jq substr() 相当于格式化一个值

Question

希望你在这些时候一切都好。

我遇到了一个问题，希望你有更多的知识。

使用 'jq' 我想更改所有日期，例如。19731013（字符串）到 1973-10-13

[
    {
        "Mail": "john@example.com",
        "Name": "Smith",
        "Employee_Number": "000555",
        "First_Name": "John",
        "Company": "ACME",
        "Department": null,
        "Employment_Status": "Retiree",
        "Start_Date": "19770516",
        "Function_Start_Date": "19770516",
        "Group_Phone": "",
        "Job_Title": "Operations Manager Warehousing",
        "Sub_Group": "Exempts",
        "Location": "Tibuktu",
        "Organizational_Unit": null,
        "Date_of_Birth": "19560719",
        "Gender": "1"
    },
    {
        "Mail": "mary@example.com",
        "Name": "Smith",
        "Employee_Number": "000777",
        "First_Name": "Mary",
        "Company": "ACME",
        "Department": null,
        "Employment_Status": "Retiree",
        "Start_Date": "19770516",
        "Function_Start_Date": "19770516",
        "Group_Phone": "",
        "Job_Title": "Manager",
        "Sub_Group": "Exempts",
        "Location": "Tibuktu",
        "Organizational_Unit": null,
        "Date_of_Birth": "19560719",
        "Gender": "2"
    }
] 

是否可以substr(.Start_Date,1,5)"-"substr(.Start_Date,6,2)"-"substr(.Start_Date,8,3)在 CSV 中将 like 与 awk一起使用？

也许我正盯着墙而错过了右边的门？

更新：感谢很多人，这就像一个魅力！

jq -r '.[].Start_Date |= "\(.[0:4])-\(.[4:6])-\(.[6:8])" | .[].Function_Start_Date |= "\(.[0:4])-\(.[4:6])-\(.[6:8])" | .[].Date_of_Birth|="\(.[0:4])-\(.[4:6])-\(.[6:8])"' employees.json > test.json

Answer 1

在 JQ 中，我们为此提供了字符串切片和字符串插值语法。

$ jq '.[].Start_Date | "\(.[0:4])-\(.[4:6])-\(.[6:8])"' file
"1977-05-16"
"1977-05-16"

Answer 2

jq使用中还有正则表达式匹配函数capture，它会发出命名的捕获组，稍后可以将其连接起来-以形成您想要的日期字符串。

jq '.[].Start_Date | capture("(?<x>[0-9]{4})(?<y>[0-9]{2})(?<z>[0-9]{2})") | join("-")'

这是假设，您的Start_Date字段至少有 8 个字符长，并且不会验证小于该长度的长度。