Solved

ADDDATE function using a variable

Posted on 2008-06-21
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Last Modified: 2012-05-05
Hi there,

I'm having trouble using ADDDATE function in PHP. I want to add 2 days to m $edt variable that I set at the top of my page.

I'm guessing it should follow these lines...

ADDDATE(date, INTERVAL 2 DAY).

I can't get the variable $edt to work tho. I get a Parse error: syntax error, unexpected T_LNUMBER

Here's a snippet of my code...
$edt2 = ADDDATE("$edt", INTERVAL 2 DAY);
 
echo $edt2;

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Question by:dm404
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8 Comments
 
LVL 48

Expert Comment

by:hernst42
ID: 21836936
That is SQL syntax in PHP you need something like this:

$edt2 = date('r', strtotime("$edt plus 2 days"));
0
 

Author Comment

by:dm404
ID: 21836959
Hi,

Yep that works although I need the value to be in a YYYY-MM-DD

Thanks in advance,

Daniel
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Author Comment

by:dm404
ID: 21836964
Sorry that didn't work.

It comes up with

Thu, 01 Jan 1970 00:00:00 +0000 !!
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LVL 48

Expert Comment

by:hernst42
ID: 21836971
What is the content of $edt?
0
 
LVL 4

Expert Comment

by:afzz
ID: 21836978
try this
$edt2 = date('Y-m-d', strtotime(strtotime($edt)." plus 2 days"));

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Author Comment

by:dm404
ID: 21837039
$edt is a variable set in two ways.

1) Form is submitted, uses a JS calendar so the value will always be in the format YYYY-MM-DD.
2) Form not submitted takes current date in format YYYY-MM-DD.

afzz your code returns 1970-01-01, which is the correct format just not the correct date.

Thanks again,#
Daniel
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LVL 48

Accepted Solution

by:
hernst42 earned 500 total points
ID: 21837048
Ok in thiscase this works:

$edt = '2008-10-03';
$edt2 = date('Y-m-d', strtotime($edt)+2*86400);
echo $edt2; // 2008-10-05
0
 

Author Comment

by:dm404
ID: 21837056
Yep that works.

Thankyou very much!
0

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