Passing a variable in strtotime () function in PHP

Question

Like this question , but there was no answer to my specific problem.

The current date is 2011-12-14, for reference, if this issue is considered in the future.

I tried this:

$maxAge = $row['maxAge']; // for example, $row['maxAge'] == 30 $cutoff = date('Ym-d', strtotime('-$maxAge days'));

And it returns the following value for $cutoff: 1969-12-31

And I tried this:

 $maxAge = $row['maxAge']; // for example, $row['maxAge'] == 30 $cutoff = date('Ym-d', strtotime('-' . $maxAge . ' days'));

And it returns the following value for $cutoff: 2011-03-14

How to successfully pass this variable to the strtotime() function so that it calculates the number of days to subtract correctly?

For example, if $maxAge == 30 and the current date is 2011-12-14, then $cutoff should be 2011-11-14

+4

Oneag Dec 14 '11 at 16:51

3 answers

Use double quotes:

 $cutoff = date('Ym-d', strtotime("-$maxAge days"));

However, if you do simple calculations like this, you can just use your code without using strtotime, for example:

 $date = getdate(); $cutoff = date('Ym-d', mktime( 0, 0, 0, $date['mon'], $date['mday'] - $maxAge, $date['year'])); echo $cutoff;

+4

nickb Dec 14 '11 at 16:54

You can use either a double quote or the heredoc string in PHP for extended variables. Single quotes and nowdoc strings do not expand variables.

+1

Sheridan bulger Dec 14 '11 at 16:58

samxli · Accepted Answer · 2011-12-14T16:54:59+0000

Use double quotes:

 date('Ym-d', strtotime("-$maxAge days"));