[英]Regex inside path parameter of Gulp.src()
How does one exclude minimised files from a gulp.src ( glob
?) file path via regex or wildcards? 如何通过正则表达式或通配符从gulp.src( glob
?)文件路径中排除最小化文件?
Or maybe even simpler, how does one move minimised files only to their min/{javascript|css} folders? 甚至更简单,如何将最小化文件仅移动到其min / {javascript | css}文件夹?
I have tried the following: 我尝试了以下方法:
var dir = "./src/",
lib_dir = "./lib/";
gulp.task("movelib", function(){
gulp.src(lib_dir + "libraries\/.*\.min\.js")
.pipe(plumber({errorHandler: function(e){console.log(e); this.emit('end');}}))
.pipe(gulp.dest("./dist/assets/min/js"));
gulp.src(lib_dir + "libraries\/.*(?!\.min)\.js")
.pipe(plumber({errorHandler: function(e){console.log(e); this.emit('end');}}))
.pipe(gulp.dest("./dist/assets/js"));
gulp.src(lib_dir + "libraries\/.*\.min\.css")
.pipe(plumber({errorHandler: function(e){console.log(e); this.emit('end');}}))
.pipe(gulp.dest("./dist/assets/min/css"));
gulp.src(lib_dir + "libraries\/.*(?!\.min)\.css")
.pipe(plumber({errorHandler: function(e){console.log(e); this.emit('end');}}))
.pipe(gulp.dest("./dist/assets/css"));
return;
});
However it does not work. 但是,它不起作用。
To exclude min.js file, how about that ? 要排除min.js文件,那又如何呢?
gulp.src(lib_dir + "libraries/!( .min.js| .min.css)") gulp.src(lib_dir +“ libraries /!(.min.js | .min.css)”)
For your second question, did you try without escaping slashes and dots ? 对于您的第二个问题,您是否尝试了不逃脱斜线和圆点的情况?
gulp.src(lib_dir + "libraries/*.min.js") gulp.src(lib_dir +“ libraries / *。min.js”)
cf: https://github.com/isaacs/node-glob cf: https : //github.com/isaacs/node-glob
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