parsing the date-time string 2018-06-20 00:00:00.

2018 matches yyyy. MM means that month should be two chars, so -0 is taken for the month. And -0 or just 0 is taken to be the month before month 1 of 2018, that is, December 2017. Finally 6 is taken as the day of month. It should have been two chars too, but since there is only one digit, SimpleDateFormat settles with that. The remainder of the string is tacitly ignored.

Exactly the same thing happens when parsing the other string.

It’s the long outdated SimpleDateFormat class in a nutshell: In its attempts to be friendly and helpful it produces the most unpleasant and confusing surprises. Where you would have wished it would tell you something is wrong, it just pretends that everything is fine. It’s one of the main reasons that this class is considered troublesome, and why the replacement for the old classes came out with Java 8 more than 4 years ago. So just never use SimpleDateFormat again.

Instead look into java.time and its DateTimeFormatter.

Also don’t get date values as strings from your database. Depending on the datatype that the query returns get either a LocalDateTime or a LocalDate object. This will free you completely from parsing.

Link: Oracle tutorial: Date Time explaining how to use java.time.

Wednesday, September 21, 2022

chandrakanth

One approach would be to use SimpleDateFormat to parse the strings to Date objects and then do the comparisons through the Date API.

Edit for change to Question:

Below is what I have tried.

SimpleDateFormat dateFormat = new SimpleDateFormat("mm/dd/yy");  
TaskDate = dateFormat.parse(LinkStorage[i][1]);  
if (todayDate.after(TaskDate)) {  
    System.out.println("")  
}

Unfortunately not getting desired output

The problem is with your format string, you are asking for Minutes and not Months. Change "mm/dd/yy" to "MM/dd/yy" and you should have better luck.

Wednesday, August 10, 2022

astha_garg

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Converting string to Date and DateTime

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