# Year function related questions, online score-php Tutorial

Source: Internet
Author: User
Year function problems, online score for the number of years that have passed \$ year, for example, the year, month, and day starting from 5 \$ staff_join, for example, I want to add the number of years to the start year, month, and day, for example, what expression is used for + 5 =? for code that can be run, thank you, brother. (\$ Begin (date (& quot; Y & quot;, \$ staff_join) + \$ year) error. output 1 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)

------ Solution --------------------
PHP code
```  ------ Solution --------------------The value assignment here is incorrect.\$ Staff_join = "20010-05-02 ";Change to \$ staff_join = "2010-05-02 ";------ Solution --------------------

PHP code

"; Echo \$ arrdate [0]; // 2013?>------ Solution --------------------
\$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);?>------ Solution --------------------

PHP code

------ Solution --------------------\$ Year = 5;\$ Staff_join = '2017-2008 ';\$ Begin = date ("Y-m-d", strtotime ("+ \$ year \$ staff_join ));------ Solution --------------------Change it to the boss upstairs.
\$ Year = 5;\$ Staff_join = '2017-2008 ';\$ Begin = date ("Y-m-d", strtotime ("+ \$ year \$ staff_join "));Echo \$ Begin;?>------ Solution --------------------
\$ Year = 5;\$ Staff_join = '2017-2008 ';\$ Begin = date ("Y-m-d", strtotime ("+ \$ year, \$ staff_join "));Echo \$ Begin;?>------ Solution --------------------It seems that the solution has been solved.------ Solution --------------------Strtotime------ Solution --------------------Dude, let's take a look at the strtotime function, which is described in detail in the manual.

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