json - 将 JSON 对象转换为 URL 查询字符串的命令行

Question

我正在寻找将 JSON 对象转换为 URL 查询字符串的 CLI:

{
  "action_name": "Example",
  "idsite": 1,
  "rec": 1,
  "r": 345007,
  "h": 22,
  "m": 58,
  "s": 47,
  "url": "http://example.com",
  "_id": "44ebfeb79a5bb5e6",
  "_idts": 1539806328,
  "_idvc": 1,
  "_idn": 0,
  "_refts": 0,
  "_viewts": 1539806328,
  "send_image": 1,
  "cookie": 1,
  "res": "412x732",
  "gt_ms": 266,
  "pv_id": "bQkkXv"
}

结果会是这样的:

curl -Gi 'http://127.0.0.1:8000/piwik.php?action_name=Example&idsite=1&rec=1&r=345007&h=22&m=58&s=47&url=http%3A%2F%2Fexample.com%2F&_id=44ebfeb79a5bb5e6&_idts=1539806328&_idvc=1&_idn=0&_refts=0&_viewts=1539806328&send_image=1&cookie=1&res=412x732&gt_ms=266&pv_id=bQkkXv'

Answer 1

您可以使用 jq使用以下过滤器

[to_entries[] | (@uri "\(.key)"+ "="+ @uri "\(.value)")] |加入(“&”)

$ jq -r '[to_entries[] | (@uri "\(.key)" + "=" + @uri "\(.value)")] | join("&")' << JSON
{
    "var1":"value 1",
    "another var":"Some value with special & = [ ] chars"
}
JSON
var1=value%201&another%20var=Some%20value%20with%20special%20%26%20%3D%20%5B%20%5D%20chars

观看直播 https://jqplay.org/s/lu0TAWF2HHf

过滤空值/空值

在 jq 过滤器前加上 with_entries(select(.value | length !=0)) |

https://jqplay.org/s/6jQpPs5EepJ