在gulp任务中根据文件名切换输出文件夹 [英] Switch output folder based on filename in gulp task
问题描述
在我的src文件夹中有不同的 *。scss
文件,我希望将一个文件编译到它自己的单独文件夹中。
让我们假设我有文件 normalFile_1.scss
, specialFile.scss
, normalFile_2.scss
。我想要将两个普通文件编译到文件夹 Public / Css
,然而特殊文件应该放在文件夹 Public / Css / Special
。
我试图用 gulp-tap
获取当前文件名,这很好。(文件,t){
文件名= path.basename(file.path) ;
console.log(filename); //输出normalFile_1.css,specialFile.css,normalFile_2.css
)))
然后使用 gulp-if
然后我想根据 filename 来切换输出文件夹。 code> variable(PATHS.dist是输出root文件夹
Public
):
.pipe($。if(filename =='specialFile.css',gulp.dest(PATHS.dist +'/ Css / Special'),gulp.dest(PATHS.dist +'/ Css )));
但一切仍然以 Public / Css $ c $结尾c>文件夹。为什么这不起作用?这是甚至是一个很好的方式来实现,或有更好的方法吗?
有两种方法可以显示如下:
var gulp = require(gulp);
var sass = require(gulp-sass);
var rename = require(gulp-rename);
var path = require('path');
$ b $ gulp.task('sass',function(){
return gulp.src('src / *。scss')
.pipe(sass ).on('error',sass.logError))
.pipe(重命名(函数(路径){
if(path.basename ==specialFile){
b.d.dname =Special;
}
}))
.pipe(gulp.dest('Public / Css'))
// .pipe(gulp.dest(function(file){
// var temp = file.path.split(path.sep);
// var baseName = temp [temp.length - 1] .split('。')[0];
// console.log(baseName);
// if(baseName ==specialFile){
// return' Public / Css / Special';
//}
//否则返回'Public / Css';
//}))
});
gulp.task('default',['sass']);
显然我建议重命名版本。
[为什么一个简单的file.stem或file.basename不适用于我在gulp.dest(function(file){}版本中我不知道 - 这当然会更容易,但我只是未定义。]
I have different *.scss
files in my src folder and I want one file to be compiled in its own separate folder.
Lets assume I have the files normalFile_1.scss
, specialFile.scss
, normalFile_2.scss
. I want the two normal files to be compiled to the folder Public/Css
, the special file however should end up in the folder Public/Css/Special
.
I have tried to get the current filename in the task with gulp-tap
, which works fine.
.pipe($.tap(function (file, t) {
filename = path.basename(file.path);
console.log(filename); //outputs normalFile_1.css, specialFile.css, normalFile_2.css
}))
And with gulp-if
I then wanted to switch the output folder based on the filename
variable (PATHS.dist is the output "root" folder Public
):
.pipe($.if(filename == 'specialFile.css', gulp.dest(PATHS.dist + '/Css/Special'), gulp.dest(PATHS.dist + '/Css')));
But everything still ends up in the Public/Css
folder. Why does this not work? Is this even a good way of trying to accomplish that or are there better methods?
There are two ways to do this shown below:
var gulp = require("gulp");
var sass = require("gulp-sass");
var rename = require("gulp-rename");
var path = require('path');
gulp.task('sass', function () {
return gulp.src('src/*.scss')
.pipe(sass().on('error', sass.logError))
.pipe(rename(function (path) {
if (path.basename == "specialFile") {
path.dirname = "Special";
}
}))
.pipe(gulp.dest('Public/Css'))
// .pipe(gulp.dest(function(file) {
// var temp = file.path.split(path.sep);
// var baseName = temp[temp.length - 1].split('.')[0];
// console.log(baseName);
// if (baseName == "specialFile") {
// return 'Public/Css/Special';
// }
// else return 'Public/Css';
// }))
});
gulp.task('default', ['sass']);
Obviously I suggest the rename version.
[Why a simple file.stem or file.basename doesn't work for me in the gulp.dest(function (file) {} version I don't know - that would certainly be easier but I just get undefined.]
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