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我在views_php創建以下代碼:在views_php一個db_query使用ENTITY_ID(Drupal的7)
$userseeking = db_query("SELECT field_seeking_tid FROM {field_data_field_seeking} WHERE entity_id = 6")->fetchAll(PDO::FETCH_ASSOC);
$userseekingcount = count($userseeking);
return $userseekingcount
在上面的例子中,我使用的 '6' 的特定值ENTITY_ID。我收到此查詢的預期結果。
但是,entity_id需要是取決於登錄用戶的配置文件ID(連接到其用戶ID)的變量。
因此我引入3行來從當前用戶檢索PID($ currentpid)。我已經確認這3行返回$ currentpid值6.當我修改db_query來選擇entity_id = $ currentpid的值時,雖然它們應該產生相同的結果,但看起來是失敗的,請參閱下面的代碼:
global $user;
$currentuser = $user->uid;
$currentpid = db_query('SELECT pid FROM {profile} WHERE uid = ' . $currentuser . ' limit 1')->fetchField();
$userseeking = db_query("SELECT field_seeking_tid FROM {field_data_field_seeking} WHERE entity_id = ' . $currentpid . '")->fetchAll(PDO::FETCH_ASSOC);
$userseekingcount = count($userseeking);
return $userseekingcount
我在濫用如何在此查詢中引用entity_id嗎?
感謝您的幫助!
PS這可能不是最乾淨的代碼,所以我對先進的道歉! :)