在PHP中从sql数据库创建下拉列表

There are couple of things, you to verify and correct.

First of all, your database connection code, that doesn't seems to be correct. I didn't see any condition on which you are suppose to invoke die.

$connect = mysql_connect('localhost', 'root');
{
die  ("Unable to connect to database<br>");
}

From the above code, below line will execute all the time

 die  ("Unable to connect to database<br>");

You should correct your database connection code to below :

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
mysql_close($link);

You can refer mysql_connect for the usage.

Also, verify your file extension is .php

You might want to try to use this ..The dept_id and the description u have to change according to your database .Hope this will help you

 <?php 
  $sql = mysql_query("SELECT dept_id ,description FROM department 
  ORDER BY dept_id ASC");
  db_select($sql,"dept_id",$dept_id,"","-select Department-","","");
  ?>

There could be few issues:

• Use mysqli_connect instead of mysql_connect because mysql is depreciated
• Make 3rd parameter as empty `mysqli_connect('localhost', 'root', '');//as per standards
• Debug your code for example: (print_r($connect), execute your query in phpmyadmin, print_r($result) and then in loop print_r($row).

At end, I am sure you will get your desired result and also you will know in detail that what was happening.

Mainly your problem is database connection. Try this -

<?php
mysql_connect("localhost", "root", "") or
    die("Could not connect: " . mysql_error());
mysql_select_db("your_db");

$result = mysql_query("SELECT * FROM your_table");

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    print_r($row);
    echo "<br>";
}

mysql_free_result($result);
?>

So there are many problems with your approach. First of all you are using a deprecated mysql_* functions which is bad idea and second you are not debugging your database connection properly :

$connect = mysql_connect('localhost', 'root');
   {
     die  ("Unable to connect to database<br>");
  }

In above code the die statement will always execute stopping further execution.

Also make sure the database ViviansVacations and table Destinations and column Europe exists with correct names(Follow standards and try to use all small letters for database/table/column naming)

The correct mysqli_* approach is(tested locally and the select box forms correctly) :

<?php
$db = 'ViviansVacations';

$mysqli = new mysqli('localhost', 'root', '', $db);

if($mysqli->connect_error) 
  die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());


$query = "SELECT * FROM Destinations";
$result = mysqli_query($mysqli, $query);

?>

<select name="select1">
<?php
while ($row = mysqli_fetch_array($result)) {
    echo "<option value='" . $row['Europe'] . "'>" . $row['Europe'] . "</option>";
}
?>
</select>