我試圖給get_lists方法添加一個新的變量(參數)`$ sugg_only = false'。現在,我將根據用戶是否具有登錄令牌來提取所有數據。這工作正常,並返回自定義列表中的數據數組。我的PHP類中的其他參數
如果建議列設置爲Y
,我想要做的就是從shopping_list_name中返回東西。這樣,如果他們沒有自定義列表,它會提取我們提供的建議列表。
這裏是一個什麼樣的表中的完整列表:
ID SHOPPING_LIST_NAME S SEQUENCE
1 test amnaik shopping list N
2 bonner shopping list N
3 793d7384fa4fa247d6fae07db104147d0a2dad6e Y
4 kj's shopping list N
5 kj's shopping list from 1384201636 N
6 kj's shopping list from 1384201659 N
7 kj's shopping list from 1384202055 N
8 kj's shopping list from 1384202089 N
9 kj's shopping list from 1385064064 N
10 kj's shopping list from 1385064145 N
11 kj's shopping list from 1385064150 N
12 kj's shopping list from 1385064257 N
13 kj's shopping list from 1385064825 N
14 kj's shopping list from 1385064857 N
所以,你看,這裏有一個(可怕的名字命名)的購物列表的設置作爲一個建議。
// Get a user's shopping lists
public function get_lists($clobber = false, $sugg_only = false) {
if ($this->UserShoppingList != null && !$clobber) {
return $this->UserShoppingList;
} else if ($this->get_sign_in_token()) {
global $db;
$vars = array();
$vars[] = array(':i_sign_in_token', strtoupper($this->get_sign_in_token()));
$rows = $db->get_function_ref_cursor('custom.japi_shopping_list.get_lists_for_shopper(:i_sign_in_token)', $vars);
// Turn the rows into objects and get their items.
foreach ($rows as $row) {
$list = new UserShoppingList(null, $this->sign_in_token);
$list->get_from_array($row);
$list->get_items();
$this->UserShoppingList[] = $list;
}
return $this->UserShoppingList;
} else {
return false;
}
}
API頁面:
if(!isset($_GET['token']) && !isset($_GET['suggested_only'])) {
die('Must pass-in either a \'token\' or \'suggested_only\' flag');
}
if(isset($_GET['token'])) {
$shopper = new Shopper($_GET['token'])
or die('Could not instantiate a new Shopper from the \'token\' passed-in');
$array = array();
$shopper_lists = $shopper->get_lists(true);
foreach ($shopper_lists as $list) {
$array[] = $list->json();
}
echo json_encode($array);
// echo json_encode($shopper_lists);
}
我將有可能被寫入另一個foreach
環路和公正包括無論發生什麼事,如果$ sugg_only等於TRUE?
像這樣的東西添加到我的API頁面??的底部:
if(isset($_GET['suggested_only']) && $_GET['suggested_only'] == 'true')
{
}
任何幫助將不勝感激。謝謝。