SED:删除两个模式之间的特定行

edl*_*oaa 3 icalendar design-patterns sed

我正在尝试使用过滤*.ics文件sed.该*.ics文件如下所示:

[...]

BEGIN:VEVENT
UID:0xfoo
SUMMARY:foo
DTSTART:20131212T090000
DTEND:20131212T100000
SEQUENCE:0
DTSTAMP:20131212T100000
LOCATION:foo
CATEGORIES:foo
DESCRIPTION:foo
CLASS:PUBLIC
END:VEVENT

[...]
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我想删除例如用UID或开始的行SEQUENCE,但仅限于它们之间 BEGIN:VEVENTEND:VEVENT

我尝试删除这些行:

sed '/^BEGIN:VEVENT/,/^END:VEVENT/ /^UID/d'
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但它只会返回一个错误,例如 unknown command '/'

如何删除这些行?

谢谢!

Ken*_*ent 9

试试这一行:

 sed '/^BEGIN:VEVENT/,/^END:VEVENT/{/^\(UID\|SEQUENCE\)/d}' file
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Ed *_*ton 5

sed 是一个很好的工具,可以在单行上进行简单的替换,对于其他任何事情,只需使用 awk:

awk '
/BEGIN:VEVENT/ { inBlock = 1 }
inBlock {
    if ( /END:VEVENT/ ) {
        inBlock = 0
    }
    else if ( /^(UID|SEQUENCE)/ ) {
        next
    }
}
{ print }
' file
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伪代码解释(inBlock是一个布尔变量,line只是一个虚构的字符串变量):

WHILE read line from file DO

    IF ( line contains the regexp "BEGIN:VEVENT" ) THEN
        inBlock = TRUE

    ENDIF

    IF ( inBlock == TRUE ) THEN

        IF ( line contains the regexp "END:VEVENT" ) THEN
            inBlock = FALSE

        ELSIF ( line starts with the regexp "UID" or "SEQUENCE" ) THEN
            do no further processing on this line

        ENDIF

    ENDIF

    print the current line

ENDWHILE
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