选择从数据库中读取的选项菜单并使用它的值

Question

<?php
mysql_connect("localhost", "root","") or die(mysql_error());
mysql_select_db("tnews2") or die(mysql_error());

$query = "SELECT name,id FROM categories ORDER BY ID DESC LIMIT  0,6";
$result = mysql_query($query) or die(mysql_error()."[".$query."]");?>

    <select name="categories">
<?php while ($row = mysql_fetch_array($result)){
?>
   <option value=" <?php $row['path']; ?> ">
     <?php echo $row['name']; ?>
    </option>
<?php
}
?>        
</select>?>

所以这是选择选项菜单,它是我从数据库中读取的值,但是当我尝试获取所选值时,我只得到第一个.

<?php
mysql_connect("localhost", "root","") or die(mysql_error());
mysql_select_db("tnews2") or die(mysql_error());


   if(!empty($_POST['title']) && !empty($_POST['date']) && !empty($_POST['txt']) && !empty($_POST['image'])){
$TITLE=$_POST['title'];
$DATE=$_POST['date'];
$TXT=$_POST['txt'];
$IMAGE=$_POST['image'];
$CATEGORIES=$_POST['categories'];
echo $CATEGORIES;
$ANSWER=$_POST['main'];
$MAINPAGE=0;

你能帮助我一个想法来获得所选的选项吗？

Answer 1

编辑:(减少你的代码)

<?php
mysql_connect("localhost", "root","") or die(mysql_error());
mysql_select_db("tnews2") or die(mysql_error());

$query = "SELECT name,id,path FROM categories ORDER BY ID DESC LIMIT  0,6";
$result = mysql_query($query) or die(mysql_error()."[".$query."]");
?>

<select name="categories">
<?php 
while ($row = mysql_fetch_array($result))
{
    echo "<option value='".$row['path']."'>'".$row['name']."'</option>";
}
?>        
</select>