如何用awk或sed去除每一行匹配字符后面所有字符串

sonorous

内容如下：
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\servertool.exe
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\tnameserv.exe
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\unpack200.exe
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\TestClient.exe
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\GSM_ACT\Management\Release\license\LicenseCreate.exe

要去除最后的*.exe，保留目录即可，要求效果如下：
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\GSM_ACT\Management\Release\license\

注：
1。目录层次不同，有长有短
2。其实就是匹配最后一个“\”，然后删除后面所有的内容

Perl_Er

1. dirname

复制代码

回复 1# sonorous

1. sed -n 's/\(.*\)\/.*/\1/p'

复制代码

sonorous

Perl_Er 发表于 2012-04-06 12:23
回复 1# sonorous

试过，不行。

[root@chengyh01 tmp]# cat 2.txt | tail -5
\\10.159.219.22\hangzhou_file_server\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\servertool.exe
\\10.159.219.22\hangzhou_file_server\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\tnameserv.exe
\\10.159.219.22\hangzhou_file_server\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\unpack200.exe
\\10.159.219.22\hangzhou_file_server\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\TestClient.exe
\\10.159.219.22\hangzhou_file_server\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\GSM_ACT\Management\Release\license\LicenseCreate.exe
[root@chengyh01 tmp]# cat 2.txt | tail -5 | sed -n 's/\(.*\)\/.*/\1/p'
[root@chengyh01 tmp]#

yinyuemi

1. sed -n 's#\(.*\)\\.*#\1#p'

复制代码

Perl_Er

回复 3# sonorous

丢失了一个\,
    sed -n 's/\(.*\)\\.*/\1/p'

jason680

回复 1# sonorous

1. sed -r 's/[^\\]+$//'

复制代码

pxbfeiniao

#!/bin/awk -f

BEGIN{
        FS="\\"
}

{
        for(x=1;x<NF;x++) {
                if (x==NF-1)
                        printf "%s\\\n", $x
                else
                        printf "%s\\", $x
        }
}


./tmp.awk tmp1
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\jre\bin\
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\DIMETRA_RP_DMS\TestClient\ForTest\
\\ZCH66_BR476_DEV\TRANSFERED_PROJECTS\GSM_ACT\Management\Release\license\

dn833

1. awk 'BEGIN{FS="\\";OFS="\\"}{$NF="";print}'

复制代码

写个awk的

serapy

sed 's/\(^[A-Za-z0-9_\\]\{0,\}\\\)[^\\]\{0,\}$/\1/g'

看起来复杂了点儿，不过能得到正确答案