Run Submit Button on inital page load - php

Question

I have a table using the variables from the below drop down menu. When the user selects the option in the drop down the table pulls the information based on the variable tied to the selection. How ever when the page first loads it errors out saying the query is missing the variable from the drop down. If I make a selection from the drop down it refreshes the page and resolves the issue. I need the drop down to initially submit the data or whatever is necessary for the query to get its variable on intial page load.
$selected = 'selected = "selected" ';
$Country =$ID_SOCIEDAD;
echo "<form name='country_list' method='POST' action='http://opben.com/colombia/familias-de-carteras' >";
echo "<select name='Country' tabindex='1' >";
while($row = mysql_fetch_array($result))
{
echo " <option ".($row['Fund_Manager_Company_Code'] == $Country? $selected : '')."value='". $row['Fund_Manager_Company_Code'] ."'>". $row['Fund_Manager_Company_Name'] ."</option>";
}
echo " </select>
<input type='submit' value='Filter' />";
echo " </form>
Here is the sql query for the drop down menu options:
$result = mysql_query("
SELECT
ID_SOCIEDADADM as Fund_Manager_Company_Code,
DES_SOCIEDAD_CORTO as Fund_Manager_Company_Name
FROM dr_lista_rentabilidad_diaria
GROUP BY ID_SOCIEDADADM
")
or die(mysql_error());
Here is the query for the table:
$result = mysql_query("
SELECT
ID_CARTERA as Fund_ID,
DES_CARTERA_CC as Fund_Name,
DES_CARTERACLASE as Class_Name,
DES_CARTERACLASE_ESP as Special_Class_Name,
FORMAT(POR_RENTCARTERA_C1,2) AS Yield_1month
FROM dr_lista_rentabilidad_diaria
WHERE COD_PAIS = $COD_PAIS
AND ID_SOCIEDADADM = $ID_SOCIEDAD
AND `ID_COLUMNA_C1`= $ID_COLUMNA
ORDER BY DES_CARTERA_CC ASC
")
or die(mysql_error());

Answer 1

You didn't post your query, so it is hard to give an appropriate answer. However, my guess is that you have the variable $_POST['Country'] somewhere in that query. So why don't you do this:
$country = $_POST['Country'];
if (!isset($country)) { $country = 'US'; }
$query = 'SELECT * FROM table WHERE country = ' . $country;
EDIT: As BenM (correctly) pointed out, this code snippet is absolutely not save. You should NOT include this in your production code. It was just meant to give you an idea on how to avoid the SQL error you are getting by always providing a valid value for the country.