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如何用 \n 替换双引号外的所有空格？

tag-icon • tag-icon bash text-processing scripting
如何用 \n 替换双引号外的所有空格？

我有一个$variable有许多用空格分隔的双引号路径

echo $variable

"/home/myuser/example of name with spaces" "/home/myuser/another example with spaces/myfile"

我的变量上的路径数量可能会有所不同，并且不受控制。例如，它可以类似于以下示例：

example 1: "path1" "path2" "path3" "path4"
example 2: "path1" "path2" "path3" "path4" "path5" "path6" path7" "path8"
example 3: "path1" "path2" "path3" 
example 4: "path1" "path2" "path3" "path4" "path5" "path6"

我想将双引号外的所有空格替换为新行 ( \n)，同时保留引号内的空格。使用echo $variable | tr " " "\n"像这答案对我不起作用，因为它用新行替换了所有空格。我该怎么做？

答案1

如果元素是总是双引号，那么您可以将 quote-space-quote 替换为 quote-newline-quote：

$ sed 's/" "/"\n"/g' <<< "$variable"
"/home/myuser/example of name with spaces"
"/home/myuser/another example with spaces/myfile"

或（使用 shell 参数替换）

$ printf '%s\n' "${variable//\" \"/\"$'\n'\"}"
"/home/myuser/example of name with spaces"
"/home/myuser/another example with spaces/myfile"

但如果您可以修改脚本以使用数组，那就更简单了：

$ vararray=("/home/myuser/example of name with spaces" "/home/myuser/another example with spaces/myfile")
$ printf '"%s"\n' "${vararray[@]}"
"/home/myuser/example of name with spaces"
"/home/myuser/another example with spaces/myfile"

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