perl - 替换多行纯文本

Question

假设您有大量的 yaml 文件(或任何类似的文件)并且您想要为具有给定名称的所有对象添加描述，例如

- name: alan
  age: 8

- name: britney # some comment about britney
  hobbies: ["painting", "CS"]
  age: 21

- name: charles # some comment about charles
                # that spans over multiple lines
  age: 42

我有一个对象列表，其中的名称需要描述，例如

britney: teamblue
charles: foobar

如何添加带有描述的行以达到以下目的:

- name: alan
  age: 8

- name: britney # some comment about britney
  hobbies: ["painting", "CS"]
  age: 21
  description: teamblue

- name: charles # some comment about charles
                # that spans over multiple lines
  age: 42
  description: foobar

到目前为止，我已经很接近了，但我一直无法用另一个替换多行纯文本:

s=$(awk "/${name}/" RS= ./*.yml)
r=$(awk "/${name}/" RS= ./*.yml && echo "  description: ${desc}")

我需要以某种方式寻找 $s 并将其替换为 $r，但我无法使其正常工作。我尝试了以下两个的多种变体:

sed "s/$s/$r/" ./*.yml
perl -i -0pe "s/$s/$r/" ./*.yml

但不知何故，yaml 中的特殊字符(换行符、双引号等)破坏了它们，我要么收到一条错误消息，如 unterminated substitute pattern，要么输出相同但没有任何变化匹配。

也可能与 sed 相关，我使用的是 macOS。

Answer 1

$ cat tst.awk
NR==FNR {
    sub(/:/,"",$1)
    map[$1] = $2
    next
}
$3 in map {
    $0 = $0 "\n  description: " map[$3]
}
{ print }

.

$ awk -f tst.awk list RS= ORS='\n\n' foo.yaml
- name: alan
  age: 8

- name: britney # some comment about britney
  hobbies: ["painting", "CS"]
  age: 21
  description: teamblue

- name: charles # some comment about charles
                # that spans over multiple lines
  age: 42
  description: foobar

上面使用了这些输入文件:

$ cat list
britney: teamblue
charles: foobar

.

$ cat foo.yaml
- name: alan
  age: 8

- name: britney # some comment about britney
  hobbies: ["painting", "CS"]
  age: 21

- name: charles # some comment about charles
                # that spans over multiple lines
  age: 42