如果这是有效的json，一个解决方案是使用适当的json解析工具，例如jq。

cat test.json
[{
    "Music": "1410",
    "Countryid": "0115",
    "History": "20220621",
    "Legend": "/api/legacysbo/bondue",
    "Sorting": "/api/dmplus/test",
    "Nick": "hinduja",
    "Scenario": ["K", "A", "S", "F", "D"]
}, {
    "Music": "1466",
    "Countryid": "1312",
    "History": "20221012",
    "Legend": "/api/legacysbo/grenob",
    "Sorting": "/api/dmplus/prod",
    "Nick": "Grenoble",
    "Scenario": ["K", "A", "S", "F", "D"]
}]

jq -r '.[] | "Countryid: \(.Countryid) History: \(.History)"' < test.json
Countryid: 0115 History: 20220621
Countryid: 1312 History: 20221012

假设：

• OP想要打印 * 所有 * Countryid/History对
• Countryid条目可以在History条目之前或之后出现
• 每个Countryid条目必须具有匹配的History条目
• 所有集都有一个Music条目，该条目位于Countryid/History条目之前

一个awk创意：

awk -F '"' '

function print_pair() {
    if (countryid && history)                                      # if both variables are non-empty then ...
       printf "Countryid: %s History: %s\n", countryid, history

    countryid=histor=""                                            # reset variables
}

$2 == "Music"     { print_pair() }                                 # print previous pair to stdout
$2 == "Countryid" { countryid=$4 }
$2 == "History"   { history=$4   }
END               { print_pair() }                                 # print last pair to stdout
' input.dat

如果我们可以假设Countryid总是在History之前，那么我们可以将代码简化为：

awk -F '"' '
$2 == "Countryid" { countryid=$4 }
$2 == "History"   { printf "Countryid: %s History: %s\n", countryid, h$4   }
' input.dat

这两种方法都会产生：

Countryid: 0115 History: 20220621
Countryid: 1312 History: 20221012