Question

无法开玩笑地掩盖以下陈述。

initListener = (logout, refreshToken) => {
  document.body.addEventListener("click", () => this.reset(logout, refreshToken));
  document.body.addEventListener("onkeydown", () => this.reset(logout, refreshToken));
  document.body.addEventListener("onkeyup", () => this.reset(logout, refreshToken));
  document.body.addEventListener("onkeypress", () => this.reset(logout, refreshToken));
};

和

logout = (logout, refreshToken) => {
  logout();
  redirect(appConstants.LOGIN, []);
  document.body.removeEventListener("click", () => this.reset(logout, refreshToken));
  document.body.removeEventListener("keydown", () => this.reset(logout, refreshToken));
  document.body.removeEventListener("keyup", () => this.reset(logout, refreshToken));
  document.body.removeEventListener("keypress", () => this.reset(logout, refreshToken));
  storageClearAll();
};

Answer 1

我同意@lonut，这不是做事的一种反应方式。

尽管如此，是否需要测试……

测试前：

{
    faultcode: 500,
    faultstring: 'Invalid XML',
    detail: 'Error: Unquoted attribute value\nLine: 75\nColumn: 14\nChar: c',
    statusCode: 500
}

在测试中，可以检查是否添加了侦听器并模仿这样的事件

document.body.addEventListener = jest.fn((event, cb) => {
      map[event] = cb;
    });

如何在笑话中覆盖addeventlistener

1 个答案: