我想在这里做的是向表中插入一些值,但该值来自另一个表。
但是,有一些价值可以发布。
$reference_no = $_POST ['reference_no']
$location_id = $_SESSION ['location_id']
INSERT INTO infistall_location_stock_collect_update (package, quantity, location_id)
SELECT product_id, quantity, '...'
FROM orders_history
WHERE reference_no = '$reference_no'
如何将$ location_id放入SELECT查询?
由于
答案 0 :(得分:1)
INSERT INTO infistall_location_stock_collect_update (package, quantity, location_id)
SELECT product_id, quantity, '$location_id'
FROM orders_history
WHERE reference_no = '$reference_no'
答案 1 :(得分:0)
尝试引用列
$reference_no = $_POST ['reference_no']
$location_id = $_SESSION ['location_id']
INSERT INTO infistall_location_stock_collect_update (package, quantity, location_id)
SELECT product_id, quantity, '(".$location_id.")' as location_id
FROM orders_history
WHERE reference_no = '$reference_no'
快乐编码..