I want to display $row->depositdate in dd-mm-yyyy format.
If the date column in database is null the date displayed is : 01-01-1970
echo "".date('d-m-Y', strtotime($row->depositdate))." "; If the date is null in database it should display nothing , otherwise the date in dd-mm-yyyy format should be displayed.
Thanks in advance
Sandeep
NULL is interpreted as 0 by strtotime, since it want to be passed an integer timestamp. A timestamp of 0 means 1-1-1970.
So you'll have to check for yourself if $row->depositdate === NULL, and if so, don't call strtotime at all.
NULL is converted to 0 - the epoch (1-1-1970)
Do this instead
echo "".($row->depositdate ? date('d-m-Y', strtotime($row->depositdate)) : '')." "; You need to check if $row->depositdata is_null previously or check for 0 after strtotime if the value of $row->depositdata is unrecognizable for strtotime.
echo ""; if (!is_null($row->depositdate)) { $jUnixDate = strtotime($row->depositdate)); if ($jUnixDate > 0) { echo date('d-m-Y', $jUnixDate); } } echo " "; strtotime expects to be given a string containing an English date format and will try to parse that format into a Unix timestamp (the number of seconds since January 1 1970 00:00:00 UTC), relative to the timestamp given in now, or the current time if now is not supplied.
more about unixtime and Y2K38 problem: http://en.wikipedia.org/wiki/Year_2038_problem
Oh! I know why this happens? Simply you have not included "depositdate" in your SELECT query. Firstly Change SQL query to select all with wild card sign as shown here
$sql = "SELECT * FROM `yourtable`";