linux - 按文本查找行并从该行删除到 bash 中的其他行

Question

使用 linux bash 或命令行，我需要在文件中找到特定的字符串并删除该行和直到出现空格的行。

我的文件:

geo_ws:
    resource: "@GeoWSBundle/Controller/"
    type:     annotation
    prefix:   /

mappublic:
    resource: "@MapPublicBundle/Controller/"
    type:     annotation
    prefix:   /map/ws

webpublic:
    resource: "@WebPublicBundle/Controller/"
    type:     annotation
    prefix:   /public

map:
    resource: "@MapBundle/Controller/"
    type:     annotation
    prefix:   /map

geo:
    resource: "@GeoBundle/Controller/"
    type:     annotation
    prefix:   /geo

ws:
    resource: "@WSBundle/Controller/"
    type:     annotation
    prefix:   /ws

web:
    resource: "@WebBundle/Controller/"
    type:     annotation
    prefix:   /

所以我需要找到“ map :”，并清除所有这 4 行，直到下一个白行。

我试过使用种子，但我就是想不通。

Answer 1

空记录分隔符表示记录由空行分隔:

$ awk -v RS='' -v ORS='\n\n' '!/^map/' file
geo_ws:
    resource: "@GeoWSBundle/Controller/"
    type:     annotation
    prefix:   /

webpublic:
    resource: "@WebPublicBundle/Controller/"
    type:     annotation
    prefix:   /public

geo:
    resource: "@GeoBundle/Controller/"
    type:     annotation
    prefix:   /geo

ws:
    resource: "@WSBundle/Controller/"
    type:     annotation
    prefix:   /ws

web:
    resource: "@WebBundle/Controller/"
    type:     annotation
    prefix:   /