Unable to post results of selected droplist item using database ....

[robinhood]

Can any 1 help me on my mistakes.I suppose i had declare the :".$_POST['drop1']; wrongly .i am trying to display the selected items form the drop list using post method but it failed. 

 

//populating rank/name from admin table to drop down list

$query="SELECT name FROM admin where timecounter='1'";

 

/*order by clause to the sql statement if the names are to be displayed in alphabetical order */

 

$result = mysql_query ($query);

echo "<select name='drop1'Select Name>";

   

// printing the list box select command

 

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt

echo "<option value=$nt[id]>$nt[name]</option>";

 

/* Option values are added by looping through the array */

}

echo "</select>";// Closing of list box

?>

  </p>

  <p>

    <input type="submit" name="Submit" value="Submit" >

 

  </p>

</form>

</body>

<?php echo "Name is:".$_POST['drop1']; ?>

<p></p>

<p></p>

<?php echo "Product Description:".$_POST['jobdescription']; ?>

</html>

[gerkintrigg]

put the </body> tag at the bottom and seperate the echo statement from the rest of the code with :

 

if ($submit){

 

[robinhood]

i am using 2 post methods but only  of drop list was not displaying output.

 

how to go about using

if ($submit){

 

[GingerRobot]

Something like:

 

<?php
$query="SELECT name FROM admin where timecounter='1'";
/*order by clause to the sql statement if the names are to be displayed in alphabetical order */
$result = mysql_query ($query) or die(mysql_error());
echo "<select name='drop1'";
     
// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt
echo '<option value='.$nt['id'].'>'.$nt['name'].'</option>';

/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box 
?>
  </p>
  <p>
    <input type="submit" name="Submit" value="Submit" >
   
  </p>
</form>
<?php
if(isset($POST('Submit'))){
echo 'Name is: '.$_POST['drop1'].'<br />Product Description: '.$_POST['jobdescription'];
}
</body>
</html>

 

Edit: You do however realise that you will be getting the ID from the drop down box and not the name? Since you set the value of each option to the id rather than the name, that is what you will receive. This is probably what you want if you are doing anything with a database, but not if you want to display the name.