PHP和MySQL：我如何使選定的選項保持選擇在s Combo後提交

Question

我正在尋找stackoverflow和唯一的東西我得到的是像this。我想這個代碼

<select class="input_select" name='nombre_compania'><? 
    msqlcon_catering(); //which is the function i made that connects to the database 
    $query = "SELECT * FROM compania ORDER BY id DESC"; 
    $result = mysql_query($query); 
    while ($r = mysql_fetch_array($result)) 
    { 
     echo "<option value=\"". $r['id'] ."\">". $r['compania_nombre'] ."</option>"; 
    }?> 
</select> 

，當用戶選擇某個選項，留如果他錯過的東西或他把什麼問題提交表單後選做。 感謝

Answer 1

<?php 
$selected = null; 
if(isset($_POST['nombre_compania'])) 
{ 
    $selected = $_POST['nombre_compania']; 
} 

?> 

<select class="input_select" name='nombre_compania'><? 
    msqlcon_catering(); //which is the function i made that connects to the database 
    $query = "SELECT * FROM compania ORDER BY id DESC"; 
    $result = mysql_query($query); 
    while ($r = mysql_fetch_array($result)) 
    { 
     echo "<option value=\"". $r['id'] ."\" ".($selected == $r['id'] ? 'selected=\"selected\"': '').">". $r['compania_nombre'] ."</option>"; 
    }?> 
</select> 

Answer 2

<select class="input_select" name='nombre_compania'><? 
    msqlcon_catering(); //which is the function i made that connects to the database 
    $query = "SELECT * FROM compania ORDER BY id DESC"; 
    $result = mysql_query($query); 
    while ($r = mysql_fetch_array($result)) 
    { 
     echo "<option value=\"". $r['id'] ."\"" . (isset($_REQUEST["nombre_compania"]) && $_REQUEST["nombre_compania"] == $r['id'] ? " selected='selected'" : "") . ">". $r['compania_nombre'] ."</option>"; 
    }?> 
</select>