How to populate a dropdown menu in ahtml form - php

Question

I'm trying to populate a dropdown field inside a html form, but the field doesn't show any value, only a blank one.
I'm trying the following code:
<select name="Select" class="textfields" id="prods">
<option id="0">--Producto--</option>
<?php
require("conectdb.php");
$allproducts = mysql_query("SELECT * FROM Productos");
while ($viewallproducts = mysql_fetch_array($allproducts)){
?>
<option id="<?php echo $viewallproducts["ID"];?>"><?php echo $viewallproducts["CODIGO"];?></option>
<?php } ?>
</select>
I have changed the quotes and ;... But still nothing, here is the code of connection to the Database (conectdb.php):
<?php
$con=mysqli_connect("localhost","xxxxxx","xxxxxx","xxxxxx");
// Check connection
if (mysqli_connect_errno())
echo "Failed to connect to MySQL: " . mysqli_connect_errno();
mysqli_close($con);
?>
It seems to be a problem of the database connection, Im trying now the following code to see what happens:
<?php
$username = "xxxxx";
$password = "xxxxx";
$hostname = "xxxx";
$dbhandle = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
echo "Connected to MySQL<br>";
$allproducts = mysql_query("SELECT * FROM PRODUCTOS");
while ($viewallproducts = mysql_fetch_array($allproducts)){
echo ($viewallproducts);
}
?>
But, on this php, I receive the following error:
Warning: mysql:fetch_array() expects parameter 1 to be resource.
I have also tried with mysql_fetch_row() and gives me the same error.

Answer 1

1) You have error in require line(put file name inside quotation )
2) you don't put value attribute in option.
Try this:
<select name="Select" class="textfields" id="prods">
<option id="0" value="o">--Producto--</option>
<?php
require("conectdb.php");
$allproducts = mysql_query("SELECT * FROM Productos");
while ($viewallproducts = mysql_fetch_array($allproducts)){
?>
<option id="<?php echo $viewallproducts['ID'];?>" value="<?php echo $viewallproducts['ID'];?>"><?php echo $viewallproducts['DESCRIPCION'];?></option>
<?php } ?>
</select>

Answer 2

I think you made a mistake because you think ID in html is ID in database. You need to use "value" attribute for option
chance this line
<option id="<?php echo $viewallproducts['ID'];?>"><?php echo $viewallproducts['DESCRIPCION'];?></option>
to
<option value="<?php echo $viewallproducts['ID'];?>"><?php echo $viewallproducts['DESCRIPCION'];?></option>
You have also forget about:
require argument in ''
; after require
Function to get rows should be 'mysql_fetch_row' instead of 'mysql_fetch_array'
I would also suggest you using better DB engine for example PDO or MySQLi because MySQL_* functions are depracated.
Code formatted:
<select name="Select" class="textfields" id="prods">
<option value="0">--Producto--</option>
<?php
require('conectdb.php'); // be sure that file exist you forgot about '' and ;
$allproducts = mysql_query("SELECT * FROM `Productos`");
while ($viewallproducts = mysql_fetch_row($allproducts))
echo '<option value="'.$viewallproducts['ID'].'">'.$viewallproducts['DESCRIPCION'].'</option>';
?>
</select>

Answer 3

You forget to put the semicolon after the line
require(conectdb.php)
And you should add a quotation
require("conectdb.php");
Maybe is that

Answer 4

Some edit on your code:
<select name="Select" class="textfields" id="prods">
<option id="0">--Producto--</option>
<?
require("conectdb.php");
$allproducts = mysql_query("SELECT * FROM Productos")
or die(mysql_error());
while ($viewallproducts = mysql_fetch_assoc($allproducts)){
?>
<option id="<?=$viewallproducts['ID'];?>" value="<?=$viewallproducts['ID'];?>">
<?=$viewallproducts['DESCRIPTION'];?></option>
<?php } ?>
</select>
You should use mysql_fetch_assoc instead of mysql_fetch_array. You should also check if the sql query succeed first.