[RESOLVED] Need Help with date function

[stebut05]

I have a real difficult problem and would really, really apreciate as much halp as possible, thanks.

I am building an ecommerce site where a user can pick a delivery date and delivery zone. However, if a person lives or want their goods shipped to a particular zone then delivery schedules are different. Also, if a user orders before 10am the are eligible for next day delivery if they live in a 'delivery zone' that offers next day delivery:

zone 1 : next day delivery, suturday delivery
zone 2 : next day delivery, suturday delivery
zone 3 : standard deliery (3 days)

Users can pick dates in the future, i.e for delivery in a month etc.

What i need to do is take the delivery date they have choosen then add some sort of function,

delivery date + standard delivery (3days)
delivery date + next day delivery (1 day)

then insert it into the database as a dispatch date.

Hope this makes sense, think i have nw confused myself!!, thanks for any help!!

Regards

Steven

[toplay]

Let's say you figured out which delivery zone they're in and have put the number of days in a variable called $days. Then you can use the DATE_ADD() MySQL function, like so:

$sql = "INSERT INTO table_name SET delivery_schedule_date = DATE_ADD(CURDATE(), INTERVAL $days DAY)";

If you don't want to count Sunday as a delivery date, then you can use the DAYOFWEEK() function too. Example (Sunday is represented by the value 1):

$sql = "INSERT INTO table_name SET delivery_schedule_date =
IF(DAYOFWEEK(DATE_ADD(CURDATE(), INTERVAL $days DAY)) = 1,
DATE_ADD(CURDATE(), INTERVAL " . $days + 1 . " DAY),
DATE_ADD(CURDATE(), INTERVAL $days DAY))";

FYI - There's also the PHP strtotime() function, but I recommend to use MySQL date and time functions whenever possible.

hth.

[bailo81]

I think i can help here, hold on tight while i try to explain!

Firstly you would need some kind of set the amount of time for delivery depending on what they had selected. I assume it will be posted from the previous page so (you would need to set the value that it passed as a number for this to work, so 1 for next day and 3 for standard)...

$delivery_duration = $_POST['delivery_duration'];
$delivery_day = $POST['delivery_day'];
$delivery_month= $POST['delivery_month];
$delivery_year= $POST['delivery_year];

Then you need to work out the date, obviously this is a problem, you cant just add 3 or 1 because if its the end of the month or the year it will mess things up, what you need to do is convert it to days.

Now PHP can do this using the cal_to_jd function. It doesnt convert it to just a day of this year, but that doesnt matter because there's a function to convert it back so first convert it:

$deliv_in_days = cal_to_jd(CAL_GREGORIAN,$delivery_month,$delivery_year,$delivery_year);

Now you have the days in the julian day format so add the delivery time:

$dispatch_in_days = $deliv_in_days + $delivery_duration;

Then convert back, this returns an array:

$dispatch_date_array = cal_from_jd($dispate_in_days,CAL_GREGORIAN;

So then get your dispatch date from the array:

$dispatch = $dispatch_date_array[date];

Then you just need to save it into your db.

More info is available about the functions and how they work in the manual

[stebut05]

Thanks.

However, on DATE_ADD(CURDATE(), function. Is that the current date or the delivery date they have selected. If the user selects a delivery date in say 2 weeks then it would not be current date plus $days. Thanks for your help

Regards,

Steven

[bailo81]

Sorry it took me so long to type that out someone had already replied!

Thinking about it i also think that if they selected a delivery date then you would need to subtract either 1 or 3 to get the date you needed to dispatch

[stebut05]

In reply to bailo81,

I think that might be the solution. However, i am passing the delivery date in a date format ("d/m/Y"), rather than a variable for day, month, year. Is it still possible to use the cal_to_jd function??

Thanks for your help

Regards,

Steven

[toplay]

stebut05 wrote:

Thanks.

However, on DATE_ADD(CURDATE(), function. Is that the current date or the delivery date they have selected. If the user selects a delivery date in say 2 weeks then it would not be current date plus $days. Thanks for your help

Regards,

Steven

Yes, if the user is specifying a delivery date, then you would replace CURDATE() with that actual date (after you've validated it's a valid date of course). Make sure it's in 'YYYY-MM-DD' format though.

FYI:
http://us2.php.net/manual/en/function.checkdate.php

🙂

[toplay]

stebut05 wrote:

In reply to bailo81,

I think that might be the solution. However, i am passing the delivery date in a date format ("d/m/Y"), rather than a variable for day, month, year. Is it still possible to use the cal_to_jd function??

Thanks for your help

Regards,

Steven

Please note that in order to use those Calendar functions on a UNIX/Linux server PHP must have been compiled with --enable-calendar. So, if you're using a web host provider, make sure this is set using phpinfo(). For Windows, the support for these functions is already built in.

You would simply use substr() to get at the different parts of the date (or use regular expressions for more control).

.

[bailo81]

It needs to be entered in the format of (calendar,month,day,year)

so if you can get it to pass it as the full date but (m/d/Y) rather than (d/m/Y)

you could use str_replace to chop out the forward slashes and replace them with commas. For example, your date is the 3rd of December 2005 in the format 12/02/2005 and saved in a variable called $date:

$date = str_replace('/',',',$date);

then $date can be used in the function

[toplay]

If you insist on still doing it in PHP, then you don't necessarily need to use those Calendar functions. PHP's built in timestamp functions will do. Example:

$input_date = '22/09/2005'; // Example input

$input_day = substr($input_date, 0, 2);
$input_month = substr($input_date, 3, 2);
$input_year = substr($input_date, -4);

if (checkdate($input_month, $input_day, $input_year)) {
    $ts = mktime(0, 0, 0, $input_month, $input_day, $input_year);

$days = 3; // Set depending on zone or whatever
$ts = strtotime("+$days day", $ts);

if (0 == date('w', $ts)) {  // Is it Sunday?
    $ts = strtotime('+1 day', $ts); // Make it Monday
}

// Format final date for database YYYY-MM-DD
$schedule_date = date('Y-m-d', $ts);

} else {
    $schedule_date = '';
    echo 'invalid date entered. Correct the date.';
    // Display form again
}

The $schedule_date will be set to '2005-09-26' in this example since the 25th is a Sunday.

hth.

[stebut05]

Thanks for all your help, really appreciate it!! Problem Solved!!

Regards,

Steven