从 BigQuery 列中提取 Json 值

Posted 2023-03-25

【中文标题】从 BigQuery 列中提取 Json 值

【英文标题】：Extract Json value from BigQuery column

【发布时间】：2021-06-14 09:05:11

【问题描述】：

我有一列包含 JSON，例如：

    "overview": 
      "resourceName": "VM-test",
      "recommendedAction": "Change machine type",
      "resource": "//compute.googleapis.com/projects/test/zones/europe-west2-c/instances/VM-test",
      "currentMachineType": 
        "cpuMilliVcores": 8000.0,
        "memoryBytes": 3.221225472E10,
        "name": "n1-standard-8",
        "memoryMb": 30720.0,
        "guestCpus": 8.0
      ,
      "location": "europe-west2-c",
      "recommendedMachineType": 
        "cpuMilliVcores": 4000.0,
        "name": "custom-4-23552",
        "memoryBytes": 2.4696061952E10,
        "memoryMb": 23552.0,
        "guestCpus": 4.0
      
    

我想将"resourceName": 及其值提取到另一个名为resourceName 的列中，并将"resource": "//compute.googleapis.com/projects/test/europe-west2-c/instances/VM-test" 提取到另一个名为ProjectName 的列中，该列将保存位于 Uri 内的项目名称。我查看了 JSON_EXTRACT 但无法弄清楚。

任何帮助将不胜感激。

谢谢

【参考方案1】：

考虑下面

select 
  json_value(text, '$.overview.resourceName') as resourceName,
  regexp_extract(json_value(text, '$.overview.resource'), r'/projects/([^/]+)') as projectName
from `project.dataset.table`         

如果应用于您问题中的样本数据 - 输出是

【参考方案2】：

试试json_value:

with mytable as (
  select '''
  
    "overview": 
      "resourceName": "VM-test",
      "recommendedAction": "Change machine type",
      "resource": "//compute.googleapis.com/projects/test/zones/europe-west2-c/instances/VM-test",
      "currentMachineType": 
        "cpuMilliVcores": 8000.0,
        "memoryBytes": 3.221225472E10,
        "name": "n1-standard-8",
        "memoryMb": 30720.0,
        "guestCpus": 8.0
      ,
      "location": "europe-west2-c",
      "recommendedMachineType": 
        "cpuMilliVcores": 4000.0,
        "name": "custom-4-23552",
        "memoryBytes": 2.4696061952E10,
        "memoryMb": 23552.0,
        "guestCpus": 4.0
      
    
  ''' as text
)
select 
  json_value(text, '$.overview.resourceName') as resourceName,
  json_value(text, '$.overview.resource') as projectName,
from mytable

【讨论】：

嗨，谢尔盖，非常感谢您的回复，您知道我可以从 uri 中提取“test”值并将其放在 projectName 列下吗？

是的，试试regexp_extract