>我正在尝试更新表中的字段以反映在HTML中所做的更改,数据被发布到php中,但是尽管没有显示错误,但更改并未反映在数据库中
<?php
session_start();
$db = mysqli_connect ('localhost', 'root', '', 'car_rental') or die ("SQL is Off");
$sku = $_POST['firstsku'];
$days = $_POST['days'];
$user = $_SESSION["userEmail"];
mysqli_select_db($db,"car_rental");
$query = " UPDATE products SET rented=1 WHERE sku='$sku'";
$query1 = "UPDATE `users` SET `rented` = 'pie 32' WHERE `users`.`email` = 'user'";
$result1 = mysqli_query($db, $query1);
$result = mysqli_query($db, $query);
?>
第二个查询结果 1 是我在数据库中没有获得任何数据更改的结果
查询
中缺少$user
。用这个更改你的第二个查询:
$query1 = "UPDATE `users` SET `rented` = 'pie 32' WHERE `users`.`email` = '$user'";
查询应该是这样的
$query1 = "UPDATE `users` SET `rented` = 'pie 32' WHERE `users`.`email` = '".$user."' ";
试试这个:
$combine = $days.'-'.$sku;
$query1 = "UPDATE users SET rented = '$combine' WHERE users.email = '$user'";
使用 -
合并两个值,然后尝试更新。