Linux Grep 时间命令输出
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Grep time command output
提问by jettisamba
Using time ls, I have the following output:
使用time ls，我有以下输出：
$ time ls -l
total 2
-rwx------+ 1 FRIENDS None 97 Jun 23 08:59 location.txt
-rw-r--r--+ 1 FRIENDS None 10 Jun 23 09:06 welcome
real 0m0.040s
user 0m0.000s
sys 0m0.031s
Now, when I try to greponly the realvalue line, the actual result is:
现在，当我grep只尝试真正的价值线时，实际结果是：
$ time ls -l | grep real
real 0m0.040s
user 0m0.000s
sys 0m0.031s
My question is, how to get only the real value as output? In this case, 0m0.040s.
我的问题是，如何只获得真正的价值作为输出？在这种情况下，0m0.040s。
采纳答案by rici
timewrites its output to stderr, so you need to pipe stderr instead of stdout. But it's also important to remember that timeis part of the syntax of bash, and it times an entire pipeline. Consequently, you need to wrap the pipeline in braces, or run it in a subshell:
time将其输出写入 stderr，因此您需要通过管道传输 stderr 而不是 stdout。但同样重要的是要记住这time是 bash 语法的一部分，它对整个管道进行计时。因此，您需要将管道包装在大括号中，或在子 shell 中运行它：
$ { time ls -l >/dev/null; } 2>&1 | grep real
real 0m0.005s
With Bash v4.0 (probably universal on Linux distributions but still not standard on Mac OS X), you can use |&to pipe both stdoutand stderr:
使用 Bash v4.0（可能在 Linux 发行版上通用，但在 Mac OS X 上仍然不是标准），您可以使用|&管道stdout和stderr：
{ time ls -l >/dev/null; } |& grep real
Alternatively, you can use the timeutility, which allows control of the output format. On my system, that utility is found in /usr/bin/time:
或者，您可以使用该time实用程序，它允许控制输出格式。在我的系统上，该实用程序位于/usr/bin/time：
/usr/bin/time -f%e ls -l >/dev/null
man timefor more details on the timeutility.
man time有关该time实用程序的更多详细信息。
回答by Clyde
(time ls -l) 2>&1 > /dev/null |grep real
This redirects stderr (which is where time sends its output) to the same stream as stdout, then redirects stdout to dev/null so the output of ls is not captured, then pipes what is now the output of time into the stdin of grep.
这会将 stderr（时间发送其输出的地方）重定向到与 stdout 相同的流，然后将 stdout 重定向到 dev/null 以便不捕获 ls 的输出，然后将现在的时间输出通过管道传输到 grep 的 stdin 中。
回答by doubleDown
If you just want to specify the output format of timebuiltin, you can modify the value of TIMEFORMATenvironment variable instead of filtering it with grep.
如果只想指定timebuiltin的输出格式，可以修改TIMEFORMAT环境变量的值，而不是用grep.
In you case,
在你的情况下，
TIMEFORMAT=%R
time ls -l
would give you the "real" time only.
只会给你“真实”的时间。
Here's the linkto relevant information in Bash manual (under "TIMEFORMAT").
这是Bash 手册中相关信息的链接（在“TIMEFORMAT”下）。
Thisis a similar question on SO about parsing the output of time.
这是关于解析time.
回答by VeLKerr
I think, it can be made a little easier:
我认为，它可以变得更容易一些：
time ls &> /dev/null | grep real
回答by Immotus
Look out.. bash has a built-in "time" command. Here are some of the differences..
注意.. bash 有一个内置的“时间”命令。以下是一些差异。
# GNU time command (can also use $TIMEFORMAT variable instead of -f)
bash> /usr/bin/time -f%e ls >/dev/null
0.00
# BASH built-in time command (can also use $TIME variable instead of -f)
bash> time -f%e ls >/dev/null
-f%e: command not found
real 0m0.005s
user 0m0.004s
sys 0m0.000s
