My $date
output is in the foreach loop
09/25/11, 02/13/11, 09/15/10, 06/11/10, 04/13/10, 04/13/10, 04/13/10, 09/24/09, 02/19/09, 12/21/08
My mysql query(PHP) is as follows
("INSERT INTO table_name(`field1`, `field2`,`date`) VALUES ('".$value1."','".$value2 ."','".$date."')");
Question: In my database all the dates stores as 0000-00-00 00:00:00
. But 4th date (06/11/10) is stored as 2006-11-10 00:00:00
.
I tried with date('Y-m-d H:i:s', $date);
but no help.
Note: My database field is datetime type. Any idea?
MySQL Date Data Types DATE - format YYYY-MM-DD. DATETIME - format: YYYY-MM-DD HH:MI:SS. TIMESTAMP - format: YYYY-MM-DD HH:MI:SS. YEAR - format YYYY or YY.
Introduction to MySQL DATE data type MySQL uses yyyy-mm-dd format for storing a date value. This format is fixed and it is not possible to change it.
Example: For converting datetime to format – dd:mm:yyyy, you can use the following query: SELECT DATE_FORMAT('2020-03-15 07:10:56.123', '%d:%m:%Y');
You're on the right track with your date('Y-m-d H:i:s',$date);
solution, but the date() function takes a timestamp as its second argument, not a date.
I'm assuming your examples are in American date format, as they look that way. You can do this, and it should get you the values you're looking for:
date('Y-m-d H:i:s', strtotime($date));
The reason it's not working is because it expects the date in the YYYY-MM-DD format, and tries to evaluate your data as that. But you have MM/DD/YY, which confuses it. The 06/11/10 example is the only one that can be interpreted as a valid YYYY-MM-DD date out of your examples, but PHP thinks you mean 06 as the year, 11 as the month, and 10 as the day.
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