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mysql - 如何使用查询解决 sql 导入错误 #1064 (joomla)

转载 作者:行者123 更新时间:2023-11-29 23:43:07 25 4
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我想将我的sql导入到我的主机,为此我删除数据库和joomla,然后上传新的joomla网站并昨晚创建新的数据库和用户...

当我上传 sql 时,我收到以下查询的错误 #1064:

CREATE TABLE IF NOT EXISTS `vnmbl_assets` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT COMMENT 'Primary Key',
`parent_id` int(11) NOT NULL DEFAULT '0' COMMENT 'Nested set parent.',
`lft` int(11) NOT NULL DEFAULT '0' COMMENT 'Nested set lft.',
`rgt` int(11) NOT NULL DEFAULT '0' COMMENT 'Nested set rgt.',
`level` int(10) unsigned NOT NULL COMMENT 'The cached level in the nested tree.',
`name` varchar(50) NOT NULL COMMENT 'The unique name for the asset.\n',
`title` varchar(100) NOT NULL COMMENT 'The descriptive title for the asset.',
`rules` varchar(5120) NOT NULL COMMENT 'JSON encoded access control.',
PRIMARY KEY (`id`),
UNIQUE KEY `idx_asset_name` (`name`),
KEY `idx_lft_rgt` (`lft`,`rgt`),
KEY `idx_parent_id` (`parent_id`)
) TYPE=MyISAM AUTO_INCREMENT=360 ;

MySQL said: Documentation
#1064 - You have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version
for the right syntax to use
near 'TYPE=MyISAM AUTO_INCREMENT=360' at line 14

当我TYPE:MyISAM”更改为“ENGINE:MyIsam”时,错误会减少一些:

CREATE TABLE IF NOT EXISTS `vnmbl_finder_tokens` (
`term` varchar(75) NOT NULL,
`stem` varchar(75) NOT NULL,
`common` tinyint(1) unsigned NOT NULL DEFAULT '0',
`phrase` tinyint(1) unsigned NOT NULL DEFAULT '0',
`weight` float unsigned NOT NULL DEFAULT '1',
`context` tinyint(1) unsigned NOT NULL DEFAULT '2',
`language` char(3) NOT NULL DEFAULT '',
KEY `idx_word` (`term`),
KEY `idx_context` (`context`)
) TYPE=MEMORY

请帮我解决这个问题,我的主机现在死了......

最佳答案

打开 .sql 文件,搜索“TYPE=MEMORY”并将全部替换为“ENGINE=MEMORY”

关于mysql - 如何使用查询解决 sql 导入错误 #1064 (joomla),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26074010/

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