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我有一个具有这种扁平结构的 JSON：

[{
    "PK": "1111",
    "SOURCE_DB": "Oracle",
    "CONTACT_TYPE": "Phone",
    "CONTACT_SUBTYPE": "Work",
    "EMAIL": null
    "PHONE_COUNTRY_CODE": "44",
    "PHONE_NUMBER": "12345678",
    "PHONE_EXT": "907643",
    "STATUS": "Active"
    }, {
    "PK": "1111",
    "SOURCE_DB": "Oracle",
    "CONTACT_TYPE": "Phone",
    "CONTACT_SUBTYPE": "Home",
    "EMAIL": null
    "PHONE_COUNTRY_CODE": "353",
    "PHONE_NUMBER": "87654321",
    "PHONE_EXT": null,
    "STATUS": "Active"
    }, {
    "PK": "1111",
    "SOURCE_DB": "",
    "CONTACT_TYPE": "Email",
    "CONTACT_SUBTYPE": "Personal",
    "EMAIL": "me@mail.com"
    "PHONE_COUNTRY_CODE": null,
    "PHONE_NUMBER": null,
    "PHONE_EXT": null,
    "STATUS": "Active"
    },
    {
    "PK": "2222",
    "SOURCE_DB": "DB2",
    "CONTACT_TYPE": "Phone",
    "CONTACT_SUBTYPE": "Home",
    "EMAIL": null
    "PHONE_COUNTRY_CODE": "44",
    "PHONE_NUMBER": "98761234",
    "PHONE_EXT": null,
    "STATUS": "Inactive"
    }, {
    "PK": "2222",
    "SOURCE_DB": "DB2",
    "CONTACT_TYPE": "Email",
    "CONTACT_SUBTYPE": "Work",
    "EMAIL": "you@mail.co.uk"
    "PHONE_COUNTRY_CODE": null,
    "PHONE_NUMBER": null,
    "PHONE_EXT": null,
    "STATUS": "Active"
    }
]

然后，我想将它们分组，首先按 Key (PK)，然后在每个条目中，ContactMethods 将被分组在一起。这是输出：

{
    "Accounts": [{
            "Reference": {
                "Key": "1111",
                "System": "Oracle"
            },
            "ContactMethods": {
                "Phone": [{
                        "Subtype": "Work",
                        "CountryCode": "44",
                        "Number": "12345678",
                        "Extension": "907643",
                        "Active": true
                    }, {
                        "Subtype": "Home",
                        "CountryCode": "353",
                        "Number": "87654321",
                        "Extension": null,
                        "Active": true
                    }
                ],
                "Email": [{
                        "Subtype": "Personal",
                        "EmailAddress": "my@mail.com",
                        "Active": true
                    }
                ]
            }
        }, {
            "Reference": {
                "Key": "2222",
                "System": "DB2"
            },
            "ContactMethods": {
                "Phone": [{
                        "Subtype": "Home",
                        "CountryCode": "44",
                        "Number": "98761234",
                        "Extension": null,
                        "Active": false
                    }
                ],
                "Email": [{
                        "Subtype": "Work",
                        "EmailAddress": "you@mail.co.uk",
                        "Active": true
                    }
                ]
            }
        }
    ]
}

我可以按 PK 对其进行分组，但我在第二部分遇到了困难，即如何在嵌套结构中进行分组。你能展示一个样本规格并做一些解释吗？

4

1 回答 1

1

可能但确实令人费解/冗长。这推动了 Jolt 应该做的事情的界限。

一个枢轴和一些重新映射是可维护的，但这足够复杂，如果出现问题/您的数据很奇怪，将很难调试。

需要5个步骤。两个将 STATUS 从字符串修复为布尔值。两个来透视和子透视数据。最后一个将所有内容放在正确的最终位置。

我建议检查每个步骤检查 Jolt 演示站点的自己的选项卡/副本中的每个步骤，以查看/了解每个步骤在做什么。

规格

[
  {
    // ninja in a true and false value so that 
    // Status "Active" / "Inactive" can be "mapped" to booleans
    "operation": "default",
    "spec": {
      "*": {
        "FALSE": false,
        "TRUE": true
      }
    }
  },
  {
    // fix STATUS
    "operation": "shift",
    "spec": {
      "*": {
        // 
        "STATUS": {
          // Match "Active" as make STATUS be true
          "Active": {
            "@(2,TRUE)": "[&3].STATUS"
          },
          // Everything else set to false
          "*": {
            "@(2,FALSE)": "[&3].STATUS"
          }
        },
        // match and discard TRUE and FALSE
        "TRUE|FALSE": null,
        // pass everything else thru
        "*": "[&1].&"
      }
    }
  },
  {
    // now, group by PK value
    "operation": "shift",
    "spec": {
      // top level array
      "*": {
        "PK": {
          "*": { // match any value of PK
            // go back up and grab the whole block and write
            // it to the ouput where the key, is the value of PK
            "@2": "&1[]"
          }
        }
      }
    }
  },
  {
    // sub group by CONTACT_TYPE, with the complication of 
    // pulling one entry off to serve as the "Reference"
    "operation": "shift",
    "spec": {
      "*": { // pk value
        "0": { // special case the Zeroth item so that
          // we can pull off once copy to serve as the 
          // Reference
          "@": "&2.Reference",
          // sub group by CONTACT_TYPE
          "CONTACT_TYPE": {
            "*": {
              "@2": "&4.ContactMethods.&1[]"
            }
          }
        },
        "*": { // all the rest of the array indicies
          // sub group by CONTACT_TYPE
          "CONTACT_TYPE": {
            "*": {
              "@2": "&4.ContactMethods.&1[]"
            }
          }
        }
      }
    }
  },
  {
    // Data fixing and Grouping done, now put everything
    // in its final place
    "operation": "shift",
    "spec": {
      "*": { // top level pk
        "Reference": {
          "PK": "Accounts[#3].Reference.Key",
          "SOURCE_DB": "Accounts[#3].Reference.System"
        },
        "ContactMethods": {
          "Phone": {
            "*": {
              "CONTACT_SUBTYPE": "Accounts[#5].ContactMethods.Phone[&1].Subtype",
              "PHONE_COUNTRY_CODE": "Accounts[#5].ContactMethods.Phone[&1].CountryCode",
              "PHONE_NUMBER": "Accounts[#5].ContactMethods.Phone[&1].Number",
              "PHONE_EXT": "Accounts[#5].ContactMethods.Phone[&1].Extension",
              "STATUS": "Accounts[#5].ContactMethods.Phone[&1].Active"
            }
          },
          "Email": {
            "*": {
              "CONTACT_SUBTYPE": "Accounts[#5].ContactMethods.Email[&1].Subtype",
              "EMAIL": "Accounts[#5].ContactMethods.Email[&1].EmailAddress",
              "STATUS": "Accounts[#5].ContactMethods.Email[&1].Active"
            }
          }
        }
      }
    }
  }
]
于 2017-08-25T14:03:44.077 回答