CREATE TABLE IF NOT EXISTS `admin_menu` (
`id` int unsigned NOT NULL,
`parent_id` int NOT NULL DEFAULT '0',
`order` int NOT NULL DEFAULT '0',
`title` varchar(50) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci NOT NULL,
`icon` varchar(50) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci NOT NULL,
`uri` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
`permission` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
`created_at` timestamp NULL DEFAULT NULL,
`updated_at` timestamp NULL DEFAULT NULL
) ENGINE=InnoDB AUTO_INCREMENT=14 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;
INSERT INTO `admin_menu` (`id`, `parent_id`, `order`, `title`, `icon`, `uri`, `permission`, `created_at`, `updated_at`) VALUES
(1, 0, 1, 'Dashboard', 'fa-bar-chart', '/', NULL, NULL, NULL),
(2, 0, 2, 'Admin', 'fa-tasks', '', NULL, NULL, NULL),
(3, 2, 3, 'Users', 'fa-users', 'auth/users', NULL, NULL, NULL),
(4, 2, 4, 'Roles', 'fa-user', 'auth/roles', NULL, NULL, NULL),
(5, 2, 5, 'Permission', 'fa-ban', 'auth/permissions', NULL, NULL, NULL),
(6, 2, 6, 'Menu', 'fa-bars', 'auth/menu', NULL, NULL, NULL),
(7, 2, 7, 'Operation log', 'fa-history', 'auth/logs', NULL, NULL, NULL),
(8, 0, 0, 'Food Category', 'fa-certificate', '/food-types', '*', '2021-11-08 04:58:13', '2022-01-01 00:16:59'),
(10, 0, 0, 'Users', 'fa-bars', '/users', '*', '2021-11-08 05:38:59', '2021-11-08 05:38:59'),
(11, 0, 0, 'Food', 'fa-beer', '/foods', '*', '2021-11-17 01:37:06', '2022-01-01 00:06:15'),
(12, 0, 0, 'Orders', 'fa-first-order', '/orders', '*', '2022-02-09 18:27:25', '2022-02-09 18:27:25'),
(13, 0, 0, 'Payment Setting', 'fa-dollar', '/business-settings', '*', '2022-02-27 22:24:23', '2022-02-27 22:24:23');
1条答案
按热度按时间zzzyeukh1#
在我看来，您的
admin_menu表可能已经存在，所以您的CREATE TABLE IF NOT EXISTS语句不会做任何事情。而且，现有的表很可能对第一列有这样的定义。它与CREATE表中的定义不同。
而且现有表似乎已经有一行
id=1。因此，您向我们展示的INSERT的第一行在重复的键上失败。这就是你的错误。您似乎正在执行某种正在失败的转储/恢复操作。如果不更多地了解你正在遵循的工作流程，就很难在这方面给你具体的建议。