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Rafael Muynarsk
Asked: 2018-06-29 16:30:00 +0800 CST

如何用\n替换双引号外的所有空格？

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我有一个$variable由空格分隔的许多双引号路径

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，它可以像以下示例：

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。在这个答案中使用echo $variable | tr " " "\n"like对我不起作用，因为它用新行替换了所有空格。我该怎么做？

bash text-processing

1 个回答

  1. Best Answer

    如果元素总是双引号，那么你可以用quote-newline-quote替换quote-space-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"

    但是，如果您可以修改脚本以使用数组会更简单：

    $ 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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