UNIX timestamps

[Dean C]

I worked out thanks to someone here at .org how to put a unix timestamp into a readable format with the date() function

Now say if im inserting time() into a field in a table

e.g.

[sql]
UPDATE timer SET time='".time()."';
[/sql]

How would i adapt that query to add a year to the current time?

Thanks in Advance

- miSt

[Dean C]

WTF bug in the sql code alert

It was meant to read:

UPDATE timer SET time='"time()'";

- miSt

[Xenon]

erm. '". isn't a sql query, you are trying to mix up php and mysql so you should use php or code tags

updateing timefield to add a year:
[sql]UPDATE timer SET time=time+3600*24*356[/sql]

[SiGmA_X]

Hang on, time() doesn't have a year in it? It says it does, and in practice seems to, because it records the amount of seconds from the Epoch, which would include a year. Right? I think miSt's idea of just using time() is correct.

But that's just my 2 cents

[filburt1]

time() returns a date in epochs which implicitly has the year.

[Dean C]

I had to do this in the end

PHP Code:

$DB_site->query("UPDATE secrettable SET secretfield=(time()+31536000) WHERE userid=$userid"); 


- miSt

[SiGmA_X]

What does the +3153600 do? Being the time is from the Epoch, it will have year.. I am still unsure why you are adding the extra numbers in :\

[Dean C]

It simply adds a year to the current timestamp outputted from the time() function

- miSt