silly me
upon further looking
<INPUT TYPE="checkbox" NAME="<? echo $search_result2['user_cd'];?>"
VALUE="<? echo $search_result2[a.retired_flag];?>"
<?if ($search_result2[a.retired_flag] == 1) echo " CHECKED"; ?>>
<?echo $search_result2[a.retired_flag]?>
>
----- Original Message -----
From: "Amanda McComb" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, August 06, 2003 2:47 PM
Subject: [PHP] bug in code - can't find it!
> Ok, after all of the discussion on posting, I'm afraid to post.
> Unfortunately, no one I know is a programmer, and I am a beginner. I can
> normally figure out what I've done wrong, but I'm missing something here.
>
> Ok, here is my query and the bit of code I can't figure out:
>
> $query = "SELECT * from apt_user_t a, apt_company_t b ";
> $query .= "WHERE a.user_cd = b.user_cd ";
> $query .= "ORDER BY a.username";
>
> $search_results = mysql_query($query) or die("Select
> Failed!");
> while ($search_result2 = mysql_fetch_array($search_results))
> {
> <INPUT
> TYPE="checkbox"
> NAME="<? echo $search_result2['user_cd'];?>"
> SIZE="20"
> MAXLENGTH="50"
> VALUE="<?if ($search_result2[a.retired_flag] == 1){?>CHECKED<?}?>"> <?echo
$search_result2[a.retired_flag]?>
> }
> Nothing shows up with the echo or the value. I only included this
> checkbox, but all of the other values show up fine. Can someone give me a
> hint?
>
>
>
>
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