mysql - SET FOREIGN_KEY_CHECKS = 0 在 Sequelize 迁移中不起作用

Question

这是我的功能:

up: queryInterface => {
  return queryInterface.sequelize.query('SET FOREIGN_KEY_CHECKS = 0')
   .then(() => queryInterface.dropTable('my_table'));
},

此查询在表中有效:

SET FOREIGN_KEY_CHECKS = 0; 
DROP TABLE IF EXISTS `my_table`;

错误是

Cannot delete or update a parent row: a foreign key constraint fails

Answer 1

如果你想在 Sequelize 迁移中删除一个约束/foreignKey，你需要使用这个:

// an example of table projects and users
await queryInterface.removeConstraint('projects', 'projects_ibfk_1'); // tableName, constraintId
await queryInterface.dropTable('users');

最好执行'SET FOREIGN_KEY_CHECKS = 0'。

如果您不知道约束 ID(在我的例子中是 projects_ibfk_1)，请使用:

const refs = await queryInterface.getForeignKeyReferencesForTable('projects');
console.log(refs);