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Code
$sql = "INSERT INTO tbl_product (cat_id, pd_name, pd_description, pd_price, pd_qty, pd_date)
VALUES ('$catId', '$name', '$description', $price, $qty, NOW())";
$result = mysqli_query($dbConn, $sql) or die(mysqli_error($dbConn));
$lastID = mysqli_insert_id();
พอรันแล้วมัน error แบบนี้ครับ
Code
mysqli_insert_id() expects exactly 1 parameter, 0 given
Tag : PHP
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ประวัติการแก้ไข 2017-03-19 13:16:45
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Date :
2017-03-19 11:35:12 |
By :
mmc01 |
View :
1578 |
Reply :
2 |
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