Question

我试图添加一个包含2列的外键。

以下是引用外键的表的DDL：

queue:work

拥有外键的表的DDL：

CREATE TABLE IF NOT EXISTS `sf_file_category` (
  `id_file_category`        INT                         NOT NULL AUTO_INCREMENT,
  `name`                    VARCHAR(45)
                            CHARACTER SET 'latin1'
                            COLLATE 'latin1_general_ci' NOT NULL,
  `file_type`               ENUM('document', 'image', 'video', 'archive')
                            CHARACTER SET 'latin1'
                            COLLATE 'latin1_general_ci' NULL,
  `id_file_category_parent` INT UNSIGNED                NULL,
  PRIMARY KEY (`id_file_category`),
  INDEX `fk_sf_file_category_sf_file_category1_idx` (`id_file_category_parent` ASC),
  INDEX `fk_sf_file_category_sf_file_idx` (`id_file_category` ASC, `file_type` ASC)
)
  ENGINE = InnoDB;

尝试执行以下外键语法：

CREATE TABLE IF NOT EXISTS `sf_file` (
  `id_file`           INT UNSIGNED                NOT NULL AUTO_INCREMENT,
  `fullpath`          VARCHAR(100)
                      CHARACTER SET 'latin1'
                      COLLATE 'latin1_general_ci' NOT NULL,
  `basename`          VARCHAR(45)
                      CHARACTER SET 'latin1'
                      COLLATE 'latin1_general_ci' NOT NULL,
  `accesskey`         CHAR(8)
                      CHARACTER SET 'latin1'
                      COLLATE 'latin1_general_ci' NOT NULL,
  `file_type`         ENUM('document', 'image', 'video', 'archive')
                      CHARACTER SET 'latin1'
                      COLLATE 'latin1_general_ci' NULL,
  `name`              VARCHAR(45)
                      CHARACTER SET 'latin1'
                      COLLATE 'latin1_general_ci' NULL,
  `description`       VARCHAR(255)
                      CHARACTER SET 'latin1'
                      COLLATE 'latin1_general_ci' NULL,
  `id_aircraft_image` SMALLINT UNSIGNED           NULL,
  `id_aircraft`       SMALLINT UNSIGNED           NULL,
  `id_file_category`  INT UNSIGNED                NULL,
  PRIMARY KEY (`id_file`),
  INDEX `fk_sf_file_sf_file_category1_idx` (`id_file_category` ASC, `file_type` ASC),
  INDEX `fk_sf_file_sf_aircraft1_idx` (`id_aircraft` ASC),
  INDEX `fk_sf_file_sf_aircraft2_idx` (`id_aircraft_image` ASC)
)
  ENGINE = InnoDB
  DEFAULT CHARACTER SET = latin1
  COLLATE = latin1_general_ci;

但是我收到了这个错误：错误：错误1822：无法添加外键constaint。缺少约束索引＆f; fk_sf_file_sf_file_category1＆＃39;在引用的表格中＆＃39; sf_file_category＆＃39;。

我认为他是指已在表sf_file_category中创建的INDEX ALTER TABLE `sf_file` ADD CONSTRAINT `fk_sf_file_sf_file_category1` FOREIGN KEY (`id_file_category` , `file_type`) REFERENCES `sf_file_category` (`id_file_category` , `file_type`) ON DELETE NO ACTION ON UPDATE NO ACTION;（fk_sf_file_sf_file_category1_idx ASC，id_file_category ASC）。

是否有任何特定方法可以创建我缺少的多个字段外键？

Answer 1

就像提到的@Pankaj Pandey一样，我的字段声明不完全匹配，id_file_category在表上是NULL而在另一个表上NOT NULL会阻止外键创建。

SQL错误1822：无法添加外键constaint。缺少约束索引但是索引EXISTS

1 个答案: