fix(cron): don't show error for empty cron schedule

Makefile

@@ -77,8 +77,8 @@ v/v:
 clean:
 	rm -rf 'data' 'vieter' 'dvieter' 'pvieter' 'vieter.c' 'dvieterctl' 'vieterctl' 'pkg' 'src/vieter' *.pkg.tar.zst 'suvieter' 'afvieter' '$(SRC_DIR)/_docs'
 
-.PHONY: docs
-docs:
+.PHONY: api-docs
+api-docs:
 	rm -rf '$(SRC_DIR)/_docs'
 	cd '$(SRC_DIR)' && v doc -all -f html -m -readme .
 

src/build/build.v

@@ -114,7 +114,6 @@ pub fn build_repo(address string, api_key string, base_image_id string, repo &gi
 			break
 		}
 
-		// Wait for 5 seconds
 		time.sleep(1 * time.second)
 	}
 

src/cron/daemon/daemon.v

@@ -132,12 +132,17 @@ pub fn (mut d Daemon) run() ? {
 
 // schedule_build adds the next occurence of the given repo build to the queue.
 fn (mut d Daemon) schedule_build(repo_id string, repo git.GitRepo) ? {
-	ce := parse_expression(repo.schedule) or {
+	ce := if repo.schedule != '' {
+		parse_expression(repo.schedule) or {
 			// TODO This shouldn't return an error if the expression is empty.
 			d.lerror("Error while parsing cron expression '$repo.schedule' ($repo_id): $err.msg()")
 
 			d.global_schedule
 		}
+	} else {
+		d.global_schedule
+	}
+
 	// A repo that can't be scheduled will just be skipped for now
 	timestamp := ce.next_from_now() ?