在生产模式下,处于react组件状态的变量名称不会由webpack 4更改。
我尝试了"uglifyjs-webpack-plugin":"^2.1.3",但没有什么不同。
例如,
class Counter extends React.Component {
constructor(props) {
super(props);
this.state = {count: props.initialCount};
}
// ...
}我想在生产中,州名count被提升了。我希望是这样的,
this.state = {c:props.initialCount};webpack.config.js如下,//加载包
const path = require('path');
const webpack = require('webpack');
const UglifyJsPlugin = require('uglifyjs-webpack-plugin');//定义路径
const STATIC_DIR = path.resolve(__dirname,
'static',
'app', 'js');
const SOURCE_DIR = path.resolve(STATIC_DIR, 'src');module.exports = {
mode: "production",
devtool: 'source-map',//定义条目
entry: {
plugin: path.resolve(SOURCE_DIR, 'index.js')
},//定义输出
output: {
filename: '[name]-2.4.0.js',
path: STATIC_DIR
},这里是重要的部分,
module: {
rules: [
{
test: /\.jsx?/,
include: SOURCE_DIR,
use: {
loader: 'babel-loader',
options: {
// compact: false,
// presets: ["es2015", "react"],
plugins: ['transform-class-properties']
}
}
}
]
},
externals: {
jquery: 'jQuery'
}
};发布于 2019-06-27 17:40:40
不幸的是,我不相信这会发生。丑化器不能知道this.state是如何使用的。例如,您可以以动态方式对其进行索引:
this.state = {count: 8};
console.log(this.state['c' + 'ount']); // 8即使它可以理解在组件中不会发生这种情况,您也可以这样做
somethingElse(this.state);无论somethingElse()是什么,如果它的键被缩小,它可能无法理解this.state。
我不推荐这样做,但一种间接的方法是使用键和computed property names的字符串变量-
const COUNT_KEY = (process.env.NODE_ENV === "production" ? 'c' : 'count');
this.state = {[COUNT_KEY]: 8};
console.log(this.state[COUNT_KEY]);在生产中,它可能会被丑化成这样:
const t = "c";
this.state = {[t]: 8};
console.log(this.state[t]);发布于 2019-06-27 17:18:42
你可以像这样设置条件
if (process.env.NODE_ENV === "production") {
this.state = {c:props.initialCount}; //execute in production time
} else {
this.state = {count: props.initialCount};
}https://stackoverflow.com/questions/56787583
复制相似问题