O Joy Multiple Select and PHP!!!

bakaneko

Hi,

I am trying to pass a multiple select form into my php, but no value is showing up. I tried looking online but there isnt much help out there about these two subjects combined. Can someone plz help. I dont know what i am doing wrong. i believe it is a fairly simple fix but yea simplicity is very deceptive.

For my form page I have:

function multiple_list($tablename, $field) {

    echo "<select multiple name = '$deletename'>";
    echo "<option value= '$value'>$name\n";

    echo "<input type = 'hidden' name ='delete'  value = '$deletename'>";
echo "</select>";
return;

}

/ I have take some php stuff out of here for simplicity sake /

For my php page I have:

$tablename = $_POST['delete'];
echo "$tablename";

/ plus lots of other php stuff /

I just dont see where I am going wrong. I use the input type in the top to just test but it doesnt work without it as well.

Thanks for any help,

Bakaneko :p

HalfaBee

try putting some debug statements in to see what is happening.

echo '<PRE>';
print_r( $_POST );

is a good start.

HalfaBee

tbach2

what is the value of '$deletename'? (I'm not asking what you think it should be, but what is it really?)

answer that and we can discuss what it needs to be 🙂

bakaneko

o... hehe, i see. it is nothing. but, how with out a post or input type how can it be submitted to the other php page? 😕

tbach2

You have a problem with variable scope: variables inside a function are local to that function. In order to get the variable into the function, you must either pass it in or global $variable_name;

get that working, we'll talk more...

Frederick

You have conflicting tags in your select.

echo "<form action=\"where\" method=\"post\">
<select  name\"$deletename\" size=\"1\">";

$sql = mysql_query("SELECT * FROM $tablename")
or die(mysql_error());
while ($a = mysql_fetch_array($sql)) {
    echo "<option value=\"$value\"></option>";
}

echo "</select>&nbsp&nbsp
<input type=\"submit\" name=\"delete\" value=\"Delete\">
</form>";