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Answer» What I'm trying to do is select once from the dropdown and use the selection
to update the numbersdb database via the submit button. Thanks for your HELP.
Code: [Select]<?php
$dbconnect=mysqli_connect('localhost','root','');
mysqli_select_db($dbconnect,'numbersdb')ordie("Unabletoselectdatabase");
$taxrate=(isset($_POST['submit']))?mysqli_real_escape_string($dbconnect,$_POST['taxrate']):'';
$id=(isset($_POST['id']))?mysqli_real_escape_string($dbconnect,$_POST['id']):'';
$result=mysqli_query($dbconnect,"SELECT*FROMnumbdata");
if(!empty($_POST['update_taxrate'])){
$update=mysqli_query($dbconnect,"UPDATEnumbdataSETtaxrate='$taxrate'WHEREid='$id'");
echo"Taxratehasbeenset...";
}
?>[/color]
Code: [Select][color=red]<!DOCTYPE html>
<html>
<HEAD>
<title>Select taxrate</title>
<style type="text/css">
body {
background: #cff;
}
form {
text-align: center;
}
</style>
</head>
<body>
<form name="taxset" action="<?phpecho$_SERVER['PHP_SELF'];?>" method="post">
<p><label>Select state/rate</label><p>
<select name="taxrate">
<option value="0.04000" SELECTED>4% Alabama</option>
<option value="0.05600">5.6% Arkansas</option>
</select>
</p>
<!--<p><label>Update taxrate</label>
<input type="text" name="update_taxrate">-->
<p><input type="submit" name="submit" value="update"></p>
</form>
</body>
</html>
[/color]
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