You should do date arithmetic using the Unix date format (number of
seconds since 1970).
I think these string format dates are not appropriate.
Mario
Merlin wrote:
Hi there,
I am trying to create a date which is 25 years back from today. The
purpose of this is to be able to query a mysql databas
Merlin wrote:
Hi there,
I am trying to create a date which is 25 years back from today. The
purpose of this is to be able to query a mysql database "date" field for
columns smaller than this date.
I tried this:
$years = 25;
$start_from = date("Y-m-d",strtotime("- ".$years." year"));
Somehow it a
On Wednesday 15 December 2004 03:21, Merlin wrote:
> I am trying to create a date which is 25 years back from today. The purpose
> of this is to be able to query a mysql database "date" field for columns
> smaller than this date.
>
> I tried this:
> $years = 25;
> $start_from = date("Y-m-d",strtoti
Merlin wrote:
Hi there,
I am trying to create a date which is 25 years back from today. The
purpose of this is to be able to query a mysql database "date" field for
columns smaller than this date.
I tried this:
$years = 25;
$start_from = date("Y-m-d",strtotime("- ".$years." year"));
Somehow it a
Merlin wrote:
Hi there,
I am trying to create a date which is 25 years back from today. The
purpose of this is to be able to query a mysql database "date" field for
columns smaller than this date.
I tried this:
$years = 25;
$start_from = date("Y-m-d",strtotime("- ".$years." year"));
Somehow it a
Richard Lynch wrote:
Merlin wrote:
I am trying to create a date which is 25 years back from today. The
purpose of
this is to be able to query a mysql database "date" field for columns
smaller
than this date.
I tried this:
$years = 25;
//> $start_from = date("Y-m-d",strtotime("- ".$years." year"));
Merlin wrote:
> I am trying to create a date which is 25 years back from today. The
> purpose of
> this is to be able to query a mysql database "date" field for columns
> smaller
> than this date.
>
> I tried this:
> $years = 25;
//> $start_from = date("Y-m-d",strtotime("- ".$years." year"));
>
> S
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