If I have a BASH variable:
如果我有一个BASH变量:
Exclude="somefile.txt anotherfile.txt"
How can I get the contents of a directory but exclude the above to files in the listing? I want to be able to do something like:
如何获取目录的内容,但将上述内容排除在列表中的文件中?我希望能够做到这样的事情:
Files= #some command here
someprogram ${Files}
someprogram should be given all of the files in a particular directory, except for those in the ${Exclude} variable. Modifying someprogram is not an option.
someprogram应该被赋予特定目录中的所有文件,除了$ {Exclude}变量中的那些文件。修改someprogram不是一个选项。
4 个解决方案
#1
I'm not sure if you were taking about unix shell scripting, but here's a working example for bash:
我不确定你是否正在使用unix shell脚本,但这是一个bash的工作示例:
#!/bin/bash
Exclude=("a" "b" "c")
Listing=(`ls -1Q`)
Files=( $(comm -23 <( printf "%s\n" "${Listing[@]}" ) <( printf "%s\n" "${Exclude[@]}"
) ) )
echo ${Files[@]}
Note that I enclosed every filename in Exclude with double quotes and added parenthesis around them. Replace echo with someprogram, change the ls command to the directory you'd like examined and you should have it working. The comm program is the key, here.
请注意,我使用双引号括起Exclude中的每个文件名,并在它们周围添加括号。将echo替换为someprogram,将ls命令更改为您要检查的目录,并且应该让它工作。 comm程序是关键,在这里。
#2
You can use find. something like:
你可以使用find。就像是:
FILES=`find /tmp/my_directory -type f -maxdepth 1 -name "*.txt" -not -name somefile.txt -not -name anotherfile.txt`
where /tmp/my_directory is the path you want to search.
其中/ tmp / my_directory是您要搜索的路径。
You could build up the "-not -name blah -not -name blah2" list from Excludes if you want with a simple for loop...
如果你想要一个简单的for循环,你可以从Excludes中建立“-not -name blah -not -name blah2”列表...
#3
Here's a one liner for a standard Unix command line:
这是标准Unix命令行的一个内容:
ls | grep -v "^${Exclude}$" | xargs
ls | grep -v“^ $ {Exclude} $”| xargs的
It does have one assumption. ${Exclude} needs to be properly escaped so charaters like period aren't interpreted as part of the regex.
它确实有一个假设。 $ {Exclude}需要被正确转义,因此像句点这样的字符不会被解释为正则表达式的一部分。
#4
Assuming filenames with no spaces or other pathological characters:
假设文件名没有空格或其他病态字符:
shopt -s extglob
Files=(!(@(${Exclude// /|})))
#1
I'm not sure if you were taking about unix shell scripting, but here's a working example for bash:
我不确定你是否正在使用unix shell脚本,但这是一个bash的工作示例:
#!/bin/bash
Exclude=("a" "b" "c")
Listing=(`ls -1Q`)
Files=( $(comm -23 <( printf "%s\n" "${Listing[@]}" ) <( printf "%s\n" "${Exclude[@]}"
) ) )
echo ${Files[@]}
Note that I enclosed every filename in Exclude with double quotes and added parenthesis around them. Replace echo with someprogram, change the ls command to the directory you'd like examined and you should have it working. The comm program is the key, here.
请注意,我使用双引号括起Exclude中的每个文件名,并在它们周围添加括号。将echo替换为someprogram,将ls命令更改为您要检查的目录,并且应该让它工作。 comm程序是关键,在这里。
#2
You can use find. something like:
你可以使用find。就像是:
FILES=`find /tmp/my_directory -type f -maxdepth 1 -name "*.txt" -not -name somefile.txt -not -name anotherfile.txt`
where /tmp/my_directory is the path you want to search.
其中/ tmp / my_directory是您要搜索的路径。
You could build up the "-not -name blah -not -name blah2" list from Excludes if you want with a simple for loop...
如果你想要一个简单的for循环,你可以从Excludes中建立“-not -name blah -not -name blah2”列表...
#3
Here's a one liner for a standard Unix command line:
这是标准Unix命令行的一个内容:
ls | grep -v "^${Exclude}$" | xargs
ls | grep -v“^ $ {Exclude} $”| xargs的
It does have one assumption. ${Exclude} needs to be properly escaped so charaters like period aren't interpreted as part of the regex.
它确实有一个假设。 $ {Exclude}需要被正确转义,因此像句点这样的字符不会被解释为正则表达式的一部分。
#4
Assuming filenames with no spaces or other pathological characters:
假设文件名没有空格或其他病态字符:
shopt -s extglob
Files=(!(@(${Exclude// /|})))