jdmonthname function 3

[jisoo23]

Ok so I might be going a little nuts here...

I'm using jdmonthname($var,0) to pick up the abbreviated month names like "Jan", "Feb", etc. Here's the problem:

I feed it the number 11 for the first argument and I end up getting "Dec"(?!). It's very strange...I replaced that variable with a 2 and ended up getting "Nov". Has anyone run into this issue before? My web host company is running PHP 4.4.1 and has calendar support enabled...

Thanks,
Jisoo23

 

[BabyJeffy]

0 = Jan, 1 = Feb ... 11 = Dec

But that doesn't explain why 2 gave you November!

Cheers,
Jeff

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What is Javascript? faq216-6094

 

[sleipnir214]

It could be a bug. Since Julian dates are the number of days and fractions of days since noon GMT on January 1, 4713 BC, both values of "11" and "2" should have produced "Jan".

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[BabyJeffy]

Another situation where my understanding of PHP is unable to rely on prior experience from javascript, java, asp or jsp I've not done *any* date stuff in php... looks like I may have found a reason to start a new project!

Cheers,
Jeff

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[/tt]

What is Javascript? faq216-6094

 

[jisoo23]

That's strange because when I tossed in a "0" and echoed, it didn't print out anything in the html. Looks like I'm going to have to use a case statement for this. How annoying =P

 

[jisoo23]

I just realized why I'm getting this problem, first argument is a julian DAY, not month. I could just drive my head through my desk right now =/ lol. Does anyone know if there's a proper function that will take a month number and give me the corresponding month name?

 

[DRJ478]

Something like this should work:

echo monthName(1);

function monthName($month){
  return date("M", mktime(0, 0, 0, $month, 1, 2000));
}

 

[vacunita]

Can't you use a simple array:

$nameofonth[1]="Jan";
$nameofonth[2]="Feb";
$nameofonth[3]="Mar";
$nameofonth[4]="Apr";
$nameofonth[5]="May";
$nameofonth[6]="Jun";
$nameofonth[7]="Jul";
$nameofonth[8]="Aug";
$nameofonth[9]="Sep";
$nameofonth[10]="Oct";
$nameofonth[11]="Nov";
$nameofonth[12]="Dec";

When you need a name of month just do:

echo $nameofmonth['somenumber'];

You can even create a small function that checks if the number provided is larger than 12 or less than 1. so as to not have any Errors.



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[jisoo23]

DRJ478's suggestion would probably work well. Yeah I could use an array, just thought there'd be a function something as useful as this though. I was thinking along the lines of DR's function anyway...thanks everyone for their suggestions!

 

[SimonSellick]

Could you use your original function but pass to it the month number * 29 as the first argument? That should land somewhere in the month that you want.