在wget bash命令中转义 [英] Escaping in wget bash command

问题描述

wget -q -T 60 --retry-connrefused -t 5 --waitretry = 60 --user = ftp2.company.com | company2013 --password =！company2013 -N -P数据/ parser / company / ftp://ftp2.company.com/Production/somedata.zip

此命令遇到问题，因为密码包含感叹号。我尝试使用进行转义，尝试使用单引号，它要么给出输出：

I'm having trouble with this command, because the password contains an exclamation mark. I tried escaping with \, tried single quotes, and it either gives the output:

wget：缺少URL

或

bash：！company2013：未找到事件

这确实令人沮丧...

This is really demotivating...

推荐答案

也许需要对此部分加引号，以防止将其视为到另一个命令的管道。

Perhaps this part needs to be quoted to prevent it from being seen as a pipe to another command.

--user='ftp2.company.com|company2013'

这也是为了防止使用！：

And this one too to prevent history expansion with !:

--password='!company2013'

最终：

wget -q -T 60 --retry-connrefused -t 5 --waitretry=60 --user='ftp2.company.com|company2013' --password='!company2013' -N -P data/parser/company/ ftp://ftp2.company.com/Production/somedata.zip

如果以后还会有空格，则引用其他部分也是一个好主意：

And it's also a good idea to quote the other parts if on later time they have spaces:

wget -q -T 60 --retry-connrefused -t 5 --waitretry=60 --user='ftp2.company.com|company2013' --password='!company2013' -N -P "data/parser/company/" "ftp://ftp2.company.com/Production/somedata.zip"

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