搜索文件中的单词，然后替换其他文件中的文本

Question

我有两个文件。File1包含用户名和密码，如下所示：

[reader]
label                         = anylabel
protocol                      = cccam
device                        = some.url,13377
user                          = Username1
password                      = password1
password2                     = password2
inactivitytimeout             = 30
group                         = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
cccversion                    = 2.3.2
ccckeepalive                  = 1

File2包含如下一行：

http://link:port/username/password/12345

现在我有了这个“代码”来更改File2中的用户名/密码

UsernameOLD=Username1
PasswordOLD=password1
UsernameNEW=Username2
PasswordNEW=password2

sed -i -e "s/\/$UsernameOLD\/$PasswordOLD/\/$UsernameNEW\/$PasswordNEW/" /etc/enigma2/file2.cfg

现在我有了不同的用户名，它们总是在File1中更新。现在我正在寻找一个解决方案，将用户名和Password2从File1写入一个变量，然后在File2中设置这个新的用户名和密码。

因此，作为菜鸟，psuedocode应该是这样的：

find "username" & "password1" in file1 
set "username" as $UsernameNEW and
    "password1" as $PasswordNEW and
then just execute my sed command.

有人能帮忙吗？我想我可以用grep来做这个？但是老实说，我很高兴我得到了这个带有变量的sed命令:D

Answer 1

这里有一些可以让你开始。

oscam.conf

[reader]
label                         = anylabel
protocol                      = cccam
device                        = some.url,13377
user                          = xxx1
password                      = password1
inactivitytimeout             = 30
group                         = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
cccversion                    = 2.3.2
ccckeepalive                  = 1


[reader]
label                         = anylabel
protocol                      = cccam
device                        = test.url,13377
user                          = yyy1
password                      = password1
inactivitytimeout             = 30
group                         = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
cccversion                    = 2.3.2
ccckeepalive                  = 1

对于密码文件，我已经更改了一些格式，但可以使用原始格式。

passwd (格式oldUser，newUser，oldPass，newUser)

xxx1,xxx2,passxxx1,passxxx2
yyy1,yyy2,passyyy1,passyyy2

Awk命令

awk -F, 'FNR==NR {usr[$1]=$2;pass[$1]=$4;next} FNR!=NR{FS=" = "}  /^user/ {t=$2;$2="= "usr[$2]} /^password/ {$2="= "pass[t]} 1' passwd oscam.reader

结果

[reader]
label                         = anylabel
protocol                      = cccam
device                        = some.url,13377
user                          = xxx2
password                      = passxxx2
inactivitytimeout             = 30
group                         = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
cccversion                    = 2.3.2
ccckeepalive                  = 1


[reader]
label                         = anylabel
protocol                      = cccam
device                        = test.url,13377
user                          = yyy2
password                      = passyyy2
inactivitytimeout             = 30
group                         = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
cccversion                    = 2.3.2
ccckeepalive                  = 1

Answer 2

快速和肮脏--将旧密码加载到环境参数中：

set -- $(grep -m1 -A1 '^user' File1)
sed -i -e "s#/${UsernameOLD}/${PasswordOLD}#/$3/$6#;T;q" /etc/enigma2/file2.cfg

工作原理：grep输出六个空格分隔的项，set将其转化为命令行参数$1 $2 $3 ... $6。我们只需要$3和$6。