2011-07-05 97 views
1

我有一組似乎無法工作的PDO語句。基本上我試圖更新1表中的「等待」值,然後選擇相同的行並將其插入到另一個表中。PDO查詢不起作用

$statement = $db->prepare("UPDATE waiting SET wait = :status WHERE id = :id"); 
$statement->bindValue(':status', 0); 
$statement->bindParam(':id', $id); 
$statement->execute(); 
$statement = $db->prepare("INSERT INTO approved (fname, lname, student_id, email, type) (SELECT fname, lname, student_id, email, type FROM waiting WHERE id = :id)"); 
$statement->bindParam(':id', $id); 
$statement->execute(); 

我也試着設置$statement爲null我做其他查詢之前,但沒有工作,要麼:

$statement = $db->prepare("UPDATE waiting SET wait = :status WHERE id = :id"); 
$statement->bindValue(':status', 0); 
$statement->bindParam(':id', $id); 
$statement->execute(); 
$statement = null; 
$statement = $db->prepare("INSERT INTO approved (fname, lname, student_id, email, type) (SELECT fname, lname, student_id, email, type FROM waiting WHERE id = :id)"); 
$statement->bindParam(':id', $id); 
$statement->execute(); 

爲什麼這是行不通的任何想法?

+4

你是否收到任何錯誤消息? – Wabbitseason

+0

定義「沒有工作」。兩個查詢都沒有正確執行?你是否收到'$ statement-> errorInfo()'的錯誤信息? –

+0

您的[INSERT-Query(MySQL)](http://dev.mysql.com/doc/refman/5.1/en/insert.html)錯誤:INSERT INTO已批准(...)** VALUES ** (...)' – Saxoier

回答

1

您的插入查詢語法錯誤。刪除選擇周圍的支架,它應該工作:

INSERT INTO approved (fname, lname, student_id, email, type) 
SELECT fname, lname, student_id, email, type FROM waiting WHERE id = :id 
+0

謝謝,這似乎有竅門! – Aaron