[SOLVED] From 2007-11-07 to just 2007

[globex]

I have a variable with a date, for example: 2007-11-07

I want to just get the year, in this case: 2007.

 

Is there a way to do this with some function? The only way I can figure it out is to to use substr to remove the last 6 digits. But then the variable is no longer a number, but a string and I can't perform operands on it. In which case is there a way to convert a string to an integer?

[trq]

Did you look at the strtotime() function?

[phpQuestioner]

Did you look at the strtotime() function?

 

left your link out there thrope.

 

http://us2.php.net/manual/en/function.strtotime.php

[globex]

Err, not quite sure where to go from here. So I've converted it to a Unix timestamp... how do I get the year?

[trq]

<?php

  $v = '2007-11-07';
  echo date("Y",strtotime($v));

?>

[globex]

But this doesn't work:

 

<?php

$v = '2007-11-07';
$d = date("Y",strtotime($v));
echo $d - 1;

?>

[neilfurry]

Hi,

 

i think you can make use of this format:

 

<?

$year = date('Y');

  echo $year ;

?>

 

for this will print the current year which is 2008.

 

Hope this help.

[globex]

<?php

$year = date('Y');

?>

 

Gives me the current year. I want it to give me the year from my string.

[neilfurry]

This work fine:

 

<?

$s ="2007-11-07";

$r =date('Y',strtotime($s));

print $r;

?>

[kenrbnsn]

<?php
$v = '2007-11-07';
$d = date("Y",strtotime($v));
echo $d - 1;?>

 

works fine for me.

 

What are you attempting to do. If you want to get the previous year, you can do:

<?php
$v = '2007-11-07';
$d = date("Y",strtotime($v . '-1 year'));
echo $d;
?>

 

Ken

[neilfurry]

<?

$s ="2007-11-07";

$r =date('Y',strtotime($s));

print $r;

?>

[globex]

Thanks Ken, it worked. I was just an idiot and running wrong functions afterwards.