Google Security Command Center -如何以JSON格式导出调查结果

Question

使用此示例：

from google.cloud import securitycenter

# Create a client.
client = securitycenter.SecurityCenterClient()

# organization_id is the numeric ID of the organization. e.g.:
# organization_id = "111122222444"
org_name = "organizations/{org_id}".format(org_id=organization_id)
# The "sources/-" suffix lists findings across all sources.  You
# also use a specific source_name instead.
all_sources = "{org_name}/sources/-".format(org_name=org_name)
finding_result_iterator = client.list_findings(all_sources)
for i, finding_result in enumerate(finding_result_iterator):
    print(
        "{}: name: {} resource: {}".format(
            i, finding_result.finding.name, finding_result.finding.resource_name
        )
    )

我想将所有查找结果导出为'google.cloud.securitycenter_v1.types.Finding‘数组，但是类型(finding_result.finding)返回: class JSON

使用json.dumps(finding_result.finding)会导致错误，表明它是不可序列化的。

使用gcloud SDK可以通过指定"--format json“来实现。

Answer 1

我想通了。

您必须添加以下导入from google.cloud.securitycenter import ListFindingsResponse

然后在循环中添加以下代码行

ListFindingsResponse.ListFindingsResult.to_json(finding_result)

因此，解决方案应该如下所示

from google.cloud import securitycenter
from google.cloud.securitycenter import ListFindingsResponse

# Create a client.
client = securitycenter.SecurityCenterClient()

# organization_id is the numeric ID of the organization. e.g.:
# organization_id = "111122222444"
org_name = "organizations/{org_id}".format(org_id=organization_id)
# The "sources/-" suffix lists findings across all sources.  You
# also use a specific source_name instead.
all_sources = "{org_name}/sources/-".format(org_name=org_name)
finding_result_iterator = client.list_findings(all_sources)
for i, finding_result in enumerate(finding_result_iterator):
    print(
        ListFindingsResponse.ListFindingsResult.to_json(finding_result)
    )
    print(
        "{}: name: {} resource: {}".format(
            i, finding_result.finding.name, finding_result.finding.resource_name
        )
    )