heidisql

Connect to localDB with windows authentication

由 ↘锁芯ラ 提交于 2020-06-27 06:18:05
问题 This is the first time that I try to create a local DB to use for my spring boot application so it might be an easy thing to fix that Ive missed. What ive done so far: Downloaded SQLExpress Logged with SSMS using windows authentication Created a database Connected to the new database using HeidiSQL After all this steps ive tried to setup a connection with my spring boot app using the following properties: spring.datasource.url= jdbc:sqlserver://localhost:3306 spring.datasource.username=

Centos7 安装Mysql5.7

由 房东的猫 提交于 2020-04-28 08:58:31
我们经常需要在服务器上安装mysql，各种文档都有，但是很多都是一部分，我现在总结了一下，放到一起，以后大家不用一篇一篇查询了。 1.安装yum repo 由于CentOS 的yum源中没有mysql，需要到mysql的官网下载yum repo配置文件。 wget https://dev.mysql.com/get/mysql57-community-release-el7-9.noarch.rpm 然后进行repo的安装： rpm -ivh mysql57-community-release-el7-9.noarch.rpm 执行完成后会在 /etc/yum.repos.d/ 目录下生成两个repo文件 mysql-community.repo mysql-community-source.repo 2. 安装MySQL 使用yum命令即可完成安装 yum install mysql-server 启动msyql： systemctl start mysqld #启动MySQL 配置MySQL 获取安装时的临时密码： grep 'temporary password' /var/log/mysqld.log 登录： mysql -u root -p 登录成功后修改密码： set password=password("yourpassword"); 其他命令： systemctl

SQL: How do I group by a unique combination of two columns?

由 女生的网名这么多〃 提交于 2020-03-23 05:53:29
问题 Context: A table message has the columns from_user_id and to_user_id The user should see the recent conversations with the last message displayed A conversation consists of multiple messages, that have the same combination of user IDs (user sends messages, user receives messages) Table content: +-------------------------------------------------+--------------+------------+ | text | from_user_id | to_user_id | +-------------------------------------------------+--------------+------------+ | Hi

How to view table foreign keys (FKs) in HeidiSQL

由 ぐ巨炮叔叔 提交于 2020-02-27 12:48:29
In this tutorial I'd like to show you how to view table foreign keys with HeidiSQL . To see table FKs first select table in the object explorer, then select Table tab, and under this tab select Foreign keys tab. All table foreign keys will be visible in the list on the upper part of the window. 来源： oschina 链接： https://my.oschina.net/ciet/blog/3159788

Disable the natively generated identity value feature in grails

由 穿精又带淫゛_ 提交于 2020-01-16 23:14:44
问题 The following is my Domain class details. class Age { String agetype static constraints = { } } I am using HeidiSQL. I want to drop the id column that is generated automatically.And set primary key as 'agetype'. what I can I do? 回答1: Identifier can be customized easily inside the mapping block, if the default id is not required. class Age { String agetype static mapping = { id name: 'agetype', column: 'AGE_TYPE', // if the column name is AGE_TYPE generator: 'assigned' // Unique String should

Disable the natively generated identity value feature in grails

由 风流意气都作罢 提交于 2020-01-16 23:14:02
问题 The following is my Domain class details. class Age { String agetype static constraints = { } } I am using HeidiSQL. I want to drop the id column that is generated automatically.And set primary key as 'agetype'. what I can I do? 回答1: Identifier can be customized easily inside the mapping block, if the default id is not required. class Age { String agetype static mapping = { id name: 'agetype', column: 'AGE_TYPE', // if the column name is AGE_TYPE generator: 'assigned' // Unique String should

Adding a populated column to an existing table

由 我们两清 提交于 2020-01-16 19:40:14
问题 I have 2 table that have table counts in them. Something like Table 1: New_Counts 100 Table 2: Old_Counts 97 I want to create a single QC table so I can eventually check that the counts are increasing: New_Counts|Old_Counts 100|97 Any ideas how to do this in SQL? A join won't work since the counts are never the same. 回答1: select newcount,oldcount from (select (select count(*) newcount from table1) newcount, (select count(*) oldcount from table2) oldcount from dual); 来源： https://stackoverflow

Couldn't find the toolbar option for stored procedure

由 末鹿安然 提交于 2020-01-03 06:18:30
问题 This may be a silly question but I'm unable to find the stored procedure toolbar in my HeidiSQL. Could anyone tell me how can I do that? I'm using HeidiSQL 7 version. The below link demonstrate one stored routine editor but I couldn't find it in Version 7. http://www.databasejournal.com/features/mysql/article.php/3826986/Creating-Stored-Procedures-in-MySQL-Using-HeidiSQL-4146s-Stored-Routine-Editor.htm 回答1: I'm using HeidiSQL version 8. In the left pane, right-click on a database or a table

如何快速重命名MySQL数据库（更改模式名称）？

由 99封情书 提交于 2019-12-16 10:38:48
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 在MySQL手册 的MySQL 涵盖这一点。 通常我只是转储数据库并使用新名称重新导入它。 对于非常大的数据库，这不是一个选项。 显然是 RENAME {DATABASE | SCHEMA} db_name TO new_db_name; RENAME {DATABASE | SCHEMA} db_name TO new_db_name; 做坏事，只存在于少数几个版本中，总体上是一个坏主意 。 这需要与 InnoDB 一起使用， InnoDB的 存储方式与 MyISAM 完全不同。 #1楼 使用以下几个简单的命令： mysqldump -u username -p -v olddatabase > olddbdump.sql mysqladmin -u username -p create newdatabase mysql -u username -p newdatabase < olddbdump.sql 或者根据 @Pablo Marin-Garcia的建议减少I / O使用： mysqladmin -u username -p create newdatabase mysqldump -u username -v olddatabase -p | mysql -u username -p -D

trying to get the rooms of hotels that got the most profit in a query

由 ε祈祈猫儿з 提交于 2019-12-13 03:58:18
问题 I'm trying to implement a query which give me the sum of most profitable room in each hotel(25 hotels) Below is my query: SELECT hotels.hotel_id,rooms.room_id,hotel_name,room_number,sum(rooms.room_price) AS profit,COUNT(rooms.room_id) AS count FROM hotels,rooms,bookings WHERE hotels.hotel_id=rooms.hotel_id AND rooms.room_id=bookings.room_id GROUP BY rooms.room_id and this is the closest outcome i got.. ignore the hotel name language This is the outcome that I've reached so far, hotels rooms
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