这是cmd提示符上的错误: C:\xampp\mysql\bin>mysql -p -u root --default_character_set utf8 recipes < recipes.sql
输入密码:****
ERROR 1064 (42000) at line 20: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '"admin" (
"admin_id" int(10) NOT NULL,
"f_name" text NOT NULL,
"l_name" te' at line 1
这是我尝试在phpmyadmin“error static analysis:
分析中发现4处错误。
A symbol name was expected! (near ""admin_id"" at position 383)
At least one column definition was expected. (near ""admin_id"" at position 383)
Unexpected beginning of statement. (near "20" at position 398)
Unrecognized statement type. (near "NOT NULL" at position 402)
SQL query:
-- phpMyAdmin SQL Dump -- version 4.8.3. -- http://www.phpmyadmin.net -- -- Host: localhost -- Generation Time: August 29, 2018 at 09:56 AM -- Server version: 5.1.37 -- PHP Version: 7.2.9 -- -- Database: `recipes` -- -- -------------------------------------------------------- -- -- Table structure for table `admin` -- CREATE TABLE IF NOT EXISTS "admin" ( "admin_id" int(20) NOT NULL, "f_name" text NOT NULL, "l_name" text NOT NULL, "email" varchar(250) NOT NULL, "c_email" varchar(250) NOT NULL, "pass" varchar(250) NOT NULL, "c_pass" varchar(250) NOT NULL, "u_name" varchar(150) NOT NULL, "address" text NOT NULL, "date_of_birth" varchar(50) NOT NULL, "city" text NOT NULL, "state" text NOT NULL, "country" text NOT NULL, "gender" text NOT NULL, "p_o" int(50) NOT NULL, PRIMARY KEY ("admin_id") ) AUTO_INCREMENT=1
mysql说:文档
1064-您的sql语法有错误;检查与您的mariadb服务器版本相对应的手册,了解使用“admin”附近的正确语法(
“admin\u id”int(20)not null,“f\u name”text not null,“l\u name”'在第20行这是我的代码`
1条答案
按热度按时间kfgdxczn1#
删除“”
这应该是查询