查找属性值匹配的项，不包括Sequelize中的另一个值

Question

我有这样的代码:

module.exports.MyFunction= async (req, res) => {
let token = req.body.token;
let decoded = jwt_decode(token);
let email = decoded.email;
let data = req.body;
let searchUser = data.user;
try {
let user = await User.findAll({
where: {
[Op.or]: [
{ firstName: searchUser },
{ lastName: searchUser },
{ email: searchUser },
{ publicKey: searchUser },
],
},
attributes: ["firstName", "lastName", "email", "publicKey", "avatar"],
}).then((response) => {
return response;
});
res.json({ user });
} catch (err) {
res.json({ err });
}
};

如果我运行该代码，我将获得与searchUser传递的值匹配的所有用户。

我要做的是排除具有特定电子邮件的user对象。

例如，如果有多个名为Michel的用户，我希望获得所有电子邮件地址与函数顶部声明的电子邮件变量不同的用户，即使他们的firstname匹配。

Answer 1

问题解决了:

module.exports.MyFunction = async (req, res) => {
let token = req.body.token;
let decoded = jwt_decode(token);
let loggedUserEmail = decoded.email;
let data = req.body;
let searchUser = data.user;
console.log(searchUser);
try {
let user = await User.findAll({
where: {
[Op.or]: [
{
firstName: {
[Op.startsWith]: searchUser,
},
},
{
lastName: {
[Op.startsWith]: searchUser,
},
},
{
email: {
[Op.startsWith]: searchUser,
},
},
{
publicKey: {
[Op.startsWith]: searchUser,
},
},
],
email: {
[Op.ne]: loggedUserEmail,
},

},
attributes: ["firstName", "lastName", "email", "publicKey", "avatar"],
}).then((response) => {
return response;
});
res.json({ user });
} catch (err) {
res.json({ err });
}
};