bash 使用 shell 变量查找具有多个文件名的 -name
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find -name with multiple filenames using shell variable
提问by ring bearer
I have a findcommand that finds files with name matching multiple patterns mentioned against the -nameparameter
我有一个find命令可以查找名称与-name参数中提到的多个模式匹配的文件
find -L . \( -name "SystemOut*.log" -o -name "*.out" -o -name "*.log" -o -name "javacore*.*" \)
This finds required files successfully at the command line. What I am looking for is to use this command in a shell script and join this with a tarcommand to create a tar of all log files. So, in a script I do the following:
这将在命令行成功找到所需的文件。我正在寻找的是在 shell 脚本中使用此命令并将其与tar命令结合以创建所有日志文件的 tar。因此，在脚本中，我执行以下操作：
LIST="-name \"SystemOut*.log\" -o -name \"*.out\" -o -name \"*.log\" -o -name \"javacore*.*\" "
find -L . \( ${LIST} \)
This does not print files that I am looking for.
这不会打印我正在寻找的文件。
First - why this script is not functioning like the command? Once it does, can I club it with cpioor similar to create a tarin one shot?
首先 - 为什么这个脚本不像命令那样运行？一旦它出现，我可以用它cpio或类似的方法tar进行一次击球吗？
回答by lunixbochs
Looks like findfails to match *in patterns from unquoted variables. This syntax works for me (using bash arrays):
看起来find无法匹配*来自未加引号的变量的模式。此语法适用于我（使用 bash 数组）：
LIST=( -name \*.tar.gz )
find . "${LIST[@]}"
Your example would become the following:
您的示例将变为以下内容：
LIST=( -name SystemOut\*.log -o -name \*.out -o -name \*.log -o -name javacore\*.\* )
find -L . \( "${LIST[@]}" \)
回答by h0tw1r3
eval "find -L . \( ${LIST} \)"
回答by bta
When you have a long list of file names you use, you may want to try the following syntax instead:
当您使用的文件名列表很长时，您可能需要尝试以下语法：
# List of file patterns
Pat=( "SystemOut*.log"
"*.out"
"*.log"
"javacore*.*" )
# Loop through each file pattern and build a 'find' string
find $startdir \( -name $(printf -- $'\'%s\'' "${Pat[0]}") $(printf -- $'-o -name \'%s\' ' "${Pat[@]:1}") \)
That method constructs the argument sequentially using elements from a list, which tends to work better (at least in my recent experiences).
该方法使用列表中的元素顺序构造参数，这往往效果更好（至少在我最近的经验中）。
You can use find's -execoption to pass the results to an archiving program:
您可以使用 find 的-exec选项将结果传递给归档程序：
find -L . \( .. \) -exec tar -Af archive.tar {} \;
回答by dpp
You could use an eval and xargs,
你可以使用 eval 和 xargs，
eval "find -L . \( $LIST \) " | xargs tar cf 1.tar
回答by n. 'pronouns' m.
LIST="-name SystemOut*.log -o -name *.out -o -name *.log -o -name javacore*.*"
The wildcards are already quoted and you don't need to quote them again. Moreover, here
通配符已被引用，您无需再次引用它们。此外，这里
LIST="-name \"SystemOut*.log\""
the inner quotes are preserved and findwill get them as a part of the argument.
保留内部引号find并将它们作为参数的一部分。
