英文:
Unable to replace the file contents
问题
我正在尝试使用 perl 命令在另一个 shell 脚本中替换文件的内容：
文件名：INFA.param
内容：
$ParamUser=SOMEUSERNAME
$ParamPswd=SOMEPASSWORD
$$StartTime=2021-06-26 08:31:51:1195079
$$EndTime=2021-06-26 09:31:51:1195079
当我在另一个名为 "pp.sh" 的 shell 脚本中打开上述文件时
cat INFA.param > temp.param
. ./temp.param
pmcmdpassword=$(pmcmd $password -e CRYPTDATA)
perl -pe -e "s#ParamPswd=\Q$ParamPswd\E#ParamPswd=$pmcmdpassword#g" INFA.param
我得到一些错误：
=SOMEUSERNAME: not found
=SOMEPASSWORD: not found
31850942StartTime=2021-06-26 08:31:51:1195079: not found
31850942EndTime=2021-06-26 09:31:51:1195079: not found
你能否建议我应该如何替换密码。
英文:
I am trying to replace the contents of file using perl command in another shell script:
Filename: INFA.param
Contents:
$ParamUser=SOMEUSERNAME
$ParamPswd=SOMEPASSWORD
$$StartTime=2021-06-26 08:31:51:1195079
$$EndTime=2021-06-26 09:31:51:1195079
When I open the above file in another shell script "pp.sh"
cat INFA.param > temp.param
. ./temp.param
pmcmdpassword=$(pmcmd $password -e CRYPTDATA)
perl -pe -e "s#ParamPswd=\Q$ParamPswd\E#ParamPswd=$pmcmdpassword#g" INFA.param
I am getting some errors:
=SOMEUSERNAME: not found
=SOMEPASSWORD: not found
31850942StartTime=2021-06-26 08:31:51:1195079: not found
31850942EndTime=2021-06-26 09:31:51:1195079: not found
Could you please suggest me how I should replace the password.
答案1
得分: 1
你只能在文件被编写为有效的shell源代码时才能安全使用.或source。这不适用于你这里的文件：shell赋值不以$开头。
如果你尝试通过重新加密来修改现有密码，这就是你的代码试图将原始文件读入shell解释器的原因，那么你可能想要：
#!/usr/bin/env bash
# ^^^^- 不是sh，仅限bash
declare -A variables=( )
while IFS='=' read -r key value; do
variables[$key]=$value
done < INFA.param
oldPassword=${variables['$ParamPswd']}
newPassword=$(pmcmd "$oldPassword" -e CRYPTDATA)
in="$oldPassword" out="$newPassword" \
perl -pi -e 's/\Q$ENV{"in"}/$ENV{"out"}/g' INFA.param
如果你的新密码是输入的话，那就更简单了；那么你根本不需要在第一次阅读原始文件时：
newPassword=$(pmcmd "${password?Password undefined}" -e CRYPTDATA)
out="$newPassword" \
perl -pi -e 's/^$ParamPswd=.*$/$ENV{"out"}/g' INFA.param
英文:
You can only safely use . or source when a file was written to be valid shell source code. That is not true of the file you have here: shell assignments don't start with $.
If you're trying to modify the existing password by reencrypting it, and that's why your code tries to read the original file into the shell interpreter, then you might want:
#!/usr/bin/env bash
# ^^^^- NOT sh, ONLY bash
declare -A variables=( )
while IFS='=' read -r key value; do
variables[$key]=$value
done < INFA.param
oldPassword=${variables['$ParamPswd']}
newPassword=$(pmcmd "$oldPassword" -e CRYPTDATA)
in="$oldPassword" out="$newPassword" \
perl -pi -e 's/\Q$ENV{"in"}/$ENV{"out"}/g' INFA.param
It's even easier if your new password is input; then you don't need to read the original file at all in the first place:
newPassword=$(pmcmd "${password?Password undefined}" -e CRYPTDATA)
out="$newPassword" \
perl -pi -e 's/^[$]ParamPswd=.*$/$ENV{"out"}/g' INFA.param
答案2
得分: 0
作为`bash`脚本执行以下内容：
```shell
$ParamUser=SOMEUSERNAME
$ParamPswd=SOMEPASSWORD
$$StartTime=2021-06-26 08:31:51:1195079
$$EndTime=2021-06-26 09:31:51:1195079
与Perl不同，bash仅使用$来扩展变量。例如：
a='some text'
echo "$a"
在bash中输出：
some text
当您的脚本启动时，bash变量ParamUser为空，因此$ParamUser=SOMEUSERNAME这一行将被扩展为=SOMEUSERNAME。这不是一个有效的命令，因此bash会报错：
=SOMEUSERNAME: not found
对于bash
$$StartTime=2021-06-26 08:31:51:1195079
存在多个不同的问题，其中包括间距。另外，$$会扩展为您当前进程的PID，显然是31850942。
因此，一个解决方案是：
ParamUser=$(sed -n 's/ParamUser=//p' INFA.param)
（其他变量类似）
请注意，行
pmcmdpassword=$(pmcmd $password -e CRYPTDATA)
使用了在问题的脚本部分中未设置的变量password。
<details>
<summary>英文:</summary>
You are executing the following as `bash` script:
$ParamUser=SOMEUSERNAME
$ParamPswd=SOMEPASSWORD
$$StartTime=2021-06-26 08:31:51:1195079
$$EndTime=2021-06-26 09:31:51:1195079
Unlike Perl, bash uses the `$` only to expand variables. For example:
a='some text'
echo "$a"
gives in bash:
some text
When your script starts, the bash variable `ParamUser` is empty, so the line `$ParamUser=SOMEUSERNAME` will be expanded to `=SOMEUSERNAME`. That is not a valid command, and therefore bash complains
=SOMEUSERNAME: not found
For bash
$$StartTime=2021-06-26 08:31:51:1195079
contains multiple different problems, among which the spacing. Also, `$$` expands to the PID of your current process, which is 31850942, apparently.
So, a solution would be
ParamUser=$(sed -n 's/ParamUser=//p' INFA.param)
(same for the rest)
Note that the line
pmcmdpassword=$(pmcmd $password -e CRYPTDATA)
uses a variable `password` that is not set in the script-part in the question.
</details>
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