Question

在我的应用程序中，我有一个文件夹（见证），其index.js包含下面提到的代码片段。当我在index.js中导出所有的Sagas，Reducer，容器，组件，然后想在app.js中导入它们

import * as actions from './actions';
import * as components from './Components';
import * as constants from './constants';
import * as containers from './Containers';
import reducer from './reducers';
import * as sagas from './sagas';

export default {
    actions,
    components, 
    containers,
    constants,
    reducer, 
    sagas
};

现在，我该如何导入，比如说app.js中的容器以及我写的时候

import Testimonial from './ignitus-Testimonial';

这是我导入对象并将其console.log()放在app.js上时的样子

Testimonail {操作：{…}，组件：{…}，容器：{…}，常数：{…}，减速器：ƒ，...}

但是我只想在我完成的所有导出中的containers中使用app.js。

Answer 1

如果希望单独使用导出，则不应将它们作为default导出导出。

它们可以重新导出为：

import * as actions from './actions';
import * as components from './Components';
import * as constants from './constants';
import * as containers from './Containers';
import reducer from './reducers';
import * as sagas from './sagas';

export {
    actions,
    components, 
    containers,
    constants,
    reducer, 
    sagas
};

它们可以单独导入为：

import { sagas } from './ignitus-Testimonial';

或一起作为：

import * as testimonial from './ignitus-Testimonial';

这种方法的问题在于，它会扼杀摇树优化，因此不建议将其用于输出大小很重要的用途，即客户端应用程序。

Answer 2

使用如下所示的命名导入

// run this code to check random raindrop colors
// this project or codes are create Rain(loop) with different positions and colors.

// Show the positions of raindrop(x,y)
var xPositions = [200,100,250,350];
var yPositions = [0,-10,-30,-60];

// Draw function to repeat every raindrop position
draw = function() {
  background(204, 247, 255);

  // Condition to start back y position to do animation
  if (mouseIsPressed) {
    xPositions.push(mouseX);
    yPositions.push(0);
  }

  // Loop the raindrop codes to repeat with some change
  for (var i = 0; i < xPositions.length; i++) {
    noStroke();

    // fill random colors to the raindrop
    fill(random(0,255),random(0,314),random(0,255));

    // Raindrop
    ellipse(xPositions[i], yPositions[i], 10, 10);

    // Increase y position up to down everytime +=5
    yPositions[i] += 5;

    // Drops start back at the top once they've reached the bottom
    if(yPositions[i] > 400) {
        yPositions[i] = (0); 
        xPositions[i] = random(0,400);
    }
  }
};

Answer 3

我自己解决了问题：)这样写就可以解决问题。 <Testimonial.containers.TestimonialContainer />

进口/出口Sagas，容器，异径管

3 个答案: