替换文件名 bash 中的部分字符串
我有一个正在从文件夹 test 中读取的文本文件列表，如下所示：
file_list="$(ls ~/Desktop/test |
while read path; do basename "$path"; done)"
这将生成这些文件的列表：
test_1.txt test_2.txt
我想更改名称中的特定字符串，特别是将 test 更改为 this，这样列表中就会包含如下文件：
this_1.txt this_2.txt
我想直接在 file_list 中执行此操作，但不想对计算机上文件夹中的实际文件执行此操作。
逐一循环是最有效的方法吗？
I have a list of text files I am reading in like this from a folder test like this:
file_list="$(ls ~/Desktop/test |
while read path; do basename "$path"; done)"
This will produce a list of these files:
test_1.txttest_2.txt
I want to change particular strings in the name, specifically test to this so the list would then have files like this:
this_1.txtthis_2.txt
I would like to do this directly in file_list I don't want to do it on the actual files in the folder on the computer.
Is looping through one by one the most efficient way to do this?
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您不需要循环或外部命令（例如
basename、find和sed）。试试这个 Shellcheck - 干净的代码：shopt -s nullglob使 glob 扩展为空当没有文件与模式匹配时。如果没有它，当没有任何匹配时，全局变量会扩展为（相当于）垃圾。files=( ~/Desktop/test/* )使用~/Desktop/ 中所有文件（和目录）的路径填充名为目录（files的数组test(~/Desktop/test/test_1.txt ...)）。请注意，名称以点 (.) 开头的文件将被排除。可以通过在程序早期运行shopt -s dotglob来包含它们。bases=( "${files[@]##*/}" )使用files中文件的基本名称填充bases数组> 数组 (( test_1.txt ... ))。有关## 的信息，请参阅 参数扩展 [Bash Hackers Wiki]正在做。.txt扩展名，您可以向该过程添加一个额外的阶段：stems=( "${bases[@]%. txt}")。在 Bash 中不可能同时执行多个字符串操作（例如##和%）。this_list="${bases[*]//test/this}"使用bases中出现的所有条目填充this_list字符串每个中的test替换为this("this_1.txt ...")。再次，请参阅参数扩展 [Bash Hackers Wiki] 了解其工作原理的详细信息。列表中的条目由空格分隔。问题列表中的条目由换行符分隔。您可以通过在执行this_list=...赋值之前设置IFS=$'\n'来对this_list执行此操作。请参阅在构建和数组时修改 bash 中的 IFS、IFS=$'\n'的确切含义是什么？和这是“备份”$IFS变量的合理方法吗？。当使用"${arrayname[*]}"将数组转换为字符串时，IFS值中的第一个字符用于分隔数组元素。declare -p this_list以明确的方式显示this_list的内容。一些一般要点：
find和sed的组合应该能够处理更多数量的文件，但 Bash 可能难以处理生成的巨大字符串（或数组）。You don't need either loops or external commands (like
basename,find, andsed). Try this Shellcheck-clean code:shopt -s nullglobmakes globs expand to nothing when no files match a pattern. Without it globs expand to (what amounts to) garbage when nothing matches.files=( ~/Desktop/test/* )populates an array calledfileswith the paths to all the files (and directories) in the~/Desktop/testdirectory ((~/Desktop/test/test_1.txt ...)). Note that files whose names begin with a dot (.) are excluded. They can be included by runningshopt -s dotglobearlier in the program.bases=( "${files[@]##*/}" )populates thebasesarray with the basenames of the files in thefilesarray (( test_1.txt ... )). See Parameter expansion [Bash Hackers Wiki] for information about what the##is doing..txtextensions, as suggested in one of the comments, you could add an extra stage to the process:stems=( "${bases[@]%.txt}" ). It's not possible to do multiple string operations (e.g.##and%) at once in Bash.this_list="${bases[*]//test/this}"populates thethis_liststring with all the entries inbaseswith all occurrences oftestin each of them replaced bythis("this_1.txt ..."). Again, see Parameter expansion [Bash Hackers Wiki] for details of how this works. The entries in the list are separated by spaces. The entries in the list in the question were separated by newlines. You can do that forthis_listby settingIFS=$'\n'before doing thethis_list=...assignment. See Modify IFS in bash while building and array, What is the exact meaning of IFS=$'\n'?, and Is it a sane approach to "back up" the $IFS variable?. The first character in the value ofIFSis used to separate array elements when converting an array to a string with"${arrayname[*]}".declare -p this_listshows the contents ofthis_listin an unambiguous way.A few general points:
lsin programs. It's for interactive use only. You might get away with using it in programs sometimes, but it will eventually bite you hard. See Why you shouldn't parse the output of ls(1) and Why not parse 'ls' (and what do to instead)?.findandsedshould be able to handle much larger numbers of files, but Bash might struggle to handle the resulting huge strings (or arrays) anyway.不，它也不是提取基本名称的最有效方法。就此而言，解析 ls 的输出也不明智，尽管这是一个相对良性的情况。如果您想处理文件名列表，那么通过一个
sed或awk进程传递整个列表是一种更好的方法。例如：find命令输出指定目录中非点文件的路径，每行一个，就像ls所做的那样。sed然后尝试对每个替换进行两次替换：第一个删除所有内容，直到并包括最后一个/字符 alabasename，第二个替换this用于test，其中后者出现在该行剩余内容的开头。另请注意，这种方法与您原来的方法一样，会出现包含换行符的文件名问题。它不存在包含其他空格的文件名的固有问题，但如果任何文件名包含空格，您将无法正确解释结果。
No, nor is it the most efficient way to to extract the base names. Nor, for that matter, is it wise to parse the output of
ls, though this is a relatively benign case. If you want to massage a list of filenames then passing the whole list through onesedorawkprocess is a better approach. For example:That
findcommand outputs paths to the non-dotfiles in the specified directory, one per line, much aslswould do.sedthen attempts two substitutions on each one: the first removes everything up to and including the last/character, alabasename, and the second substitutesthisfortestwhere the latter appears at the beginning of what's left of the line.Note also that this approach, like your original one, will have issues with filenames containing newlines. It doesn't have an inherent issue with file names containing other whitespace, but you will have trouble interpreting the results correctly if any of the file names contain whitespace.
在这里解决：
https://unix.stackexchange.com/questions/36795/find-sed-search- and-replace
您可以使用 find 和 -exec 以及多个由
;分隔的 sed 命令在一行中执行此操作：第一个 sed 命令最多可达
s/\([^\.]*\)\..*/\1/g删除第一个.之后的所有内容。第二个 sed 命令
s?users/ uname?gs://uname?g进行替换解析
ls输出是不好的做法。Solved here:
https://unix.stackexchange.com/questions/36795/find-sed-search-and-replace
You can do it one line, using find with -exec and multiple sed commands separated by
;:First sed command up to
s/\([^\.]*\)\..*/\1/gremoves everyting after first.Second sed command
s?users/uname?gs://uname?gdoes substitutionParsing
lsoutput is bad practice.