路过的bash参数到sed的删除线 [英] passing bash arguments into sed to delete lines

问题描述

我试图写一个将采取所有发送给它的参数，并通过 SED 庆典功能C>删除文件中的匹配参数的行。用例例如：

I'm trying to write a bash function which will take all the arguments sent to it and push them through sed to delete lines in a file that match the arguments. A use case example:

to_delete get the trash

我可以得到它几乎的工作是这样的：

I can get it to "almost" work like this:

function to_delete() { sed -i -e "/$@/d" /tmp/testfile; }

这样做的问题是，如果我发送一个命令行参数，将只工作：

The problem with this is it will only work if I send a single command line argument:

to_delete get

如果我发送一个以上返回这个错误：

If I send more than one it returns this error:

sed: 1: "/get": unterminated regular expression

如果我把周围的参数报价也将工作：

It will also work if I throw quotes around the arguments:

to_delete "get the trash"

但我宁愿没有这样做。

But I'd rather not have to do that.

这是如何得到这个工作任何想法？谢谢你。

Any ideas on how to get this working? Thanks.

推荐答案

您有$ @的问题是，它拓展了单独报价参数。

The problem you have with $@ is that it expands in separately quoted arguments.

您在理论上可以使用$ *来代替，因为这已经不是功能。

You could in theory use $* instead because this hasn't the feature.

BUT $ *将通过IFS的第一个字符（通常是一个空格）分隔的参数。

BUT $* will have the arguments separated by the first character of IFS (usually a space).

那么麻烦的是，如果你做一些事情：

So the trouble is if you do something:

to_delete foo   bar baz

的格局将是这样的：

The pattern will look like:

/foo bar baz/d

（只有一个空格）。

(Only one space).

如果您使用的是加引号的参数这一切混乱避免，没有错。

All this mess is avoided if you use a quoted argument, nothing wrong with this.

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