Hi Riccardo,
> if(mysql_num_rows($rty->resu)) { //result
> $rec = mysql_fetch_array($rty->resu);
> if(!isset($_SESSION["bkmks"]) ||
When you save something for the first time, the element of $_SESSION["bkmks"] is a
string and you can compare the array $rec with this string.
>!in_array($rec, $_SESSION["bkmks"])) {
Than you convert the element into an array
> $_SESSION["bkmks"][] = $rec;
and AFAIK in_array() canīt compare two arrays, correct me if i am wrong...
Greetings
Jochen
*********** REPLY SEPARATOR ***********
On 22.08.02 at 12:06 Riccardo Sepe wrote:
>Hi every1 I got this script that works fine on my local windows pc but
>on the remote server (FreeBSD)
>I get this message:
>
>Warning: Wrong datatype for first argument in call to in_array in
>/usr/local/......
>
>the script should bookmark an user choice storing it in the
>$_SESSION["bkmks"] array.
>this is the code:
>$rty->mark($id,$tab); // call to the method
>that perform a standard query on the db
> if(mysql_num_rows($rty->resu)) { //result
> $rec = mysql_fetch_array($rty->resu);
> if(!isset($_SESSION["bkmks"]) ||
>!in_array($rec, $_SESSION["bkmks"])) {
> $_SESSION["bkmks"][] = $rec;
>
>when I store for the first time no problem ... When I try to store
>another item or the same item again
>I got that awful error
>
>thanks in advance !
>
> Ricky
>
>
>
>
>
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