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。
最佳答案
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: /
关于linux - 按文本查找行并从该行删除到 bash 中的其他行，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/35557562/