This is a more advanced version of a previous question (How to pass arguments to perl when trying to change a line?) I made.
This time I am trying to pass a path, but evertything seems that the the perl script is readin the / wrongly.
Suppose the line 4 in file.txt looks like this
path_root_abs = "/path/to/thefile"
To obtain the working directory and replace it in /path/to/file I did
directory=`pwd`
perl -i -pe "s/(path_root_abs\s=\s\")(.*)(\")/\$1${directory}$3/ if \$. == 4" file.txt
And got:
Bareword found where operator expected at -e line 1, near "s/(path_root_abs\s=\s")(.*)(")/$1/scratch"
syntax error at -e line 1, near "s/(path_root_abs\s=\s")(.*)(")/$1/scratch"
Execution of -e aborted due to compilation errors.
What should I do to avoid unix reading the \ that comes after scratch as a bareword.
The first issue is that $directory contains slashes which are also being used as the delimiter for the substitution operator (s///). Basically, if $directory is /home/je_b, what Perl sees is:
perl -i -pe "s/foo//home/jb/ if \$. == 4" file.txt
It takes the / of /home as the second / of the s/// operator. The simplest solution is to use a different character instead of /:
perl -i -pe "s#foo#${directory}#" file.txt
You can also simplify in other ways though. Consider this:
perl -pe "s#(path_root_abs = \")(.*)#\1${directory}\"# if \$. == 4" file
\s when you only need to match one space, just use a space.$1 and \1 so use the latter and avoid escaping." character. If you know it's there, add it yourself.Finally, you could also get the pwd from Perl directly. Perl has access to all exported shell variables through the %ENV hash. So, you could just do:
perl -pe 's#(path_root_abs = ").*#$1$ENV{PWD}"# if $.==1' file
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