問題描述

我有一個 gulp 任務，我想獲取一些源文件并將它們復制到 build/premium 和 build/free 然后從構建/免費.

我的嘗試是這樣做的:

gulp.task("build", ["clean"], function () {gulp.src(["src/*", "!src/composer.*", "LICENSE"]).pipe(gulp.dest("build/premium")).pipe(del(["build/free/plugins/*", "!build/free/plugins/index.php"])).pipe(gulp.dest("build/free"));});

這會導致錯誤:

TypeError: dest.on 不是函數在 DestroyableTransform.Stream.pipe (stream.js:45:8)在 Gulp.<匿名>(/Users/gezim/projects/myproj/gulpfile.js:9:6)

如何在刪除端口時完成此操作?有沒有更好的方法來做到這一點?

我會使用 gulp-filter 只刪除不應該從第二個目的地復制的內容.

我將任務的意圖解釋為希望 src 中存在的所有內容都存在于 build/premium 中.但是，build/free 應該排除最初在 src/plugins 中但仍應包含 src/plugins/index.php 的所有內容.p>

這是一個有效的 gulpfile:

var gulp = require("gulp");var filter = require("gulp-filter");變種德爾=要求(德爾")；gulp.task("干凈", function () {返回德爾(構建")；});gulp.task("build", ["clean"], function () {返回 gulp.src(["src/**", "!src/composer.*", "LICENSE"]).pipe(gulp.dest("build/premium")).pipe(filter(["**", "!plugins/**", "plugins/index.php"])).pipe(gulp.dest("build/free"));});

傳遞給 filter 的模式是 relative 路徑.由于 gulp.src 模式具有 src/** 這意味著它們是相對于 src 的.

還要注意，del 不能直接傳遞給 .pipe()，因為它返回一個承諾.它可以從任務中返回，就像 clean 任務一樣.

I have a gulp task in which I want to take some source files and copy them to build/premium and build/free and then remove some extra files from build/free.

My attempt at that was doing this:

gulp.task("build", ["clean"], function () {
  gulp.src(["src/*", "!src/composer.*", "LICENSE"])
    .pipe(gulp.dest("build/premium"))
    .pipe(del(["build/free/plugins/*", "!build/free/plugins/index.php"]))
    .pipe(gulp.dest("build/free"));
});

Which results in an error:

TypeError: dest.on is not a function
    at DestroyableTransform.Stream.pipe (stream.js:45:8)
    at Gulp.<anonymous> (/Users/gezim/projects/myproj/gulpfile.js:9:6)

How do I accomplish this the deleting port? Is there a better way altogether to do this?

I would use gulp-filter to drop only what should not be copied from the 2nd destination.

I interpreted the intent of the task as wanting everything present in src to be present in build/premium. However, build/free should exclude everything which was originally in src/plugins but should still include src/plugins/index.php.

Here is a working gulpfile:

var gulp = require("gulp");
var filter = require("gulp-filter");
var del = require("del");

gulp.task("clean", function () {
  return del("build");
});

gulp.task("build", ["clean"], function () {
  return gulp.src(["src/**", "!src/composer.*", "LICENSE"])
    .pipe(gulp.dest("build/premium"))
    .pipe(filter(["**", "!plugins/**", "plugins/index.php"]))
    .pipe(gulp.dest("build/free"));
});

The patterns passed to filter are relative paths. Since the gulp.src pattern has src/** it means they are relative to src.

Note also that del cannot be passed straight to .pipe() as it returns a promise. It can be returned from a task, like the clean task does.

這篇關于在 gulp 任務中刪除文件的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！