Reply to Re: What is wrong with this if test?

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Posted by Shelly on 12/29/05 04:39

Oops, here is the output:

selHint_2=1 index=0 option value="0" selected Family member's name
index=1 option value="1" Mothers maiden



"Shelly" <sheldonlg.news@asap-consult.com> wrote in message
news:Z5mdnTRRsdh81y7eRVn-oQ@comcast.com...
> OK, I am at a toral loss after having programmed for forty years.
>
> Here are my results:
>
> selHint_2=1
> It is clear that for the second time through the loop ($i=1) that
> $index=1 and $selHint_2=1, while for the first pass through ($i=0)
> $index=0 and $selHint=1. Yet, the " selected " comes up for the first
> time and not for the second. It should be the other way around. Any
> clues?
>
> Shelly
>
>
> Here is the code:
> $numHint_2 = 2;
> $hint_2_Lookup = array(0, 1);
> $hint_2 = array();
> $hint_2[] = " Family member's name";
> $hint_2[] = " Mothers maiden name";
> $foo = ""; // put in for debugging only
> $tmpHint2 = array();
> $tmpHint2[] = '<td class="stdBold"' . $readonly . '>';
> $tmpHint2[] = '<select name="hint_1" value="" >';
> for ($i=0; $i<$numHint_2; $i++) {
> $index = $hint_2_Lookup[$i];
> if ($index == $selHint_2) {
> $sel = " selected ";
> } else {
> $sel = "";
> }
> $tmpHint2[] = '<option value="' . $index . '"' . $sel . '>' .
> $hint_2[$i] . '</option>';
> // next line put in for debugging only
> $foo .= 'index=' . $index . ' option value="' . $index . '"'
> . $sel . $hint_2[$i]. "<br>";
> }
> $tmpHint2[] = '</select></td>';
>
> $selHint_2 = $_SESSION['vmHint_2']; // put in for
> debugging only
> echo "selHint_2=" . $selHint_2 . "<br>"; // put in for
> debugging only
> echo implode("\n",$tmpHint2)."\n";
> echo $foo; // put in for debugging only
>
> index=0 option value="0" selected Family member's name
> index=1 option value="1" Mothers maiden
>
>
>
>
>

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