我有一个名为User的模型,但是Sequelize会在我试图保存到数据库中时查找表USERS。有人知道如何设置Sequelize来使用单表名称吗?谢谢。
发布于 2014-04-21 04:25:21
可以使用属性freezeTableName
的docs状态。
请看下面的示例:
var Bar = sequelize.define('Bar', { /* bla */ }, {
// don't add the timestamp attributes (updatedAt, createdAt)
timestamps: false,
// don't delete database entries but set the newly added attribute deletedAt
// to the current date (when deletion was done). paranoid will only work if
// timestamps are enabled
paranoid: true,
// don't use camelcase for automatically added attributes but underscore style
// so updatedAt will be updated_at
underscored: true,
// disable the modification of tablenames; By default, sequelize will automatically
// transform all passed model names (first parameter of define) into plural.
// if you don't want that, set the following
freezeTableName: true,
// define the table's name
tableName: 'my_very_custom_table_name'
})
发布于 2016-01-02 02:30:42
虽然公认的答案是正确的,但您可以对所有表执行一次操作,而不必为每个表单独执行此操作。您只需将一个类似的options对象传入Sequelize构造函数,如下所示:
var Sequelize = require('sequelize');
//database wide options
var opts = {
define: {
//prevent sequelize from pluralizing table names
freezeTableName: true
}
}
var sequelize = new Sequelize('mysql://root:123abc@localhost:3306/mydatabase', opts)
现在,在定义实体时,不必指定freezeTableName: true
var Project = sequelize.define('Project', {
title: Sequelize.STRING,
description: Sequelize.TEXT
})
发布于 2020-08-10 15:28:31
如果您需要为单数和复数定义使用不同的模型名称,您可以将name作为参数传递到model的选项中。
请看下面的示例:
const People = sequelize.define('people', {
name: DataTypes.STRING,
}, {
hooks: {
beforeCount (options) {
options.raw = true;
}
},
tableName: 'people',
name: {
singular: 'person',
plural: 'people'
}
});
这将在查询单个记录时将"person“作为对象返回,当我们获取多个记录时将"people”作为数组返回。
https://stackoverflow.com/questions/21114499
复制相似问题