将外键添加到现有表 [英] Add Foreign Key to existing table

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本文介绍了将外键添加到现有表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

  ALTER TABLE katalog 
添加一个外键给一个名为katalog的表。 CONSTRAINT`fk_katalog_sprache`
FOREIGN KEY(`Sprache`)
参考`Sprache`(`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL;

当我尝试这样做时,我收到以下错误消息:


 错误代码:1005.无法创建表'mytable。#sql-7fb1_7d3a'(errno:150)


INNODB状态错误:


120405 14:02:57表
mytable的外键约束出错。#sql-7fb1_7d3a:



  FOREIGN KEY(`Sprache`)
参考`Sprache`(`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL:
无法解析表名称接近:
(`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL

当我使用这个查询时,它可以正常工作,但是在删除操作上有错误:

 ALTER TABLE`katalog` 
添加FOREIGN KEY(`Sprache`)参考`sprache`(`ID`)

这两个表都是InnoDB,两个字段都是INT(11)not null。我正在使用MySQL 5.1.61。尝试在MacBook Pro上使用MySQL Workbench(最新版本)激发此ALTER查询。
$ b $ p

表创建语句:

<$ ($ b $ id` int(11)unsigned NOT NULL AUTO_INCREMENT,
```varchar(50)COLLATE utf8_unicode_ci NOT NULL,
`AnzahlSeiten` int(4)unsigned NOT NULL,
`Sprache` int(11)NOT NULL,
PRIMARY KEY(`ID`),
UNIQUE KEY`katalogname_uq`(`名称`)
)ENGINE = InnoDB AUTO_INCREMENT = 12 DEFAULT CHARSET = utf8 COLLATE = utf8_unicode_ci ROW_FORMAT = DYNAMIC $$

CREATE TABLE`sprache`(
`ID` int )NOT NULL AUTO_INCREMENT,
`Bezeichnung` varchar(45)NOT NULL,
PRIMARY KEY(`ID`),
UNIQUE KEY`Bezeichnung_UNIQUE`(`Bezeichnung`),
KEY`ix_sprache_id````
)ENGINE = InnoDB AUTO_INCREMENT = 3 DEFAULT CHARSET = utf8


解决方案

要将外键(grade_id)添加到现有表(用户),请按照以下步骤操作:

  ALTER TABLE用户ADD grade_id SMALLINT UNSIGNED NOT NULL DEFAULT 0; 
ALTER TABLE用户ADD CONSTRAINT fk_grade_id FOREIGN KEY(grade_id)REFERENCES grades(id);


I want to add a Foreign Key to a table called "katalog".

ALTER TABLE katalog 
ADD CONSTRAINT `fk_katalog_sprache` 
FOREIGN KEY (`Sprache`)
REFERENCES `Sprache` (`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL;

When I try to do this, I get this error message:

Error Code: 1005. Can't create table 'mytable.#sql-7fb1_7d3a' (errno: 150)

Error in INNODB Status:

120405 14:02:57 Error in foreign key constraint of table mytable.#sql-7fb1_7d3a:

FOREIGN KEY (`Sprache`)
REFERENCES `Sprache` (`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL:
Cannot resolve table name close to:
(`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL

When i use this query it works, but with wrong "on delete" action:

ALTER TABLE `katalog` 
ADD FOREIGN KEY (`Sprache` ) REFERENCES `sprache` (`ID` )

Both tables are InnoDB and both fields are "INT(11) not null". I'm using MySQL 5.1.61. Trying to fire this ALTER Query with MySQL Workbench (newest) on a MacBook Pro.

Table Create Statements:

CREATE TABLE `katalog` (
`ID` int(11) unsigned NOT NULL AUTO_INCREMENT,
`Name` varchar(50) COLLATE utf8_unicode_ci NOT NULL,
`AnzahlSeiten` int(4) unsigned NOT NULL,
`Sprache` int(11) NOT NULL,
PRIMARY KEY (`ID`),
UNIQUE KEY `katalogname_uq` (`Name`)
 ) ENGINE=InnoDB AUTO_INCREMENT=12 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci ROW_FORMAT=DYNAMIC$$

CREATE TABLE `sprache` (
`ID` int(11) NOT NULL AUTO_INCREMENT,
 `Bezeichnung` varchar(45) NOT NULL,
 PRIMARY KEY (`ID`),
 UNIQUE KEY `Bezeichnung_UNIQUE` (`Bezeichnung`),
KEY `ix_sprache_id` (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8

解决方案

To add a foreign key (grade_id) to an existing table (users), follow the following steps:

ALTER TABLE users ADD grade_id SMALLINT UNSIGNED NOT NULL DEFAULT 0;
ALTER TABLE users ADD CONSTRAINT fk_grade_id FOREIGN KEY (grade_id) REFERENCES grades(id);

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