bash – 相对于包含模式的行,删除n1个前面的行和n2行

script.sed的内容：

## If line doesn't match the pattern...
/pattern/ ! { 

    ## Append line to 'hold space'.
    H   

    ## Copy content of 'hold space' to 'pattern space' to work with it.
    g   

    ## If there are more than 5 lines saved,print and remove the first
    ## one. It's like a FIFO.
    /(n[^n]*){6}/ {

        ## Delete the first 'n' automatically added by previous 'H' command.
        s/^n//
        ## Print until first 'n'.
        P   
        ## Delete data printed just before.
        s/[^n]*//
        ## Save updated content to 'hold space'.
        h   
    } 

### Added to fix an error pointed out by potong in comments.
### =======================================================
    ## If last line,print lines left in 'hold space'.
    ${ 
        x   
        s/^n//
        p   
    } 
### =======================================================


    ## Read next line.
    b   
}

## If line matches the pattern...
/pattern/ {

    ## Remove all content of 'hold space'. It has the five previous
    ## lines,which won't be printed.
    x   
    s/^.*$//
    x   

    ## Read next four lines and append them to 'pattern space'.
    N ; N ; N ; N 

    ## Delete all.
    s/^.*$//
}

运行如下：

sed -nf script.sed infile