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ca: Output issuer configuration details at startup (#7582) 

As an operator, it's helpful to know what issuers a given CA has at
startup time. We already log the default issuer profile for instance.

Related to https://github.com/letsencrypt/boulder/issues/7469
This commit is contained in:
Phil Porada 2024-07-10 15:53:25 -04:00 committed by GitHub
parent 30c6e592f7
commit 861ea99cc7
No known key found for this signature in database
GPG Key ID: B5690EEEBB952194
1 changed files with 3 additions and 3 deletions
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6
cmd/boulder-ca/main.go
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@ -188,11 +188,13 @@ func main() {
cmd.FailOnError(err, "Failed to load CT Log List") cmd.FailOnError(err, "Failed to load CT Log List")
} }
clk := cmd.Clock()
issuers := make([]*issuance.Issuer, 0, len(c.CA.Issuance.Issuers)) issuers := make([]*issuance.Issuer, 0, len(c.CA.Issuance.Issuers))
for _, issuerConfig := range c.CA.Issuance.Issuers { for _, issuerConfig := range c.CA.Issuance.Issuers {
issuer, err := issuance.LoadIssuer(issuerConfig, cmd.Clock()) issuer, err := issuance.LoadIssuer(issuerConfig, clk)
cmd.FailOnError(err, "Loading issuer") cmd.FailOnError(err, "Loading issuer")
issuers = append(issuers, issuer) issuers = append(issuers, issuer)
logger.Infof("Loaded issuer: name=[%s] keytype=[%s] nameID=[%v] isActive=[%t]", issuer.Name(), issuer.KeyType(), issuer.NameID(), issuer.IsActive())
} }
if c.CA.Issuance.DefaultCertificateProfileName == "" { if c.CA.Issuance.DefaultCertificateProfileName == "" {
@ -223,8 +225,6 @@ func main() {
tlsConfig, err := c.CA.TLS.Load(scope) tlsConfig, err := c.CA.TLS.Load(scope)
cmd.FailOnError(err, "TLS config") cmd.FailOnError(err, "TLS config")
clk := cmd.Clock()
conn, err := bgrpc.ClientSetup(c.CA.SAService, tlsConfig, scope, clk) conn, err := bgrpc.ClientSetup(c.CA.SAService, tlsConfig, scope, clk)
cmd.FailOnError(err, "Failed to load credentials and create gRPC connection to SA") cmd.FailOnError(err, "Failed to load credentials and create gRPC connection to SA")
sa := sapb.NewStorageAuthorityClient(conn) sa := sapb.NewStorageAuthorityClient(conn)