Problem with arrays

[iblamemicrosoft@gmail.com]

What I'm trying to do is check if the dates in the array $holi[]
matches $evalday (not in an array)
but with limited success the data in the array is:

$holi[1] = 2006-06-20;
$holi[2] = 2006-06-28;
$holi[3] = 2006-07-19;
$holi[4] = 2008-06-24;
$holi[5] = 2006-06-30;

$evalday = date("Y-m-d", strtotime("+ 1 days");

if($evalday == $holi) {
echo("For the purpose of this excercise, $evalday is a holiday");
}

Thanks in advance.

[lorento]

Re: Problem with arrays

Try array_search() function:

if (array_search($ evalday, $holly))
echo("For the purpose of this excercise, $evalday is a holiday");


--

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http://www.sederet.com

iblamemicrosoft @gmail.com wrote:[color=blue]
> What I'm trying to do is check if the dates in the array $holi[]
> matches $evalday (not in an array)
> but with limited success the data in the array is:
>
> $holi[1] = 2006-06-20;
> $holi[2] = 2006-06-28;
> $holi[3] = 2006-07-19;
> $holi[4] = 2008-06-24;
> $holi[5] = 2006-06-30;
>
> $evalday = date("Y-m-d", strtotime("+ 1 days");
>
> if($evalday == $holi) {
> echo("For the purpose of this excercise, $evalday is a holiday");
> }
>
> Thanks in advance.[/color]

[iblamemicrosoft@gmail.com]

Re: Problem with arrays

Thank you, in_array($evald ay, $holi) worked. I should have done this
database instead for my school project.

[Bob Stearns]

Re: Problem with arrays

iblamemicrosoft @gmail.com wrote:
[color=blue]
> What I'm trying to do is check if the dates in the array $holi[]
> matches $evalday (not in an array)
> but with limited success the data in the array is:
>
> $holi[1] = 2006-06-20;
> $holi[2] = 2006-06-28;
> $holi[3] = 2006-07-19;
> $holi[4] = 2008-06-24;
> $holi[5] = 2006-06-30;
>
> $evalday = date("Y-m-d", strtotime("+ 1 days");
>
> if($evalday == $holi) {
> echo("For the purpose of this excercise, $evalday is a holiday");
> }
>
> Thanks in advance.
>[/color]
You either have to test each element separately [foreach()
if($holiday = $evalday)...] or use the builtin function
inarray($evalda y, $holi).