提取打字稿用出口来一饮而尽JSON文件 [英] Extracting typescript exports to json file using gulp

问题描述

我有几个打字稿的文件，其中一些出口一定的无功 - 名为APIS - 这是对象的数组。
我想用一饮而尽提取所有这些出口到一个JSON文件中的值，并管他们。

I have several typescript files, some of them export a certain var - named APIS - which is an array of objects. I want to extract the values of all of these exports, and pipe them to a json file using gulp.

例如，我有一个文件夹命名服务，与3个文件：service1.ts，service2.ts，service3.ts

For example, I have a folder named services, with 3 files: service1.ts, service2.ts, service3.ts.

service1.ts：

...
export const APIS = [ { "field1" : "blabla" } ];

service2.ts：

...
export const APIS = [ { "field2" : "yadayada" }, { "field3" : "yadabla" } ];

service3.ts： - 不导出API的VAR

service3.ts: - does not export the APIS var.

我想在奥得使用一饮而尽创建一个JSON文件看起来是这样的：

I want to use gulp in oder create a json file that looks something like this:

[ { "field1" : "blabla" }, { "field2" : "yadayada" }, { "field3" : "yadabla" } ]

gulpfile.js - 在？是失踪code的占位符。

gulpfile.js - the ??? is a placeholder for the missing code.

gulp.task('default', function () {
    return gulp.src('.../services/*.ts')
            .pipe(???)
            .pipe(concat('export.json'))
            .pipe(gulp.dest('./test'));
});

我是新来的两个打字稿及放大器;一饮而尽，所以我不知道如何实现这一目标...任何想法？ ：）

I'm new to both typescript & gulp, so I'm not sure how to achieve this... any ideas? :)

编辑：所以，我的理解是有这个问题没有开箱即用的解决方案，我需要作家自己的任务/插件。我真的不知道如何实现，虽然。
理想情况下，我要的是找到一个大口插件（或插件的组合），可以处理TS / js文件与属性的对象。然后，我可以提取我从文件中所需要的变种。

So, I understand that there's no OOTB solution for this, and I need to writer my own task / plugin. I'm not really sure how to achieve that, though. Ideally, what I want is to find a gulp plugin (or a combination of plugins) that can handle ts / js files as objects with properties. Then I can extract the var I need from the file.

我真的无法找到类似的东西，只是字符串操作插件 - 治疗我的TS文件作为一个字符串，并使用与正则表达式搜索似乎过于复杂了我。有什么事，我在这里丢失？有没有一种更直接的方式做到这一点？

I couldn't really find something like that, only string manipulation plugins - Treating my ts file as a string and using search with regex seems overly complicated to me. Is there something I'm missing here? is there a more straight-forward way to do this?

推荐答案

这是我落得这样做，它为我工作。我在这里并主张它的情况下，任何人发现它是有用的。 ：）

This is what I ended up doing, and it worked for me. I'm positing it here in case anyone else finds it useful. :)

相反.TS，我保存在.js文件的出口，即：

Instead of .ts, I saved the exports in .js files, i.e:

service2.export.js：

service2.export.js:

exports.APIS = [ { "field2" : "yadayada" }, { "field3" : "yadabla" } ];

基于这里给出的答案

： http://stackoverflow.com/a/36869651/3007732 我创建了一个一饮而尽任务如下：

Based on the answer given here: http://stackoverflow.com/a/36869651/3007732 I created a gulp task as following:

var gulp = require('gulp');
var concat = require('gulp-concat');
var map = require('gulp-map');
var fs = require('fs');

var allServices;

gulp.task('default', function() {

    var allServices = [];

    var stream = gulp.src('./**/*.export.js')
        .pipe(map(function(file) {
            var obj = require(file.path);
            if (obj.APIS != null) {
                allServices.push.apply(allServices, obj.APIS);
            }
            return file;
        }));

    stream.on("end", function (cb)
    {   
        fs.writeFile('./export.json', JSON.stringify(allServices), cb);
    });

    return stream;
});

，现在我得到export.json以下的输出：

and now I get the following output in export.json:

[ { "field1" : "blabla" }, { "field2" : "yadayada" }, { "field3" : "yadabla" } ]

这正是我想要的。

which is exactly what I wanted.

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