Question

在文件顶部，我有const Sequelize = require('sequelize')。此后，我做了sequelize.authenticate()，并且可以连接数据库了。然后我用它来定义我的部门表：

var Department = sequelize.define('Department', {
    departmentId: {
        type: Sequelize.INTEGER,
        primaryKey: true,
        autoIncrement: true
    },
    departmentName: Sequelize.STRING,
    }, {
        createdAt: false,
        updatedAt: false
})

然后，我有sequelize.sync()可用。但是当我尝试创建表时：

module.exports.addDepartment = function(departmentData)
{
    return new Promise(function(resolve, reject){

        for (var i in departmentData)
        {
            if (departmentData[i] == "")
                departmentData[i] = null
        }

        Department.create(
            {
                departmentId: departmentData.departmentId,
                departmentName: departmentData.departmentName
            }
        ).then(() => { 
            resolve() 
        }).catch(() => { 
            reject("unable to create department") 
        })
    })
}

它不起作用。我确保departmentData.departmentName有一个值，并且departmentData.departmentId为空。无论我尝试什么，这都是它在控制台上显示的内容：

执行（默认）：如果不存在“部门”（“ departments”（“ departmentId” SERIAL，“ departmentName” VARCHAR（255），PRIMARY KEY（“ departmentId”）），则创建表；执行（默认）：SELECT i.relname AS名称，ix.indisprimary AS主服务器，ix.indisunique AS唯一，ix.indkey AS indkey，array_agg（a.attnum）作为 column_indexes，array_agg（a.attname）AS column_names，pg_get_indexdef（ix.indexrelid）AS定义来自pg_class t，pg_class i，pg_index ix，pg_attribute WHERE t.oid = ix.indrelid AND i.oid = ix.indexrelid AND a。 attrelid = t.oid AND t.relkind ='r'和t.relname ='部门'GROUP BY i.relname，ix.indexrelid，ix.indisprimary，ix.indisunique，ix.indkey ORDER BY i.relname; 正在执行（默认）：INSERT INTO“ Departments”（“ departmentId”，“ departmentName”）值（DEFAULT，'Creative Services'）返回*; 无法创建部门

我在做什么错？有人对我可能做错了什么建议吗？

JavaScript-无法使用sequelize .create函数创建表

0 个答案: