如何处理JSON嵌套数组？实现候选邮箱与应用信息关联转换

输入JSON #

{
  "Content": [
    {
      "CandidateEmail": "john1@noexist.com",
      "Applications": [
        {
          "ApplicationId": "app1",
          "JobRefNumber": "REF1"
        },
        {
          "ApplicationId": "app2",
          "JobRefNumber": "REF2"
        }
      ]
    },
    {
      "CandidateEmail": "carl2@email.com",
      "Applications": [
        {
          "ApplicationId": "app3",
          "JobRefNumber": "REF3"
        },
        {
          "ApplicationId": "app4",
          "JobRefNumber": "REF4"
        }
      ]
    }
  ]
}

期望输出JSON #

[
  {
    "CandidateEmail": "john1@noexist.com",
    "ApplicationId": "app1",
    "JobRefNumber": "REF1"
  },
  {
    "CandidateEmail": "john1@noexist.com",
    "ApplicationId": "app2",
    "JobRefNumber": "REF2"
  },
  {
    "CandidateEmail": "carl2@email.com",
    "ApplicationId": "app3",
    "JobRefNumber": "REF3"
  },
  {
    "CandidateEmail": "carl2@email.com",
    "ApplicationId": "app4",
    "JobRefNumber": "REF4"
  }
]

原有规范的问题 #

你草拟的shift规范中，使用[&1]作为数组索引，这里的&1指向当前Applications数组的元素索引（每个Candidate下的Applications都是从0开始计数），导致第二个Candidate的Applications元素会覆盖第一个Candidate的同索引元素，最终只能得到2个结果，而非期望的4个。

修正后的Jolt规范 #

方案一：组合索引+扁平化（清晰直观） #

[
  {
    "operation": "shift",
    "spec": {
      "Content": {
        "*": {
          "Applications": {
            "*": {
              "@(2,CandidateEmail)": "[&2_&1].CandidateEmail",
              "ApplicationId": "[&2_&1].ApplicationId",
              "JobRefNumber": "[&2_&1].JobRefNumber"
            }
          }
        }
      }
    }
  },
  {
    "operation": "shift",
    "spec": {
      "*": "[]"
    }
  }
]

• 第一步shift：用&2（Content数组的索引）和&1（Applications数组的索引）组合成唯一键名（如0_0、0_1），将每个CandidateEmail与对应Application字段绑定
• 第二步shift：将所有键对应的对象扁平化到数组中

方案二：自动追加数组项（简洁高效） #

[
  {
    "operation": "shift",
    "spec": {
      "Content": {
        "*": {
          "Applications": {
            "*": {
              "@(2,CandidateEmail)": "[#4].CandidateEmail",
              "*": "[#4].&"
            }
          }
        }
      }
    }
  }
]

• #4表示当前路径中第4个层级的通配符遍历计数（从1开始数：1=根，2=Content，3=Content的*，4=Applications的*），确保每个合并后的对象对应唯一的数组位置
• *匹配ApplicationId和JobRefNumber字段，直接映射到同数组项的对应键名

内容的提问来源于stack exchange，提问作者ElielBerra