javascript - 如何在前端使用 AJAX (jquery) 警告从快速后端获取的错误

Question

当从 express 后端服务器接收数据并对其进行操作时，success 函数工作得非常好。我无法正确处理后端抛出的错误。可以获取状态码（400 或其他）和相关文本（错误请求等），但不能获取 express 后端发送的实际消息。

后台登录代码

router.post("/users/login", async (req, res) => {
  try {
    const user = await User.findByCredentails(
      req.body.email,
      req.body.password
    );
    const token = await user.generateAuthToken();
    res.send({ user, token });
  } catch (error) {
    res.status(400).json({ error: error.message }); //error.message is thrown by the database (which logs perfectly fine in the backend when some error occurs)
  }
});

发送消息的数据库代码（这里没有错误；放在这里仅供参考）

userSchema.statics.findByCredentails = async (email, password) => {
  const user = await User.findOne({ email });
  if (!user) throw new Error("Unable to find a user with the entered email");

  const isMatch = await bcrypt.compare(password, user.password);
  if (!isMatch) throw new Error("Incorrect password, try again");

  return user;
};

前端代码（成功但不会出错）

function login() {
        const data = {
          email: $("#email-field").val(),
          password: $("#password-field").val()
        };
        $.ajax({
          url: "/users/login",
          method: "POST",
          data,
          success: function(userData) {
            console.log(userData); //works just fine
          },
          error: function(jqXHR) {
            //how can I get that error message here. I'm able to get jqXHR.status and .textStatus but not the actual message that I want.
          }
        });
      }

标签： javascriptjqueryajaxmongodbexpress