来自 x509 证书的 golang 主题 dn

有没有简单的方法可以从 go 中的 x509 证书中以字符串形式获取完整的主题 DN（或颁发者 DN）？

我无法在 pkix.Name 中找到任何类似“.String()”的方法

解决方案（感谢同事）：

var oid = map[string]string{
    "2.5.4.3":                    "CN",
    "2.5.4.4":                    "SN",
    "2.5.4.5":                    "serialNumber",
    "2.5.4.6":                    "C",
    "2.5.4.7":                    "L",
    "2.5.4.8":                    "ST",
    "2.5.4.9":                    "streetAddress",
    "2.5.4.10":                   "O",
    "2.5.4.11":                   "OU",
    "2.5.4.12":                   "title",
    "2.5.4.17":                   "postalCode",
    "2.5.4.42":                   "GN",
    "2.5.4.43":                   "initials",
    "2.5.4.44":                   "generationQualifier",
    "2.5.4.46":                   "dnQualifier",
    "2.5.4.65":                   "pseudonym",
    "0.9.2342.19200300.100.1.25": "DC",
    "1.2.840.113549.1.9.1":       "emailAddress",
    "0.9.2342.19200300.100.1.1":  "userid",
}

func getDNFromCert(namespace pkix.Name, sep string) (string, error) {
    subject := []string{}
    for _, s := range namespace.ToRDNSequence() {
        for _, i := range s {
            if v, ok := i.Value.(string); ok {
                if name, ok := oid[i.Type.String()]; ok {
                    // <oid name>=<value>
                    subject = append(subject, fmt.Sprintf("%s=%s", name, v))
                } else {
                    // <oid>=<value> if no <oid name> is found
                    subject = append(subject, fmt.Sprintf("%s=%s", i.Type.String(), v))
                }
            } else {
                // <oid>=<value in default format> if value is not string
                subject = append(subject, fmt.Sprintf("%s=%v", i.Type.String, v))
            }
        }
    }
    return sep + strings.Join(subject, sep), nil
}

调用函数：

subj, err := getDNFromCert(x509Cert.Subject, "/")
if err != nil {
   // do error handling
}
fmt.Println(subj)

输出（示例）：

/C=US/O=some organization/OU=unit/CN=common name

这似乎是唯一“简单”的解决方案