Javascript Gulp - 复制和重命名文件
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Gulp - copy and rename a file
提问by Adam Rackis
I'm extremely new to Gulp. I'm basically trying to watch for a modified JavaScript file, and then make a new copy of it with a new name. (eventually there'll be some processing on it, but Rome wasn't built in a day).
我对 Gulp 非常陌生。我基本上是在尝试查看修改后的 JavaScript 文件，然后使用新名称制作它的新副本。（最终会有一些处理，但罗马不是一天建成的）。
My (naive) attempt is this:
我的（天真）尝试是这样的：
gulp.task('default', function() {
return gulp.watch('../**/**.js', function(obj){
gulp.src(obj.path)
.pipe(gulp.dest('foobar.js'));
});
});
This takes the modified file and successfully copies it into a folder now called foobar.js. Is there anything simple I can replace gulp.dest('foobar.js')with that will simply copy and rename the src file in place?
这会将修改后的文件成功复制到现在称为 foobar.js 的文件夹中。有什么简单的可以替换的东西gulp.dest('foobar.js')，它可以简单地复制并重命名 src 文件吗？
EDIT
编辑
By copy in place, I mean I want to take the modified file, and make a copy of it right where it currently iswith a new name. The equivalent of clicking the file (in windows) and hitting control-c control-v, then renaming the resulting file.
通过就地复制，我的意思是我想获取修改后的文件，并用新名称将其复制到当前所在的位置。相当于单击文件（在 Windows 中）并点击 control-c control-v，然后重命名生成的文件。
回答by knksmith57
I'm not 100% certain what you mean by
我不是 100% 确定你的意思
copy and rename ... in place
复制并重命名......就地
But, based on your current code, if you simply wish to:
但是，根据您当前的代码，如果您只是希望：
- Watch all
.jsfiles in the parentdirectory and - Copy them to the
cwd(current working directory) and - Name allcopies, regardlessof source file, the same thing
- 监视父目录
.js中的所有文件并 - 将它们复制到
cwd（当前工作目录）和 - 命名所有副本，无论源文件如何，都一样
Then you could use gulp-renameto do just that:
然后你可以使用gulp-rename来做到这一点：
var gulp = require('gulp');
var rename = require('gulp-rename');
gulp.task('default', function() {
return gulp.watch('../**/**.js', function(obj) {
gulp.src(obj.path)
.pipe(rename('newFileName.js'))
.pipe(gulp.dest('.'));
});
});
In this case, the output filename is newFileName.js
在这种情况下，输出文件名是 newFileName.js
In order to use the module, you'll need to install the gulp-renamepackage with npm (ie: npm install gulp-rename).
为了使用该模块，您需要gulp-rename使用 npm（即：）安装该软件包npm install gulp-rename。
More examples are available on the package details page on npm @ https://www.npmjs.com/package/gulp-rename#usage
在 npm @ https://www.npmjs.com/package/gulp-rename#usage上的包详细信息页面上提供了更多示例
回答by Adam Rackis
It wasn't pretty getting there, but in the end it appears this is what I want (with some ES6 transpiling in the middle).
到达那里并不是很顺利，但最终似乎这就是我想要的（中间有一些 ES6 转译）。
The key appears to be the options object with a base property in the call to src. That seems to be what's needed to maintain the path of the current file in the call to dest.
关键似乎是在调用中具有基本属性的选项对象src。这似乎是在调用dest.
var gulp = require('gulp'),
rename = require('gulp-rename'),
babel = require('gulp-babel');
gulp.task('default', function() {
return gulp.watch('../**/$**.js', function(obj){
if (obj.type === 'changed') {
gulp.src(obj.path, { base: './' })
.pipe(babel())
.pipe(rename(function (path) {
path.basename = path.basename.replace('$', '');
}))
.pipe(gulp.dest(''));
}
});
});
