Postgresql sequelize播种机的自定义查询

任何人都知道如何在sequelize seeder上自定义选择查询

我试过两种方法，但没有一种有效

第一次尝试

  up: function(queryInterface, Sequelize) {
    return queryInterface.sequelize.query(
      'SELECT * FROM "Users" WHERE username = "admin"',
      { type: queryInterface.sequelize.QueryTypes.SELECT }
    ).then(function(users) {});
    },

return queryInterface.sequelize.query('SELECT * FROM "Users" WHERE username = :admin', {
  replacement: {
    admin: 'admin'
  },
  type: queryInterface.sequelize.QueryTypes.SELECT
}).then(function(users) {
});

    return queryInterface.sequelize.query(
      'SELECT * FROM Users WHERE username = "admin"',
      { type: queryInterface.sequelize.QueryTypes.SELECT }
    ).then(function(users) {});

然后就出错了

SequelizeDatabaseError: column "admin" does not exist

我不明白为什么管理员栏在这里

第二次尝试

  up: function(queryInterface, Sequelize) {
    return queryInterface.sequelize.query(
      'SELECT * FROM "Users" WHERE username = "admin"',
      { type: queryInterface.sequelize.QueryTypes.SELECT }
    ).then(function(users) {});
    },

return queryInterface.sequelize.query('SELECT * FROM "Users" WHERE username = :admin', {
  replacement: {
    admin: 'admin'
  },
  type: queryInterface.sequelize.QueryTypes.SELECT
}).then(function(users) {
});

    return queryInterface.sequelize.query(
      'SELECT * FROM Users WHERE username = "admin"',
      { type: queryInterface.sequelize.QueryTypes.SELECT }
    ).then(function(users) {});

出现以下错误

SequelizeDatabaseError:语法错误位于或接近“：”

第三次尝试

return queryInterface.sequelize.query(
  'SELECT * FROM "Users" WHERE username = ' admin '',
  {type: queryInterface.sequelize.QueryTypes.SELECT})
.then(function(users) { })

错误：

SyntaxError: missing ) after argument list

已更新

第四次尝试

  up: function(queryInterface, Sequelize) {
    return queryInterface.sequelize.query(
      'SELECT * FROM "Users" WHERE username = "admin"',
      { type: queryInterface.sequelize.QueryTypes.SELECT }
    ).then(function(users) {});
    },

return queryInterface.sequelize.query('SELECT * FROM "Users" WHERE username = :admin', {
  replacement: {
    admin: 'admin'
  },
  type: queryInterface.sequelize.QueryTypes.SELECT
}).then(function(users) {
});

    return queryInterface.sequelize.query(
      'SELECT * FROM Users WHERE username = "admin"',
      { type: queryInterface.sequelize.QueryTypes.SELECT }
    ).then(function(users) {});

出现另一个错误：

    SequelizeDatabaseError: relation "Users" does not exist

queryInterface.sequelize.query（'SELECT*FROM“Users'）

工作正常，没有任何错误。我认为这里的问题是在哪里查询

我快发疯了：）

---

提前感谢您的帮助

仔细阅读Sequelize文档后，我找到了解决此问题的方法。如果您面临相同的问题，请尝试以下解决方案

return queryInterface.sequelize.query(
  'SELECT * FROM "Users" WHERE username = ? ', {
    replacements: ['admin'],
    type: queryInterface.sequelize.QueryTypes.SELECT
  }).then(users => {

---

在第一次尝试中，您应该删除usersIt周围的引号，因为它不起作用。请参阅我更新的Shivam:）不要在查询中使用双引号，除非要创建/使用标识符。您的查询应该写为：

“从用户名为'admin'的用户中选择*”

@JorgeCampos Sorry。但与第三次尝试相同的错误。@ToanTran关于第四次尝试。您是否尝试过将

用户

放在SQL语句的引号中？