Year function related questions, online score

Year function problems, online score for example, $ year = the number of years passed, such as 5 $ staff_join = The Start year Month Day for example I want to add the number of years to the Start year Month Day, for example, what expression is used for + 5 =? to run the code, thank you, brother. ($ Begin = (date (& quot; Y & quot ;, $ s year function problem, online score

For example, $ year = the number of years that have passed, for example, 5
$ Staff_join = the start year, month, and day, for example, 2008-05-02
Now, I want to add the year, month, and day of the beginning, for example, + 5 = What is the expression used,
You can run the code. thank you.

In this case, the younger brother ($ Begin = (date ("Y", $ staff_join) + $ year) has an error and outputs 1970)

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The value assignment here is incorrect.
$ Staff_join = "20010-05-02 ";
Change to $ staff_join = "2010-05-02 ";
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    "; Echo $ arrdate [0]; // 2013?>
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     $year = 5;
$staff_join='2008-05-02';

$join_date = mktime(0, 0, 0,  
substr($staff_join, 5, 2), substr($staff_join, -2, 2), substr($staff_join, 0, 4));
echo date('Y-m-d', $join_date), "\n";
$join_date_5_years_later =  
strtotime("+$year Year", $join_date );

echo date('Y-m-d', $join_date_5_years_later);
?>

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$ Year = 5;
$ Staff_join = '2017-2008 ';

$ Begin = date ("Y-m-d", strtotime ("+ $ year $ staff_join ));

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Change it to the boss upstairs.

      $ Year = 5;
$ Staff_join = '2017-2008 ';

$ Begin = date ("Y-m-d", strtotime ("+ $ year $ staff_join "));
Echo $ Begin;
?>
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      $ Year = 5;
$ Staff_join = '2017-2008 ';

$ Begin = date ("Y-m-d", strtotime ("+ $ year, $ staff_join "));
Echo $ Begin;
?>
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It seems that the solution has been solved.
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Strtotime
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Dude, let's take a look at the strtotime function, which is described in detail in the manual.