如何在创建数据库语句中使用变量 [英] how do I use a variable in create database statement

问题描述

我有此过程:

DELIMITER //

create DEFINER = 'root'@'localhost' procedure create_db(name TEXT) 
BEGIN
DECLARE temp TEXT;
DECLARE user TEXT;
SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = name INTO temp;
  if temp = name then
      SIGNAL SQLSTATE '45002' SET MESSAGE_TEXT = 'This database already exist';
  else
      SELECT USER() INTO user;
      create database name;
      grant all privileges on name.* to user with grant option;
  END IF;
END //

DELIMITER ;

它很好地工作，只是它提供字面意义上的名称"而不是变量名称的值.我如何告诉它名称是变量?类似于php中的$ name之类的东西.我浏览了许多文档，但是它们都使用没有前缀的变量.

it works great just it supply literally "name" instead of value of variable name. How do I tell it that name is a variable? something like $name in php or that. I browser many documentations but they all are using variable with no prefixes.

推荐答案

您只需要使用准备好的语句，这是工作代码:

You just need to use prepared statements, here is working code:

DELIMITER //

drop procedure if exists create_db //

create procedure create_db(name TEXT)
BEGIN
  DECLARE temp TEXT;
  DECLARE user TEXT;
  SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = name INTO temp;
  if temp = name then
  SIGNAL SQLSTATE '45002' SET MESSAGE_TEXT = 'This database already exist';
    else
    SELECT USER() INTO user;

    SET @s = CONCAT('CREATE DATABASE ', name);
    PREPARE stmt_create FROM @s;
    EXECUTE stmt_create;
    DEALLOCATE PREPARE stmt_create;

    SET @s = CONCAT('GRANT ALL PRIVILEGES ON ', name, '.* TO ', user, ' WITH GRANT OPTION');
    PREPARE stmt_grant FROM @s;
    EXECUTE stmt_grant;
    DEALLOCATE PREPARE stmt_grant;

  END IF;
END //

DELIMITER ;

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