Question

我试图写一个短篇的文字,在短文中,我用来搜索一个溪流,然后根据弹道功能的结果,在溪流上进行替代,这就需要争论,比如说。

#!/bin/sh

function test {
    echo "running test"
    echo $1
    }

sed -n -e "s/.*(00).*/$(test)/p" < testfile.txt

试验文件。注

(每条新线之间有新线;它们正在被贵点清除。) So ok that script work for me (output "running test”),但显然没有经过任何论证。我愿这样说:

sed -n -e "s/.*(00).*/$(test 1)/p" < testfile.txt

产出:

running test
00

So that the pattern matched by sed is fed as an argument to test. I didn t really expect the above to work, but I have tried every combination of $() brackets, backticks, and escapes I could think of, and can find no mention of this situation anywhere. Help?

Answer 1

Sed won t execute commands. Perl will, however, with the /e option on a regex command.

perl -pe  sub testit { print STDERR "running test"; return @_[0]; }; s/.*(00).*/testit($1)/e  <testfile.txt

Redirect stderr to /dev/null if you don t want to see it in-line and screw up the output.

Answer 2

这或许有助于你:

tester () { echo "running test"; echo $1; }
export -f tester
echo -e "1234
2345
3006
4567" |
sed -n  s/.*(00).*/echo "$(tester 1)"/p  | sh
running test
00

或者,如果你使用德国国籍联盟的话:

echo -e "1234
2345
3006
4567" |
sed -n  s/.*(00).*/echo "$(tester 1)"/ep 
running test
00

N.B. 你们必须记住首先出口这一职能。

Answer 3

try this:

sed -n -e "s/.*(00).*/$1$2/p" testfile.txt | sh

注:我可能有失职之处,但重要的仲裁范围正在ll上。

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