我有一个项目,其中有很多JavaScript文件放在几个文件夹中。
root
| - index.html
| - lib
| -- jquery-ui.js
| -- jquery.js
| -- html2canvas.js
|
| - js
| -- main.js
| -- app1.js
| -- app2.js
| -- app3.js
并且O需要使用webpack来分别缩小这些文件,以使每个文件都保持其路径。
我不想包含所有代码的单个文件。
我当前的配置:
const path = require('path');
module.exports = {
entry:[
'./js/index.js',
'./pages/MCF/js/index.js',
'./pages/MCF/js/refresh.js',
'./pages/MCF/lang/index.js',
'./pages/MCF/lib/scriptJs/script.js',
'./pages/MCF/lib/scriptJs/load.js'
],
output :{
path :path.resolve('./prod/js'),
filename: "app.min.js"
},
module:{
rules:[
{
test:/\.js$/,
exclude: /(node_modules|bower_components)/,
use:['babel-loader']
}
]
}
}
答案 0 :(得分:1)
You can take a reference from here for multiple-entry-points.
这是您应该进行单独文件构建的方式:
ClassLoader cl = Thread.currentThread().getContextClassLoader();
// Loop through all drivers
Enumeration<Driver> drivers = DriverManager.getDrivers();
while (drivers.hasMoreElements()) {
Driver driver = drivers.nextElement();
if (driver.getClass().getClassLoader() == cl) {
// This driver was registered by the webapp's ClassLoader, so deregister it:
try {
System.out.println("Deregistering JDBC driver " + driver);
DriverManager.deregisterDriver(driver);
} catch (SQLException ex) {
System.out.println("Error deregistering JDBC driver " + driver);
ex.printStackTrace();
}
} else {
// driver was not registered by the webapp's ClassLoader and may be in use
// elsewhere
System.out.println(
"Not deregistering JDBC driver {} as it does not belong to this webapp's ClassLoader" + driver);
}
}
module.exports = {
entry:{
main : './js/index.js',
main2 : './pages/MCF/js/index.js',
refresh : './pages/MCF/js/refresh.js',
langindex : './pages/MCF/lang/index.js',
script : './pages/MCF/lib/scriptJs/script.js',
load : './pages/MCF/lib/scriptJs/load.js'
},
output :{
path :path.resolve('./prod/js'),
filename: "[name].min.js"
},
...
};
通过此[name].min.js
,您可以将每个条目的版本保存为给定的名称(键)。输出将是:
[name]