The year function is related to the problem, the online score

Year function problem, online to divide

such as the number of years $year=, such as 5
$staff _join= began on the day of 2008-05-02
I am now going to add years to the beginning of the year, such as 2008-05-02+5=2013-05-02 with what expression Ah,
To be able to run the code, thank you

Little brother This ($Begin = (Date ("Y", $staff _join) + $year) error, output 1970)

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PHP Code

  
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This assignment is wrong.
$staff _join = "20010-05-02";
Change to $staff_join = "2010-05-02";
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   PHP Code
   
   
    "; 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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       PHP code 
      
!--? PHP $value = 5; $year = Date (' Y '); $year + = $value  echo $year. Date ('-m-d '); 
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 $year = 5; 
 $staff _join = ' 2008-05-02 '; 
 
 $Begin = Date ("y-m-d", Strtotime ("+ $year Year $staff _join)); 
  
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 to the upstairs boss. 
!--? php 
 $year = 5;  
 $staff _join = ' 2008-05-02 ';  
 
 $Begin = Date ("y-m-d", Strtotime ("+ $year year $staff _join")); 
 Echo $Begin;  
? 
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!--? php 
 $year = 5;  
 $staff _join = ' 2008-05-02 ';  
 
 $Begin = Date ("y-m-d", Strtotime ("+ $year year, $staff _join")); 
 Echo $Begin; 
? 
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 appears to have been resolved 
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 Strtotime 
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 Man, look into the Strtotime function, which is described in detail in the manual. 
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