Question

我在bash中的变量中有一个文本字符串，如下所示：

filename1.txt
filename2.txt

varname1 = v1value
$(varname1)/filename3.txt
$(varname1)/filename4.txt

varname2 = $(varname1)/v2value
$(varname2)/filename5.txt
$(varname2)/filename6.txt

我想替换所有变量，产生这个：

filename1.txt
filename2.txt

v1value/filename3.txt
v1value/filename4.txt

v1value/v2value/filename5.txt
v1value/v2value/filename6.txt

有人可以在shell中建议一个干净的方法吗？

Answer 1

在awk中：

BEGIN {
    FS = "[[:space:]]*=[[:space:]]*"
}

NF > 1 {
    map[$1] = $2
    next;
}

function replace(     count)
{
    for (key in map) {
        count += gsub("\\$\\("key"\\)", map[key])
    }

    return count
}

{
    while (replace() > 0) {}
    print
}

在lua：

local map = {}

--for line in io.lines("file.in") do -- To read from a file.
for line in io.stdin:lines() do -- To read from standard input.
    local key, value = line:match("^(%w*)%s*=%s*(.*)$")
    if key then
        map[key] = value
    else
        local count
        while count ~= 0 do
            line, count = line:gsub("%$%(([^)]*)%)", map)
        end
        print(line)
    end
end

Answer 2

我找到了使用m4的合理解决方案：

function make_substitutions() {
    # first all $(varname)s are replaced with ____varname____
    # then each assignment statement is replaced with an m4 define macro
    # finally this text is then passed through m4

    echo "$1" |\
    sed 's/\$(\([[:alnum:]][[:alnum:]]*\))/____\1____/' | \
    sed 's/ *\([[:alnum:]][[:alnum:]]*\) *= *\(..*\)/define(____\1____, \2)/' | \
    m4
}

Answer 3

也许

echo "$string" | perl -nlE 'm/(\w+)\s*=\s*(.*)(?{$h{$1}=$2})/&&next;while(m/\$\((\w+)\)/){$x=$1;s/\$\($x\)/$h{$x}/e};say$_'

打印

filename1.txt
filename2.txt

v1value/filename3.txt
v1value/filename4.txt

v1value/v2value/filename5.txt
v1value/v2value/filename6.txt

在文本字符串中替换变量

3 个答案: