On Tue, 2009-07-28 at 09:32 -0400, Miller, Terion wrote:
I need to take this:
$pastDays = strtotime(-30 days);
$past_day = date(d, $pastDays);
$past_month = date(m, $pastDays);
$past_year =date(y, $pastDays);
And make it into one var to compare to a db field that is
On Tue, Jul 28, 2009 at 6:32 AM, Miller,
Teriontmil...@springfi.gannett.com wrote:
I need to take this:
$pastDays = strtotime(-30 days);
$past_day = date(d, $pastDays);
$past_month = date(m, $pastDays);
$past_year =date(y, $pastDays);
And make it into one var to compare to a db
On 7/28/09 8:35 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
$pastDays = strtotime(-30 days);
$date = date(d/m/y, $pastDays);
Well I tried and got no results from my query and I know there results with
date ranges in the last 30 days, I basically need to count backward from now()
30
From: Ashley Sheridan
On Tue, 2009-07-28 at 09:32 -0400, Miller, Terion wrote:
I need to take this:
$pastDays = strtotime(-30 days);
$past_day = date(d, $pastDays);
$past_month = date(m, $pastDays);
$past_year =date(y, $pastDays);
And make it into one var to compare to a
On Tue, 2009-07-28 at 09:42 -0400, Miller, Terion wrote:
On 7/28/09 8:35 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
$pastDays = strtotime(-30 days);
$date = date(d/m/y, $pastDays);
Well I tried and got no results from my query and I know there results with
date ranges in
On 7/28/09 8:41 AM, Bob McConnell r...@cbord.com wrote:
From: Ashley Sheridan
On Tue, 2009-07-28 at 09:32 -0400, Miller, Terion wrote:
I need to take this:
$pastDays = strtotime(-30 days);
$past_day = date(d, $pastDays);
$past_month = date(m, $pastDays);
$past_year =date(y,
On Tue, 2009-07-28 at 09:41 -0400, Bob McConnell wrote:
From: Ashley Sheridan
On Tue, 2009-07-28 at 09:32 -0400, Miller, Terion wrote:
I need to take this:
$pastDays = strtotime(-30 days);
$past_day = date(d, $pastDays);
$past_month = date(m, $pastDays);
On 7/28/09 8:44 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
On Tue, 2009-07-28 at 09:42 -0400, Miller, Terion wrote:
On 7/28/09 8:35 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
$pastDays = strtotime(-30 days);
$date = date(d/m/y, $pastDays);
Well I tried and got no
On Tue, 2009-07-28 at 06:46 -0700, Miller, Terion wrote:
On 7/28/09 8:44 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
On Tue, 2009-07-28 at 09:42 -0400, Miller, Terion wrote:
On 7/28/09 8:35 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
$pastDays = strtotime(-30
On Tue, Jul 28, 2009 at 9:46 AM, Miller,
Teriontmil...@springfi.gannett.com wrote:
On 7/28/09 8:44 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
On Tue, 2009-07-28 at 09:42 -0400, Miller, Terion wrote:
On 7/28/09 8:35 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
$pastDays
On 7/28/09 8:52 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
On Tue, 2009-07-28 at 06:46 -0700, Miller, Terion wrote:
On 7/28/09 8:44 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
On Tue, 2009-07-28 at 09:42 -0400, Miller, Terion wrote:
On 7/28/09 8:35 AM, Ashley Sheridan
On Tue, 2009-07-28 at 09:42 -0400, Miller, Terion wrote:
On 7/28/09 8:35 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
$pastDays = strtotime(-30 days);
$date = date(d/m/y, $pastDays);
Well I tried and got no results from my query and I know there results with
date
On Jul 28, 2009, at 9:55 AM, Miller, Terion wrote:
On 7/28/09 8:52 AM, Ashley Sheridan a...@ashleysheridan.co.uk
wrote:
On Tue, 2009-07-28 at 06:46 -0700, Miller, Terion wrote:
On 7/28/09 8:44 AM, Ashley Sheridan a...@ashleysheridan.co.uk
wrote:
On Tue, 2009-07-28 at 09:42 -0400,
From: Miller, Terion
On 7/28/09 8:35 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
$pastDays = strtotime(-30 days);
$date = date(d/m/y, $pastDays);
Well I tried and got no results from my query and I know there
results with date ranges in the last 30 days, I basically need
to count
,snip
You can also do this right within MySQL without needing to create a
variable. This should work:
$sql = SELECT DISTINCT restaurants.ID, name, address, inDate FROM
restaurants, inspections WHERE restaurants.name != '' AND
datediff(curdate(),inspections.inDate)=30 GROUP BY restaurants.ID
On Tue, Jul 28, 2009 at 10:34 AM, Bob McConnellr...@cbord.com wrote:
From: Miller, Terion
On 7/28/09 8:35 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
$pastDays = strtotime(-30 days);
$date = date(d/m/y, $pastDays);
Well I tried and got no results from my query and I know there
On 7/28/09 9:40 AM, Bastien Koert phps...@gmail.com wrote:
On Tue, Jul 28, 2009 at 10:34 AM, Bob McConnellr...@cbord.com wrote:
From: Miller, Terion
On 7/28/09 8:35 AM, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
$pastDays = strtotime(-30 days);
$date = date(d/m/y, $pastDays);
Well
On Tue, Jul 28, 2009 at 10:40 AM, Miller,
Teriontmil...@springfi.gannett.com wrote:
,snip
You can also do this right within MySQL without needing to create a
variable. This should work:
$sql = SELECT DISTINCT restaurants.ID, name, address, inDate FROM
restaurants, inspections WHERE
On Tue, Jul 28, 2009 at 10:43 AM, Miller,
Teriontmil...@springfi.gannett.com wrote:
On 7/28/09 9:40 AM, Bastien Koert phps...@gmail.com wrote:
On Tue, Jul 28, 2009 at 10:34 AM, Bob McConnellr...@cbord.com wrote:
From: Miller, Terion
On 7/28/09 8:35 AM, Ashley Sheridan
Bastien Koert wrote:
On Tue, Jul 28, 2009 at 10:43 AM, Miller,
Teriontmil...@springfi.gannett.com wrote:
On 7/28/09 9:40 AM, Bastien Koert phps...@gmail.com wrote:
On Tue, Jul 28, 2009 at 10:34 AM, Bob McConnellr...@cbord.com wrote:
From: Miller, Terion
On 7/28/09 8:35 AM, Ashley Sheridan
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