如何从同一个YAML文件中的其他地方引用YAML"设置"？

Question

我有以下YAML:

paths:
  patha: /path/to/root/a
  pathb: /path/to/root/b
  pathc: /path/to/root/c

如何通过/path/to/root/从三个路径中删除并将其作为自己的设置来"规范化" ,例如:

paths:
  root: /path/to/root/
  patha: *root* + a
  pathb: *root* + b
  pathc: *root* + c

显然这是无效的,我只是弥补了.什么是真正的语法？可以吗？

Answer 1

我不认为这是可能的.您可以重用"节点"但不能重用它.

bill-to: &id001
    given  : Chris
    family : Dumars
ship-to: *id001

这是完全有效的YAML和字段given,family可以在ship-to块中重复使用.您可以以相同的方式重用标量节点,但是您无法更改内部的内容并从YAML内部添加路径的最后一部分.

如果重复打扰你那么多我建议让你的应用程序知道root属性并将其添加到看起来相对而非绝对的每个路径.

Answer 2

是的,使用自定义标签.Python中的示例,使!join标记连接数组中的字符串:

import yaml

## define custom tag handler
def join(loader, node):
    seq = loader.construct_sequence(node)
    return ''.join([str(i) for i in seq])

## register the tag handler
yaml.add_constructor('!join', join)

## using your sample data
yaml.load("""
paths:
    root: &BASE /path/to/root/
    patha: !join [*BASE, a]
    pathb: !join [*BASE, b]
    pathc: !join [*BASE, c]
""")

结果如下:

{
    'paths': {
        'patha': '/path/to/root/a',
        'pathb': '/path/to/root/b',
        'pathc': '/path/to/root/c',
        'root': '/path/to/root/'
     }
}

参数数组!join可以包含任意数据类型的任意数量的元素,只要它们可以转换为字符串,!join [*a, "/", *b, "/", *c]您所期望的也是如此.

Answer 3

另一种看待这种情况的方法是简单地使用另一个字段.

paths:
  root_path: &root
     val: /path/to/root/
  patha: &a
    root_path: *root
    rel_path: a
  pathb: &b
    root_path: *root
    rel_path: b
  pathc: &c
    root_path: *root
    rel_path: c

Answer 4

我创建了一个可在 Packagist 上使用的库，用于执行此功能：https ://packagist.org/packages/grasmash/yaml-expander

YAML 文件示例：

type: book
book:
  title: Dune
  author: Frank Herbert
  copyright: ${book.author} 1965
  protaganist: ${characters.0.name}
  media:
    - hardcover
characters:
  - name: Paul Atreides
    occupation: Kwisatz Haderach
    aliases:
      - Usul
      - Muad'Dib
      - The Preacher
  - name: Duncan Idaho
    occupation: Swordmaster
summary: ${book.title} by ${book.author}
product-name: ${${type}.title}

示例逻辑：

// Parse a yaml string directly, expanding internal property references.
$yaml_string = file_get_contents("dune.yml");
$expanded = \Grasmash\YamlExpander\Expander::parse($yaml_string);
print_r($expanded);

结果数组：

array (
  'type' => 'book',
  'book' => 
  array (
    'title' => 'Dune',
    'author' => 'Frank Herbert',
    'copyright' => 'Frank Herbert 1965',
    'protaganist' => 'Paul Atreides',
    'media' => 
    array (
      0 => 'hardcover',
    ),
  ),
  'characters' => 
  array (
    0 => 
    array (
      'name' => 'Paul Atreides',
      'occupation' => 'Kwisatz Haderach',
      'aliases' => 
      array (
        0 => 'Usul',
        1 => 'Muad\'Dib',
        2 => 'The Preacher',
      ),
    ),
    1 => 
    array (
      'name' => 'Duncan Idaho',
      'occupation' => 'Swordmaster',
    ),
  ),
  'summary' => 'Dune by Frank Herbert',
);

Answer 5

YML定义:

dir:
  default: /home/data/in/
  proj1: ${dir.default}p1
  proj2: ${dir.default}p2
  proj3: ${dir.default}p3 

在百里香的某个地方

<p th:utext='${@environment.getProperty("dir.default")}' />
<p th:utext='${@environment.getProperty("dir.proj1")}' /> 

输出: / home/data/in// home/data/in/p1

Answer 6

在某些语言中，您可以使用替代库，例如，tampax是 YAML 处理变量的实现：

const tampax = require('tampax');

const yamlString = `
dude:
  name: Arthur
weapon:
  favorite: Excalibur
  useless: knife
sentence: "{{dude.name}} use {{weapon.favorite}}. The goal is {{goal}}."`;

const r = tampax.yamlParseString(yamlString, { goal: 'to kill Mordred' });
console.log(r.sentence);

// output : "Arthur use Excalibur. The goal is to kill Mordred."

编者注：海报也是这个包的作者。