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Show a select record (field)

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Show a select record (field)

Postby kiwi » Wed Sep 08, 2004 7:58 am

this is a drop menu code, i want to select and id from the menu, after select the id, i want to show the name of the selected id, how to do it, can some one help me, thanks

<select name="select">
do {
<option value="<?php echo $row_rsPart['TypeID']?>"><?php echo $row_rsPart['TypeID']?></option>
} while ($row_rsPart = mysql_fetch_assoc($rsPart));
$rows = mysql_num_rows($rsPart);
if($rows > 0) {
mysql_data_seek($rsPart, 0);
$row_rsPart = mysql_fetch_assoc($rsPart);
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Postby gesf » Wed Sep 08, 2004 11:26 am

Code: Select all

    // Prints the 'id' of the selected option
    print $_POST['mysel'];
    // Display select menu

function ShowSel() {
   $result = mysql_query("SELECT id, name FROM table ORDER BY id");

   print '<form name="f" method="post" action="">' . "\n";   
   print '<select name="mysel">' . "\n";   
   while($row = mysql_fetch_array($result)) {
      print '<option value="' . $row['id'] . '">' . $row['name'] . '</option>' . "\n";
    print '</select>' . "\n";
    print '<input type="submit" name="DoIt" value="View">' . "\n";
    print '</form>' . "\n";


You can put the $row['name'] in the value part of the option instead of the id.

If you realy need the id, you can use javascript for that by adding a hidden input to the form:
Code: Select all
<input type="hidden" name="myname" value="">
And in your select option add something like:
Code: Select all
<select onchange="document.f.myname.value = this.options[this.selectedIndex].text">
So everytime you choose an option, your hidden input will have the respective 'name' that you can 'get' with:
Code: Select all

print $_POST['myname'];


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