我沒有做插入查詢,所以mysqli_insert_id不工作。如何從表A中獲取INT-type id並將其插入到另一個表中?在PHP
我需要選擇一個ID(主,INT)從表A將被插入到表B.
我的表結構:
Table book:
-id (Primary, Auto Increment, INT)
-title
-publisher_id (Foreign key to Table publisher.pub_id, INT)
Table publisher:
-pub_id (Primary, Auto Increment, INT)
-publisher
我想是這樣的:
$query ="SELECT pub_id FROM book WHERE publisher = '$publisher' ";
$result=$mysqli->query($query);
$pub_id=$result->fetch_assoc();
$query ="INSERT INTO book (title,publisher_id)
VALUES ('$title', '$pub_id['pub_id']')";
$mysqli->query($query);
然後,我得到錯誤信息,如:
syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting '-' or identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING)
數據類型或其他東西有問題嗎?
如何從Table Book中獲取INT-type id並將其插入到另一個表中?
哪一行的錯誤說? –
這行$ query =「INSERT INTO book(title,publisher_id) VALUES('$ title','$ pub_id ['pub_id']')」; – Rufus7
當你說'SELECT pub_id FROM book'時,你的意思是'FROM publisher'吧? –