我有一個表格,使用下面的下拉菜單中的變量。當用戶在下拉菜單中選擇該選項時,表格根據與選擇相關的變量來拉取信息。如何在頁面首次加載錯誤時指出查詢缺少下拉列表中的變量。如果我從下拉菜單中進行選擇,則刷新頁面並解決問題。我需要下拉菜單來初始提交數據或查詢所需的任何內容,以便在初始頁面加載時獲取其變量。初始頁面加載運行提交按鈕
$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>
這裏是在下拉菜單中選擇SQL查詢:
$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());
下面是表的查詢:
$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());
「...如果我從下拉菜單中進行選擇,刷新頁面並解決問題...」是通過javascript/ajax刷新還是加載另一個(不同)頁面? – Stevo 2013-02-27 13:54:42
請向我們展示您的SQL查詢。 – BenM 2013-02-27 13:58:28
我添加了sql – 2013-02-27 14:08:50