JOLT转换：重命名含@特殊字符键的规则编写及报错解决

问题场景 #

需要将JSON中带@的键Company@1重命名为CompanyTest，Age重命名为AgeTest，其余键保留，但原规则因键中@不在开头触发报错。

报错信息 #

Error running the Transform.

JOLT Chainr encountered an exception constructing Transform className:com.bazaarvoice.jolt.Shiftr at index:0.

Invalid key:Company@1 can not have an @ other than at the front.

输入JSON #

[
  {
    "C": "p",
    "ID": 1,
    "Company@1": "Tesla",
    "Age": 30.2,
    "Year": 1996,
    "Time": "22/10/1996"
  },
  {
    "C": "p",
    "ID": 2,
    "Company@1": "Facebook",
    "Age": 40.5,
    "Year": 2001,
    "Time": "22/10/2001"
  }
]

预期输出 #

[ {
  "C" : "p",
  "ID" : 1,
  "CompanyTest" : "Tesla",
  "AgeTest" : 30.2,
  "Year" : 1996,
  "Time" : "22/10/1996"
}, {
  "C" : "p",
  "ID" : 2,
  "CompanyTest" : "Facebook",
  "AgeTest" : 40.5,
  "Year" : 2001,
  "Time" : "22/10/2001"
} ]

解决方案 #

JOLT的Shiftr操作会将键名中的@识别为特殊语法，因此需要对@进行转义处理。修正后的规则如下：

[
  {
    "operation": "shift",
    "spec": {
      "*": {
        "Company\\@1": "[#2].CompanyTest",
        "Age": "[#2].AgeTest",
        "*": "[#2].&"
      }
    }
  }
]

规则说明 #

• 对Company@1中的@添加反斜杠转义（Company\\@1），避免JOLT将其解析为特殊符号
• [#2]用于保留原数组的索引位置，确保每个对象对应到输出数组的正确位置
• "*": "[#2].&"保留所有未指定重命名的键，&表示沿用原键名

内容的提问来源于stack exchange，提问作者Sumit Manna