读取原始输入行并输出单个数组 [英] read raw input lines and output single array

问题描述

我有一个包含文件的目录.我想从该文件列表中创建一个数组.我认为这很容易，例如:

I have a directory with files in it. I would like to create an array from that list of files. I thought it would be pretty easy, like:

ls mydir | jq -R '[.]'
[
  "file1"
]
[
  "file2"
]
[
  "file3"
]

我唯一能弄清楚的是:

ls mydir | jq -sR '[split("\n")[]|select(.|length>0)]'
[
  "file1",
  "file2",
  "file3"
]

有更好的方法吗?

推荐答案

在处理Unix文件名时，您必须格外小心.它们几乎可以包含文件名中的任何字符，包括空格，换行符，逗号，管道符号，以及除NUL之外几乎所有您尝试用作定界符的其他任何字符.最好的选择是用NUL字符分隔名称，这是唯一不能成为有效文件名一部分的字符，并用jq

You'd have to be extra careful in dealing with Unix filenames in general. They can contain almost any character in a filename, including whitespace, newlines, commas, pipe symbols, and pretty much anything else you'd ever try to use as a delimiter except NUL. Your best bet is to separate the names with the NUL character, which is the only character that can't be part of a valid filename and split on it with jq

使用本机外壳printf分隔\0上的条目并将其重新定界

Use the native shell printf to separate entries on \0 and delimit it back

printf '%s\0' * | jq -Rn 'inputs | split("\u0000")'

或仅用于文件

for file in *; do 
  [ -f "$file" ] && printf '%s\0' "$file"
done | jq -Rn 'inputs | split("\u0000")'

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