heres what i got so far..
im trying to go through everything to make its all right, but im missing something and its not updating the database. showing the results works in the code, just not inserting.. any ideas?
thanks
<?php
if($mode == "share1") {
$query2 = mysql_query("SELECT * FROM member WHERE member_id = '$session_member_id' and member_status = 'A'");
while($data = mysql_fetch_object($query2)) {
if(${"showclick" . $data->member_id} == N) {
$sql = "UPDATE member SET user_show='N' WHERE member_id = '$data->member_id'";
$result = mysql_query($sql);
}
if(${"showclick" . $data->member_id} == Y) {
$sql = "UPDATE member SET user_show='Y' WHERE member_id = '$data->member_id'";
$result = mysql_query($sql);
}
}
}
?>
<html>
<form name="ushow" method="post" onsubmit="return share1();" action="<?php print($PHP_SELF); ?>?mode=share1">
Allow to let others view your shared images? (default is NO)<br>
<?php if($data->user_show == "Y") { ?>
You currently have selected Yes<br>
<?php } else { ?>
You currently have selected No<br>
<?php } ?>
<br>
Would you like to change to <?php
if($data->user_show == "N") { ?>
Yes<input type="radio" name="showclick<?php print($data->member_id)?>" value="Y">
<?php } else { ?>
No<input type="radio" name="showclick<?php print($data->member_id)?>" value="N">
<?php } ?>
<input type="image" src="./images/bu.update.gif" border="0" name="update_code">
</form>
