insert multiple drop downs question

sal

Hi everyone,

I have a form with one drop down field from table1 which I insert into table2.

Here is the code I use

$sql = "SELECT category FROM catlisting";
$sql2 = mysql_query($sql);
//echo "$sql<br>";
while ($i = mysql_fetch_array($sql2))
{
$category = $i["category"];
$result .= "<OPTION value=\"$category\">$category</OPTION>";
}

Say I want to list the same drop down twice and insert both values from 2 drop downs into table2.

I know I have to loop but I already have one loop for listing the values inside the drop down.

I tried

for ($result = 0; $result < 3; $result++)

in a variety of positions but so fat no luck.

Any suggestion?

Thanks

bastien

select box? then the users can choose one ro more options...then just a small change in the code to loop through the array that the select box submits

sal

Thanks Sebastien,

Actually I want to duplicate same drop down box taking the values from the same field in table1 and insert them in table2.

Drop1 (value1,value2)
Say I select value1

Repeat same dropdown

Drop2 (value1,value2)
And I select value2

Then insert both values into table2

I know how to do the first drop down and insert. I am having problems with more than one drop downs.

bastien

then just reference the name

$val1=$POST['select1'];
$val2=$POST['select2'];

hth

sal

Here is a code I am workink on. It inserts just one record and not the other 2. I don't see where it is wrong!

$sql = mysql_query("select category from catlisting") or die("Unable to process queries at this time: " . mysql_error());
//echo "$sql<br>";
echo "$sql<br>";
while ($i = mysql_fetch_array($sql))
{
$category = $i["category"];
echo "$category<br>";
$result .= "<OPTION value=\"$category\">$category</OPTION>";
}
?>
<form action="test.php" method="post">
<?
for($i=0;$i<3;$i++){
$i = $i + 1
?>
<input name=category type=hidden class=input value="<?php echo $i[category]; ?>" size="50" maxlength="70">
<select class=select name=category[]>
<option value="" >Select category<? echo "$result"; ?>
</option>
</SELECT>

<?
}
?>
<INPUT TYPE=SUBMIT VALUE="Enter" ID="category">

</FORM>

<?
foreach $_POST["category"] as $newcat {

// if ($category !="") {
$sql1 = "INSERT INTO $mysql_cat (id, username, category) "." VALUES "." ('', '$username', "$newcat")";
echo "$sql1<br>";

$result = mysql_query($sql1) or die("ERROR: cannot INSERT data into database");
// }
}
?>

cgraz

Looking through your code, what stood out was your for loop

<? 
for($i=0;$i<3;$i++){ 
$i = $i + 1 
?>

Now you said this only inputs one record, not the other two. Well your for loop, you start out with $i equaling 0. PHP checks to see if $i < 3 is still true, if it is, it runs what is between the brackets.

well you're taking $i, and adding 1 to it. So now, $i equals 1 (you've already skipped 0). Then when the for loop is over, $i increments ($i++), so now $i equals 2. For loop starts over. $i is still less than 3, so the for loop executes what is between brackets. Well in between the brackets, you take $i (which equals 2) and add 1 to it. Now $i equals 3, but you originally specified that $i was less than 3. It was true when the for loop originally checked it, that's why it was executed, but if you're wanting to input when $i=0, $i=1, and $i=2, it's really only inputting when $i=1.

Just change that part of the for loop to

<? 
for($i=0; $i<3; $i++){ 
?>

(no need to add 1 to $i).

And another thing with your insert. is the field id set to auto increment? if so, you don't need to specify that you're inserting into that field, mysql will increment it automatically. Also, watch the double quotes around $newcat. Should be single quotes.

$sql1 = "INSERT INTO $mysql_cat (username, category) VALUES ('$username', '$newcat')";

Cgraz