What is "Notice: Undefined offset"

I have usually my notices turned off, but I thought what the hell, lets improve my code so I dont get any notices in the first
place.

I have got rid of all notices except this one. It repeats many, many times because it's in a loop:

Notice: Undefined offset: 3 in /var/www/html/project_html/search_result_summary.php on line 182 and
Notice: Undefined offset: 3 in /var/www/html/project_html/search_result_summary.php on line 184

Lines 182 and 184 are the first two 'if' lines in the following code


if (sizeof($dive_site_array[$field_number]) == 0 || !in_array($result_array[$row_number][$field_number],
$dive_site_array[$field_number]))
{
if (sizeof($dive_site_array[$field_number]) == 0)
{
$dive_site_array[$field_number][$col_number] = $result_array[$row_number][$field_number];
}
else
{
array_push($dive_site_array[$field_number], $result_array[$row_number][$field_number]);
}
}

Can anyone explain it for me? I can't find an explanation for it on the net.
Thanks.

Laurie

"laurie" <lawr_1@hotmail.com> wrote in message
news:<l4S0b.434$L15.46@newsfep4-winn.server.ntli.net>...
>
> I have got rid of all notices except this one. It repeats many, many
> times because it's in a loop:
>
> Notice: Undefined offset: 3
> in /var/www/html/project_html/search_result_summary.php on line 182 and
> Notice: Undefined offset: 3
> in /var/www/html/project_html/search_result_summary.php on line 184

Undefined offset means that you are referencing an element which
doesn't exist in an array. Say, I have this variable:

$myArray = array ('Zeroth', 'First', 'Second');

This array has three elements, which by default are numbered 0, 1,
and 2. If I attempt something like this:

$myElement = $myArray[3];

it will trigger the undefined offset error.

Cheers,
NC

It is sometimes hard to switch languages. The way to check the bounds of
a PHP array is count() not sizeof.

you can test isset()m is_array and then for a loop the break variable
should be count(array). If the array size is likely to be more than a
few, get the count into a local variable rather than calling it on every
loop for performance reasons.

When you run into this sort of problem, break the if into two pieces so
you can see which is the culprit. It sounds like you are referencing an
array element out of bounds. But the easy thing to do is to get the count
on the array and the value of $field_number and see if the latter is out
of bounds.

"laurie" <lawr_1@hotmail.com> wrote in
news:l4S0b.434$L15.46@newsfep4-winn.server.ntli.net:

> I have usually my notices turned off, but I thought what the hell,
> lets improve my code so I dont get any notices in the first place.
>
> I have got rid of all notices except this one. It repeats many, many
> times because it's in a loop:
>
> Notice: Undefined offset: 3 in
> /var/www/html/project_html/search_result_summary.php on line 182 and
> Notice: Undefined offset: 3 in
> /var/www/html/project_html/search_result_summary.php on line 184
>
> Lines 182 and 184 are the first two 'if' lines in the following code
>
>
> if (sizeof($dive_site_array[$field_number]) == 0 ||
> !in_array($result_array[$row_number][$field_number],
> $dive_site_array[$field_number])) {
> if (sizeof($dive_site_array[$field_number]) == 0)
> {
> $dive_site_array[$field_number][$col_number] =
> $result_array[$row_number][$field_number];
> }
> else
> {
> array_push($dive_site_array[$field_number],
> $result_array[$row_number][$field_number]);
> }
> }
>
> Can anyone explain it for me? I can't find an explanation for it on
> the net. Thanks.
>
> Laurie
>