Hi,
You are really messed up with this 'if' statement.
What you want is nicely described, so I won't repeat it, just make some correction...
function array2links($cat, $arr)
{
$i = 0;
$cat = rawurlencode($cat);
while ( list( $key, $val ) = each($arr))
{
$out .= "<a class=\"cats\" href=\"products.php?Cat=$cat&Code=$key\"
title=\"$val\">$val</a><br>";
if ( $i == 5 ) {
$out .= "</td><td valign=\"top\">";
$i = 1;
}
else $i++;
}
return $out;
}
----- Original Message -----
From: "Joseph Blythe" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: 2001. május 4. 08:02
Subject: [PHP] array 2 links
> Hello,
>
> I seem to be having some difficulty understanding why the following function skips
>every 5th array item,
> basically the following function takes two parameters category and an array, when
>placed inside a table it will generate
> links based on the array items splitting the table into columns when more than 4
>links have been generated.
> However as mentioned every 5th array item is skipped hmm..
>
> The solution is probably really obvious but I can't seem to see it!
>
> function array2links($cat, $arr) {
> $i = 0;
> $cat = rawurlencode($cat);
> while ( list( $key, $val ) = each($arr)) {
>
> $link = "<a class=\"cats\" href=\"products.php?Cat=$cat&Code=$key\"
>title=\"$val\">$val</a><br>";
> if ( $i < 4 ) {
> $out .= "$link";
> } else {
> $out .= "</td><td valign=\"top\">";
> $i = 0;
> continue;
> }
> $i++;
> }
> return $out;
> }
>
> Any help/ideas would be appreciated,
>
> Regards,
>
> Joseph.
>
>
>
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