使用 copyTpl 复制整个文件夹时重命名单个文件名 [英] Renaming single file names while copying entire folder using copyTpl

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本文介绍了使用 copyTpl 复制整个文件夹时重命名单个文件名的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我的 yeoman 生成器将文件从模板复制到目标路径:

My yeoman generator copies files from template to destination path:

this.fs.copyTpl(
            this.templatePath(),
            this.destinationPath(), {
                appName: this.props.appName
            });

在项目生成过程中,我需要将this.props.appName的值赋给一些文件名.

During project generation, I need to assign value of this.props.appName to some of filenames.

不幸的是,我不能像在这些文件中那样这样做:

Unfortunately I can't do this that way like I could do inside this files:

<%=appName%>-project.sln

所有需要重命名的文件的名称中都有appTemplate,所以我需要做的就是将appTemplate 替换为this.props 的值.应用名称.

All files that need to be renamed have appTemplate in their names, so what I need to do is simply replace appTemplate with value of this.props.appName.

我可以以某种方式配置 copyTpl 以在将某些文件复制到另一个目标时重命名它们吗?

Can I somehow configure copyTpl to rename some of files while copying them to another destination?

推荐答案

好的,我找到了解决方案.根据 yeoman 文档:

OK, I found a solution. According to yeoman docs:

任何生成器作者都可以注册一个transformStream来修改文件路径和/或内容.

Any generator author can register a transformStream to modify the file path and/or the content.

使用此方法:

this.registerTransformStream();

这意味着我可以通过一些脚本来管理所有生成的文件:

What that means is I can pipe all generated files through some script:

var rename = require("gulp-rename");
//other dependecies...

module.exports = yeoman.Base.extend({

    //some other things generator do...

    writing: function() {
        var THAT = this;
        this.registerTransformStream(rename(function(path) {
            path.basename = path.basename.replace(/(666replacethat666)/g, THAT.props.appName);
            path.dirname = path.dirname.replace(/(666replacethat666)/g, THAT.props.appName);
        }));
        this.fs.copyTpl(
            this.templatePath(),
            this.destinationPath(), {
                appName: this.props.appName
            });
    }
});

此脚本将通过 gulp-rename 管道所有文件,更改 666replacethat666 更智能的东西.

This script will pipe all files through gulp-rename, changing 666replacethat666 to something more intelligent.

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