[PHP] Date Problem - Last Day Of Month
Hi there, can anyone tell me how to fix my code so that on the last day of
the month, my code doesn't repeat the months...
What the code suppose to do is, makes a drop down box from which you can
select the month, and if its the current month
the box is already selected.
select name=month
?
for ($i=1; $i=12; $i++) {
if ($i == date(m)) {
print option selected value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) . /option;
} else {
print option value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) . /option;
}
}
?
/select
Thanks
Vinesh
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Re: [PHP] Date Problem - Last Day Of Month
Hi there, can anyone tell me how to fix my code so that on the last day of
the month, my code doesn't repeat the months...
I'm not sure what you mean by this, but I can be a moron on this list
sometimes and the clear answer usually comes to me about 2 seconds after I
hit the send button. Do you mean that you don't want any months in the
SELECT control when it's the last of the month? Or you don't want the
SELECTED attribute in the OPTION tags? As far as I can tell, you're not
repeating any months, but you're only listing them once.
Try this (and IMHO a *slight* *slight* improvement in clarity, but again,
that's my HO):
select name=month
?
for ($counter=1; $counter=12; $counter++) {
print option value=\. $counter . \;
if (($counter == date(m)) || (date(d) date(t))) {
print selected;
}
print . date(F,mktime(0,0,0,$counter,date(d),date(Y))) .
/option;
}
}
/select
I adjusted it so that you only print out the SELECTED attribute if it's the
current month, and I called $i as $counter, because, to me anyway, it makes
it clearer what is the $counter. If $i works for you, though, there's
nothing wrong with that, and it's not that it's unclear anyway, what with it
in the FOR loop only a few lines earlier. (I'll shut up now about $counter.)
Anyway, I added a comparison above that said if the current date is less
than the last day of the month, then print SELECTED. That's assuming I have
your question right in that you don't want to select the current month if
it's the last day of that month. Otherwise, if I'm wrong, you can take that
comparison and put it just about anywhere.
Help? Doesn't? Am I a complete goof?
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Re: [PHP] Date Problem - Last Day Of Month
change
for ($i=1; $i=12; $i++)
to
for ($i=1; $i12; $i++)
--- Vinesh Hansjee [EMAIL PROTECTED] wrote:
Hi there, can anyone tell me how to fix my code so
that on the last day of
the month, my code doesn't repeat the months...
What the code suppose to do is, makes a drop down
box from which you can
select the month, and if its the current month
the box is already selected.
select name=month
?
for ($i=1; $i=12; $i++) {
if ($i == date(m)) {
print option selected value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) .
/option;
} else {
print option value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) .
/option;
}
}
?
/select
Thanks
Vinesh
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Re: [PHP] Date Problem - Last Day Of Month
- Original Message -
From: Vinesh Hansjee [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, March 31, 2003 6:55 AM
Subject: [PHP] Date Problem - Last Day Of Month
Hi there, can anyone tell me how to fix my code so that on the last day of
the month, my code doesn't repeat the months...
What the code suppose to do is, makes a drop down box from which you can
select the month, and if its the current month
the box is already selected.
select name=month
?
for ($i=1; $i=12; $i++) {
if ($i == date(m)) {
print option selected value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) . /option;
} else {
print option value=\. $i . \ .
date(F,mktime(0,0,0,$i,date(d),date(Y))) . /option;
}
}
?
/select
Thanks
Vinesh
Vinesh, there's a fundemental flaw in your logic. The date(d) input will
fail to produce a reliable list if it is executed on any day after the 28th
of the month. Since today is the 31st obviously this is going to cause some
problems becuase any month that ends on the 30th or earlier will return plus
one the month. So February will return March, April willl return May, and
so on. If you wait until tommorrow (the 1st) your function will magically
work. :)
One way to fix this is to build your own array of month names and reference
that instead of using the date() function.
$months = array(1=January, 2=February, 3=March, 4=April,
5=May, 6= ... and so on...);
for($i=1; $i=12; $i++)
{
$selected = ;
if ($i == date(m));{$selected = SELECTED;}
echo option $selected value=\${months[$i]}\\n;
}
HTH,
Kevin
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