我在一个名为“orders”的表中从最后插入的记录中检索order_id
时遇到了一个问题。
一旦检索到,我希望将其放入我的另一个名为“报告”的表中。
我可以看到这个问题已经被问了无数次,我也提出了同样多的建议,但都没有成功。
目前,我的每个表都包含auto_increment属性,并且我的函数在“orders”表(而不是我试图实现的reports表)上执行了一次成功的插入。
function report() {
if(isset($_GET['tx'])) {
$amount = $_GET['amt'];
$currency = $_GET['cc'];
$transaction = $_GET['tx'];
$status = $_GET['st'];
$total = 0;
$item_quantity = 0;
$send_order = query(" INSERT INTO orders (order_amount, order_transaction, order_status, order_currency) VALUES('{$amount}','{$currency}','{$transaction}','{$status}')");
$last_id = last_id(); // this calls last_id in functions.php
confirm($send_order);
foreach ($_SESSION as $name => $value) {
if($value > 0 ) {
if(substr($name, 0, 8) == "product_") {
$length = strlen($name - 8);
$id = substr($name, 8 , $length);
$query = query(" SELECT * FROM products WHERE product_id = " . escape_string($id). " " );
confirm($query);
while ($row = fetch_array($query)) {
$product_price = $row['product_price'];
$sub = $row['product_price']*$value;
$item_quantity +=$value;
$insert_report = query(" INSERT INTO reports (product_id, order_id, product_price, product_quantity) VALUES('{$id}','{$last_id}','{$product_price}','{$value}')");
confirm($insert_report); //runs the confirm helper method
} // end of while loop
$total += $sub;
echo $item_quantity;
} // end of substring if statement
}// end of if value
} // end of foreach loop
session_destroy();
} else {
redirect("index.php");
} //end of if isset GET tx
} // end of report function /////////////////
functions.php
function last_id() {
global $connection;
return mysqli_insert_id($connection);
}
发布于 2018-05-28 19:40:41
你可以在你的php页面中插入查询后使用mysqli_insert_id($send_order)
。
发布于 2018-05-28 19:42:58
你应该打电话给
$last_id = last_id();
在使用之后使用
confirm($send_order);
发布于 2018-05-28 20:25:43
试着这样做:
$send_order = query(“INSERT INTO orders (order_amount,order_transaction,order_status,order_currency) VALUES('{$amount}','{$currency}','{$transaction}','{$status}')");
$last_id = $send_order->insert_id;
https://stackoverflow.com/questions/50565543
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