PHP和MySQL獲取選擇值

我正在運行以下命令以從我的MySQL數據庫中獲取類型標識和名稱。應顯示兩個結果。在選擇列表中顯示兩個選項，但id或name不存在，它們只在頁面上顯示爲空白。PHP和MySQL獲取選擇值

function get_category_selector($form_name) { 
if (isset($_SESSION['loggedin']['cur_category'])) { 
    $curr = $_SESSION['loggedin']['cur_category']; 
} else { 
    $curr = -1; 
} 
$rtn = ""; 
$rtn .= '<select id="category" name="category">'; 
$rtn .= '<option value="-1"'; 
if ($curr == -1) { 
    $rtn .= ' selected'; 
} 
$rtn .= '>Select a Category</option>'; 
$DB = new mysql_db(); 
$DB->sql_connect(); 
$q = "SELECT DISTINCT (categories.categories_id), categories.categories_name FROM categories LEFT JOIN prod2cat ON prod2cat.prod2cat_cat = categories.categories_id LEFT JOIN products ON prod2cat.prod2cat_product = products.products_id LEFT JOIN prod2type ON products_id = prod2_product LEFT JOIN type ON prod2_type = type_id WHERE type.type_id =2 order by categories.categories_name asc"; 
$DB->query($q); 
if ($DB->get_num_rows()) { 
    $results = $DB->fetch_all_rows(MYSQL_ASSOC, NULL); 
    foreach ($results as $key => $row) { 
     $rtn .= '<option value="' . $row['Categories_ID'] . '"'; 
     if ($curr == $row['Categories_ID']) { 
      $rtn .= ' selected'; 
     } 
     $rtn .= '>' . trim($row['Categories_Name']) . '</option>'; 
    } 
} 
$rtn .= '</select>'; 

return $rtn;

}

來源

2012-12-16 Joe Ainsworth

變化以小寫'$ row ['categories_id']'&&'$ row ['categories_name']' – Cyclonecode

什麼是mysql_db分類S'請告訴我們這是如何工作的（發佈類定義），謝謝。〜雷 –

訪問您的結果數組時，您必須使用相同的情況下爲您列名：

$rtn .= '<option value="' . $row['categories_id'] . '"'; 
if ($curr == $row['categories_id']) { 
    $rtn .= ' selected'; 
} 
$rtn .= '>' . trim($row['categories_name']) . '</option>';

，或者你可以改變你的選擇查詢：

$q = "SELECT DISTINCT (categories.categories_id) Categories_ID, 
     categories.categories_name Categories_Name;

來源

2012-12-16 22:32:32 Cyclonecode

PHP和MySQL獲取選擇值

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