Undefined offset: 1 - what does it mean?

[royp2000]

Hello,

I'm getting the following error when running one of my php scripts:

Undefined offset: 1 in /ftp/virt/techno/classifieds/advertisement.php on line 530

I checked the folowing line (530) and all I had in it was:

$curent_cat_id=$fourth_id_cut[1];

What can cause this error and what does it mean?

Thanks in advance,

Roy

[Distel]

I found something regarding these errors, which seem to be related to calendars. They suggested a CVS update, whatever the heck that is. ;)

[trajan]

are you sure that your array-field "$fourth_id_cut[1]" is really set?

in my opinion this notice means that the array value of $fourth_id_cut at key 1 is undefined...

[jatar_k]

Welcome to WebmasterWorld trajan,

I was thinking exactly that as well.

[royp2000]

Yea, I'm sure it is set:

529: $fourth_id_cut=explode("=",$third_id_cut);
530:$curent_cat_id=$fourth_id_cut[1];

Any other ideas?

Thanks,

Roy

[coopster]

Try this to make sure:

529: $fourth_id_cut=explode("=",$third_id_cut); 
529 1/2: print '<pre>'; print_r($fourth_id_cut); exit('</pre>'); 
530:$curent_cat_id=$fourth_id_cut[1]; 

[globay]

are you sure, that there is always a something after "=" in $third_id_cut? If $third_id_cut does not contain =, $fourth_id_cut[1] won't be set.
in $fourth_id_cut=explode("=",$third_id_cut);

And yes, Undefined offset: 1 indicates that $fourth_id_cut[1] is not set

[royp2000]

coopster: I can't enter this line, it causes errors.

globay: Now that you've mentioned it, maybe there is one chance that there is nothing after the "=".

Hmmm... I guess that if I'll enter the following line, it should sove it, isn't it:

$fourth_id_cut=explode("=",$third_id_cut);
if (isset($fourth_id_cut))
{
$curent_cat_id=$fourth_id_cut[1];
}

Thanks,

Roy.

[jatar_k]

$fourth_id_cut=explode("=",$third_id_cut);
if (isset($fourth_id_cut[1])){
$curent_cat_id=$fourth_id_cut[1];
}

I would think you want to test the 2nd element to see if it is set.

[royp2000]

o.k I'll try that.

Thanks.