Jolt转换需求：将对象数组属性展平为扁平结构

以下是实现你需求的Jolt转换规范：

[
  {
    "operation": "shift",
    "spec": {
      "citycode": "citycode",
      "branch": "branch",
      "Marks": "Marks",
      "designations": {
        "*": {
          "designation": "Occupation",
          "Language": "LANGUAGE-ID"
        }
      }
    }
  },
  {
    "operation": "modify-overwrite-beta",
    "spec": {
      "Occupation": "=split('', @(1))",
      "LANGUAGE-ID": "=split('', @(1))"
    }
  },
  {
    "operation": "shift",
    "spec": {
      "citycode": "citycode",
      "branch": "branch",
      "Marks": "Marks",
      "Occupation": {
        "*": "Occupation"
      },
      "LANGUAGE-ID": {
        "*": "LANGUAGE-ID"
      }
    }
  }
]

转换逻辑说明： #

1. 第一步Shift操作：提取顶层需要保留的字段，同时将designations数组中的每个元素的designation和Language分别映射到Occupation和LANGUAGE-ID，此时这两个字段会是数组形式（和你当前的输出一致）。
2. 第二步Modify操作：通过split('', @(1))将数组拆分成单个元素的独立项，为后续展开做准备。
3. 第三步Shift操作：遍历Occupation和LANGUAGE-ID的拆分结果，将每个元素直接映射为同名顶级字段，生成期望的平级重复键值对。

验证结果： #

输入JSON：

{"citycode": "34343","branch": "cityname","changeDatetime": 1677063272522,"Marks": "L","designations": [{"designation": "security ","Language": "de"},{"designation": "officer","Language": "fr"},{"designation": "superior","Language": "en"}]}

输出JSON：

{"citycode":"34343","branch":"cityname","Marks":"L","Occupation":"security ","LANGUAGE-ID":"de","Occupation":"officer","LANGUAGE-ID":"fr","Occupation":"superior","LANGUAGE-ID":"en"}

注意：JSON语法规范中不允许对象存在重复键，部分解析器可能会忽略后续的重复键，但Jolt支持生成这种结构以满足特定场景需求。

内容的提问来源于stack exchange，提问作者SATHK