从 ARN 中提取资源 ID

huangapple
go评论108阅读模式
英文:

Extract resource-id from ARN

问题

我在AWS文档中看到ARN格式如下：

  1. arn:partition:service:region:account-id:resource-id
  2. arn:partition:service:region:account-id:resource-type/resource-id
  3. arn:partition:service:region:account-id:resource-type:resource-id

我正在尝试从ARN中提取resource-id。

以下代码可以工作，但不够优雅...

我正在寻找如何改进它：

  1. func GetResourceNameFromARN(roleARN string) string {
  2. 	if parsedARN, err := arn.Parse(roleARN); err == nil {
  3. 		return parsedARN.Resource
  4. 	}
  6. 	return ""
  7. }
  9. func extractResourceId(arn string) string {
  10. 	resource := GetResourceNameFromARN(arn)
  11. 	switch len(strings.Split(resource, "/")) {
  12. 	case 1:
  13. 		switch len(strings.Split(resource, ":")) {
  14. 		case 2:
  15. 			return strings.Split(resource, ":")[1]
  16. 		}
  18. 	case 2:
  19. 		return strings.Split(resource, "/")[1]
  21. 	}
  22. 	return resource
  23. }
英文:

I saw in AWS documentation that ARN formats are:

  1. arn:partition:service:region:account-id:resource-id
  2. arn:partition:service:region:account-id:resource-type/resource-id
  3. arn:partition:service:region:account-id:resource-type:resource-id

I'm trying to fetch the resource-id from the ARN.

The following code works, but ugly...

I'm searching for how to improve it:

  1. func GetResourceNameFromARN(roleARN string) string {
  2. 	if parsedARN, err := arn.Parse(roleARN); err == nil {
  3. 		return parsedARN.Resource
  4. 	}
  6. 	return ""
  7. }
  9. func extractResourceId(arn string) string {
  10. 	resource := GetResourceNameFromARN(arn)
  11. 	switch len(strings.Split(resource, "/")) {
  12. 	case 1:
  13. 		switch len(strings.Split(resource, ":")) {
  14. 		case 2:
  15. 			return strings.Split(resource, ":")[1]
  16. 		}
  18. 	case 2:
  19. 		return strings.Split(resource, "/")[1]
  21. 	}
  22. 	return resource
  23. }

答案1

得分: 2

我建议使用一个简单的正则表达式：

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"regexp"
  6. )
  8. func main() {
  9. 	// 编译表达式，通常在初始化时进行
  10. 	// 使用原始字符串避免需要转义反斜杠
  11. 	var validID = regexp.MustCompile(`[^:/]*$`)
  13. 	fmt.Println(validID.FindString("arn:partition:service:region:account-id:resource-id"))
  14. 	fmt.Println(validID.FindString("arn:partition:service:region:account-id:resource-type/resource-id"))
  15. 	fmt.Println(validID.FindString("arn:partition:service:region:account-id:resource-type:resource-id"))
  16. }

在这里查看演示：链接

英文:

I would suggest a simple regular expression:

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"regexp"
  6. )
  8. func main() {
  9. 	// Compile the expression once, usually at init time.
  10. 	// Use raw strings to avoid having to quote the backslashes.
  11. 	var validID = regexp.MustCompile(`[^:/]*$`)
  13. 	fmt.Println(validID.FindString("arn:partition:service:region:account-id:resource-id"))
  14. 	fmt.Println(validID.FindString("arn:partition:service:region:account-id:resource-type/resource-id"))
  15. 	fmt.Println(validID.FindString("arn:partition:service:region:account-id:resource-type:resource-id"))
  16. }

See the demo here

huangapple
  • 本文由 发表于 2022年2月17日 17:30:02
  • 转载请务必保留本文链接：https://go.coder-hub.com/71155485.html
  • aws-sdk
  • aws-sdk-go
  • go
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