What is wrong with my code? Please help

[ChaosKnight]

Hi,

the website that I'm busy with has to be done by the end of the week.

The code is supposed to query for the countries in the hotels table,

and then generate a select dropdown based on the result, but it

returns nothing... 

 

Here is the code:

 

$db_link = mysql_connect($db_host, $db_user, $db_pass);
if ( ! $db_link )
	die( "Couldn't connect to mySql Server<br />");
//echo( "Connected to mySql Server<br />");

$db_select = mysql_select_db($db);
	if ( ! $db_select )
		die("Couldn't connect to Database<br />");
	//echo("Connected to Database<br />");

echo"<form action='' method='POST'>
	<label for='countrySelect'>Country</label>
	<select id='countrySelect' name='country'>\n";
$query = "SELECT DISTINCT `country_id` FROM `hotels` ORDER BY `country_id`";
$result = mysql_query($query, $db_link);
	or die ("couldn't execute query.");

$num_rows = mysql_num_rows($result);
  if ( $num_rows < 1 ) {
  print( "<span class=\"error_text\">NO RESULTS FOUND. Please try again...</span>" );
  }
  else{
while($row = mysql_fetch_array($result))
{
  extract($row);
  echo "<option value='$country_id'>$country_id</option>";
}
       echo "</select>";

 

I tried the rest of the code with a predefined array and it worked

perfectly, but as soon as I change it to the above it displays nothing... 

 

Thanks

[monkeypaw201]

Well, for starters, the $result line should read;

 

$result = mysql_query($query, $db_link) or die ("couldn't execute query.");

 

Let us know if there is any output now.

[ChaosKnight]

Thanks for the reply, that made my code a bit cleaner,

but still didn't fix it, so I went through everything thoroughly

and I found my problem, I forgot to put a "}" after an else statement

LOL 

 

Sometimes a make the stupidest mistakes hehe 

 

Thanks

[ChaosKnight]

But now there is another problem, the country_id returns "southafrica",

but it should display "South Africa", I tried $row['country'] which

should return the "South Africa" (I inserted the extra column in the

database, but I guess it doesn't work because I only queried for

the country_id column...

 

Please help

 

Oh and it should still use DISTINCT because there are several entries

for South Africa and I only need to display it once in the select dropdown

 

Thanks