使用jq解析嵌套JSON并提取字段生成指定格式CSV的问题求助

可用命令 #

jq -r '["顶层id", "顶层description", "project.id", "project.description", "project.url"], (inputs | . as $top | if .project | length == 0 then .project = [{}] else . end | .project[] | [$top.id, $top.description, .id, .description, .url]) | @csv' 输入文件名.json > 输出文件名.csv

命令说明 #

• -r 参数：输出原始文本格式，避免CSV内容被额外转义包裹
• 开头的数组定义为CSV表头，可根据需求自行修改字段名称
• 自动判断project数组是否为空，为空时自动填充空对象占位，保证顶层条目至少输出一行
• 自动展开project数组，每个子项单独对应一行输出
• 内置@csv方法自动处理特殊字符转义，符合标准CSV格式规范

低版本兼容命令 #

如果运行上述命令报错（多为jq版本较低不支持inputs语法），可使用slurp模式命令：

jq -s -r '["顶层id", "顶层description", "project.id", "project.description", "project.url"], (.[] | . as $top | if .project | length == 0 then .project = [{}] else . end | .project[] | [$top.id, $top.description, .id, .description, .url]) | @csv' 输入文件名.json > 输出文件名.csv

输出示例 #

运行后生成的CSV内容符合预期：

"顶层id","顶层description","project.id","project.description","project.url"
15,"package",1762,"This Red Hat Server 7 is built from the Redhat Official","some url"
15,"package",1663,"This Ubuntu 20.04 is built from the Ubuntu Official","some url"
15,"package",1557,"This Centos 7 is built from the Centos Official","some url"
22,"exe",,,
34,"brew",,,

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