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jq + 更新json文件并根据ID号附加名称

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jq + 更新json文件并根据ID号附加名称

我们要根据以下规则编辑json文件

我们需要生成 name car0X_hyb.com -undermazda 字符串（而 X 是 ID 号）

我们需要更新 Linux 机器上的 json 文件

例子 - ID=1

那么 json 将是

{
  "toyota": [
    "car1.com"
  ],
  "mazda": [
    "car01_hyb.com"
  ],
  "ford": [
    "car01_eng.com"
  ]
}

ID=4, 那么 json 将是

{
  "toyota": [
    "car1.com"
  ],
  "mazda": [
    "car01_hyb.com",
    "car02_hyb.com",
    "car03_hyb.com",
    "car04_hyb.com"
  ],
  "ford": [
    "car01_eng.com"
  ]
}

ID=6, 那么 json 将是

{
  "toyota": [
    "car1.com"
  ],
  "mazda": [
    "car01_hyb.com",
    "car02_hyb.com",
    "car03_hyb.com",
    "car04_hyb.com",
    "car05_hyb.com",
    "car06_hyb.com"
  ],
  "ford": [
    "car01_eng.com"
  ]
}

ID=10, 那么 json 将是

{
  "toyota": [
    "car1.com"
  ],
  "mazda": [
    "car01_hyb.com",
    "car02_hyb.com",
    "car03_hyb.com",
    "car04_hyb.com",
    "car05_hyb.com",
    "car06_hyb.com",
    "car07_hyb.com",
    "car08_hyb.com",
    "car09_hyb.com",
    "car10_hyb.com"
  ],
  "ford": [
    "car01_eng.com"
  ]
}

答案1

似乎没有一种不丑陋的方法来对 ID 进行零填充，但是考虑到

$ jq '.' file.json
{
  "toyota": [
    "car1.com"
  ],
  "mazda": [
    "foo.bar"
  ],
  "ford": [
    "car01_eng.com"
  ]
}

然后例如对于ID= 11：

$ jq --arg ID 11 '
    .mazda |= [range(1;($ID | tonumber)+1) as $ID | if $ID < 10 then "car0\($ID)_hyb.com" else "car\($ID)_hyb.com" end]
' file.json
{
  "toyota": [
    "car1.com"
  ],
  "mazda": [
    "car01_hyb.com",
    "car02_hyb.com",
    "car03_hyb.com",
    "car04_hyb.com",
    "car05_hyb.com",
    "car06_hyb.com",
    "car07_hyb.com",
    "car08_hyb.com",
    "car09_hyb.com",
    "car10_hyb.com",
    "car11_hyb.com"
  ],
  "ford": [
    "car01_eng.com"
  ]
}

或者（也许这更好）在 shell 中生成并格式化完整的字符串元素，并将其作为argjson:

jq --argjson IDS '['"$(seq -s, -f'"car%02.0f_hyb.com"' 11)"']' '
  .mazda |= $IDS
' file.json

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