I am unable to retrieve the value of $referral_1 from:
$new_email = mysql_escape_string ( $_POST['referral_$i'] );
why?
PHP while lopp to check if any of the fields were populated:
$i=1;
while ( $i <= 5 ) {
$new_email = mysql_escape_string ( $_POST['referral_$i'] );
if ( strlen ( $new_email ) > 0 ) {
....
}
}
The form itself:
<form method="post" action=”website”>
<p style="margin: 10px 50px 10px 50px;"><input type="text" name="email"
maxlength="60" class="referral_1" style="width: 400px;"></p>
<p style="margin: 10px 50px 10px 50px;"><input type="text" name="email"
maxlength="60" class="referral_2" style="width: 400px;"></p>
<p style="margin: 10px 50px 10px 50px;"><input type="text" name="email"
maxlength="60" class="referral_3" style="width: 400px;"></p>
<p style="margin: 10px 50px 10px 50px;"><input type="text" name="email"
maxlength="60" class="referral_4" style="width: 400px;"></p>
<p style="margin: 10px 50px 10px 50px;"><input type="text" name="email"
maxlength="60" class="referral_5" style="width: 400px;"></p>
<p style="margin: 10px 50px 10px 50px;"><input type="submit" name="submit"
value="Add New Referrals"></p>
</form>
What am I doing wrong?
Ron
The Verse of the Day
“Encouragement from God’s Word”
http://www.TheVerseOfTheDay.info
