问题描述

我正在使用 gulp 对我的 javascript 文件进行 uglify 并准备好用于生产.我所拥有的是这段代码:

I am using gulp to uglify and make ready my javascript files for production. What I have is this code:

var concat = require('gulp-concat');
var del = require('del');
var gulp = require('gulp');
var gzip = require('gulp-gzip');
var less = require('gulp-less');
var minifyCSS = require('gulp-minify-css');
var uglify = require('gulp-uglify');

var js = {
    src: [
        // more files here
        'temp/js/app/appConfig.js',
        'temp/js/app/appConstant.js',
        // more files here
    ],

gulp.task('scripts', ['clean-js'], function () {
    return gulp.src(js.src).pipe(uglify())
      .pipe(concat('js.min.js'))
      .pipe(gulp.dest('content/bundles/'))
      .pipe(gzip(gzip_options))
      .pipe(gulp.dest('content/bundles/'));
});

我需要做的是替换字符串:

What I need to do is to replace the string:

dataServer: "http://localhost:3048",

与

dataServer: "http://example.com",

在文件'temp/js/app/appConstant.js'中，

In the file 'temp/js/app/appConstant.js',

我正在寻找一些建议.例如，也许我应该复制 appConstant.js 文件，更改它(不确定如何)并将 appConstantEdited.js 包含在 js.src 中?

I'm looking for some suggestions. For example perhaps I should make a copy of the appConstant.js file, change that (not sure how) and include appConstantEdited.js in the js.src?

但我不确定 gulp 如何复制文件并替换文件中的字符串.

But I am not sure with gulp how to make a copy of a file and replace a string inside a file.

您提供的任何帮助将不胜感激.

Any help you give would be much appreciated.

推荐答案

Gulp 流式输入，执行所有转换，然后流式输出.使用 Gulp 时，在两者之间保存临时文件是 AFAIK 非惯用的.

Gulp streams input, does all transformations, and then streams output. Saving temporary files in between is AFAIK non-idiomatic when using Gulp.

相反，您正在寻找的是一种替换内容的流式传输方式.自己写东西会比较容易，或者你可以使用现有的插件.对我来说，gulp-replace 效果很好.

Instead, what you're looking for, is a streaming-way of replacing content. It would be moderately easy to write something yourself, or you could use an existing plugin. For me, gulp-replace has worked quite well.

如果您想在所有文件中进行替换，您可以像这样轻松更改任务:

If you want to do the replacement in all files it's easy to change your task like this:

var replace = require('gulp-replace');

gulp.task('scripts', ['clean-js'], function () {
    return gulp.src(js.src)
      .pipe(replace(/http://localhost:d+/g, 'http://example.com'))
      .pipe(uglify())
      .pipe(concat('js.min.js'))
      .pipe(gulp.dest('content/bundles/'))
      .pipe(gzip(gzip_options))
      .pipe(gulp.dest('content/bundles/'));
});

你也可以只对你期望模式所在的文件执行 gulp.src，并通过 gulp-replace 将它们单独流式传输，将其与 gulp.src 之后所有其他文件的流.

You could also do gulp.src just on the files you expect the pattern to be in, and stream them seperately through gulp-replace, merging it with a gulp.src stream of all the other files afterwards.

这篇关于如何使用 gulp 替换文件中的字符串?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！