Linux ksh 将命令的结果存储到变量
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ksh storing result of a command to a variable
提问by user1769538
I want to store the result of a command to a variable in my shell script. I cant seem to get it to work. I want the most recently dated file in the directory.
我想将命令的结果存储到我的 shell 脚本中的一个变量中。我似乎无法让它工作。我想要目录中最新日期的文件。
PRODUCT= 'ls -t /some/dir/file* | head -1 | xargs -n1 basename'
it wont work
它不会工作
回答by Carl Norum
回答by Barton Chittenden
The problem that you're having is that the command needs to be surrounded by back-ticks rather than single quotes. This is known as 'Command Substitution'.
您遇到的问题是命令需要用反引号而不是单引号括起来。这称为“命令替换”。
Bash allows you to use $()for command substitution, but this is not available in all shells. I don't know if it's available in KSH; if it is, it's probably not available in all versions.
Bash 允许您$()用于命令替换，但这并非在所有 shell 中都可用。不知道KSH有没有；如果是，它可能并非在所有版本中都可用。
If the $()syntax is available in your version of ksh, you should definitely use it; it's easier to read (back ticks are too easy to confuse with single quotes); back-ticks are also hard to nest.
如果该$()语法在您的 ksh 版本中可用，您绝对应该使用它；它更容易阅读（反引号太容易与单引号混淆）；反引号也很难嵌套。
This only addresses one of the problems with your command, however: lsreturns directories as well as files, so if the most recent thing modified in the specified directory is a sub-directory, that is what you will see.
然而，这仅解决了您的命令的问题之一：ls返回目录和文件，因此如果在指定目录中修改的最新内容是子目录，这就是您将看到的。
If you only want to see files, I suggest using some version of the following (I'm using Bash, which supports default variables, you'll probably have to play around with the syntax of $1)
如果你只想查看文件，我建议使用以下的一些版本（我使用的是 Bash，它支持默认变量，你可能不得不使用 的语法$1）
lastfile ()
{
find ${1:-.} -maxdepth 1 -type f -printf "%T+ %p\n" | sort -n | tail -1 | sed 's/[^[:space:]]\+ //'
}
This runs find on the directory, and only pulls files from that directory. It formats all of the files like this:
这会在目录上运行 find ，并且只从该目录中提取文件。它格式化所有文件，如下所示：
2012-08-29+16:21:40.0000000000 ./.sqlite_history
2013-01-14+08:52:14.0000000000 ./.davmail.properties
2012-04-04+16:16:40.0000000000 ./.DS_Store
2010-04-21+15:49:00.0000000000 ./.joe_state
2008-09-05+17:15:28.0000000000 ./.hplip.conf
2012-01-31+13:12:28.0000000000 ./.oneclick
sorts the list, takes the last line, and chops off everything before the first space.
对列表进行排序，取最后一行，并在第一个空格之前切掉所有内容。
回答by doniyor
you have two options, either $or backsticks`.
你有两个选择，要么 要么$backsticks `。
1) x=$(ls -t /some/dir/file* | head -1 | xargs -n1 basename)
1) x=$(ls -t /some/dir/file* | head -1 | xargs -n1 basename)
or
或者
2) x=`ls -t /some/dir/file* | head -1 | xargs -n1 basename`
2) x=`ls -t /some/dir/file* | head -1 | xargs -n1 basename`
echo $x
Edit:removing unnecessary bracket for (2).
编辑：删除（2）不必要的括号。
回答by Olivier Dulac
You need both quotes to ensure you keep the name even if it contains spaces, and also in case you later want more than 1 file, and "$(..)" to run commands in background
您需要两个引号以确保您保留名称，即使它包含空格，并且如果您以后需要超过 1 个文件，以及“$(..)”在后台运行命令
I believe you also need the '-1' option to ls, otherwise you could have several names per lines (you only keep 1 line, but it could be several files)
我相信您还需要 ls 的“-1”选项，否则每行可能有多个名称（您只保留 1 行，但可能是多个文件）
PRODUCT="$(ls -1t /some/dir/file* | head -1 | xargs -n1 basename)"
Please do not put space around the "=" variable assignments (as I saw on other solutions here) , as it's not very compatible as well.
请不要在“=”变量赋值周围放置空格（正如我在其他解决方案中看到的那样），因为它也不太兼容。
回答by Yelali
I would do something like:
我会做这样的事情：
Your version corrected:
您的版本已更正：
PRODUCT=$(ls -t /some/dir/file* | head -1 | xargs -n1 basename)
Or simpler:
或者更简单：
PRODUCT=$(cd /some/dir && ls -1t file* | head -1)
- change to the directory
- list one filename per line and sort by time/date
- grab the first line
- 切换到目录
- 每行列出一个文件名并按时间/日期排序
- 抢第一行
