Printing formatted DATE in PHP through ECHO statement

Question

Trying to print the date in "Friday, 15 November 2019" format using the following PHP code using the standard ECHO statement:

$mydate='2019-11-15';

echo "Raw mydate printed through echo is $mydate<br>";
echo "Formatted mydate through echo is ". date('l, d F Y', strtotime('$mydate'));

However, the output (produced as follows) is strangely taking the formatted date to the first date in UNIX (ie nullifying the $mydate variable value). Output is as follows:

Raw mydate printed through echo is 2019-11-15
Formatted mydate through echo is Wednesday, 31 December 1969

Does PHP ECHO statement does not recognize formatting statements? If it does, then what am I doing wrong in the above code?

PS: I have already tried using " <?php echo htmlspecialchars($row["date"], ENT_QUOTES, "UTF-8"); ?> " as mentioned in Format date within an echo , but that does not help either. Also used the " date_create " statement (like $mydate = date_create($mydate); ) both before and after the formatting statement, but in all cases the formatted date is going to 31Dec1969. What am I doing wrong here, or does ECHO not support formatting at all?

Thanks

Answer 1

Try like this:

$mydate='2019-11-15';

echo "Raw mydate printed through echo is $mydate<br>";
echo "Formatted mydate through echo is ". date('l, d F Y', strtotime($mydate));

Your mistake was the single quotes in strtotime . If you pass a variable in single quotes it is taken literally. So it actually tries to parse $mydate as a literal string, it does not take the value itself. If you want to pass the variable in a string, you should use double quotes " .

You can read more about it here .

Answer 2

Just make the following change, you are passing $mydate as string to the date function.

Remove single quote as shown below:

$mydate='2019-11-15';

echo "Raw mydate printed through echo is $mydate<br>";
echo "Formatted mydate through echo is ". date('l, d F Y', strtotime($mydate));