bash 解析 ps 的“etime”输出并将其转换为秒
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Parse ps' "etime" output and convert it into seconds
提问by Clodoaldo Neto
These are possible output formats for ps h -eo etime
这些是可能的输出格式 ps h -eo etime
21-18:26:30
15:28:37
48:14
00:01
How to parse them into seconds?
如何将它们解析为秒？
- Please assume at least 3 digits for the days part as I don't know how long it can be.
- The output will be
egrepedto one only line so no need for a loop.
- 请假设天数部分至少为 3 位数，因为我不知道它可以持续多长时间。
- 输出将
egreped只有一行，因此不需要循环。
回答by user000001
With awk:
使用 awk：
#!/usr/bin/awk -f
BEGIN { FS = ":" }
{
if (NF == 2) {
print *60 +
} else if (NF == 3) {
split(, a, "-");
if (a[2] != "" ) {
print ((a[1]*24+a[2])*60 + ) * 60 + ;
} else {
print (*60 + ) * 60 + ;
}
}
}
Run with :
运行：
awk -f script.awk datafile
Output:
输出：
1880790
55717
2894
1
And finally, if you want to pipe to the parser, you can do something like this:
最后，如果您想通过管道传输到解析器，您可以执行以下操作：
ps h -eo etime | ./script.awk
回答by Fluffy
Yet another bash solution, which works any number of fields:
另一个 bash 解决方案，适用于任意数量的字段：
ps -p $pid -oetime= | tr '-' ':' | awk -F: '{ total=0; m=1; } { for (i=0; i < NF; i++) {total += $(NF-i)*m; m *= i >= 2 ? 24 : 60 }} {print total}'
ps -p $pid -oetime= | tr '-' ':' | awk -F: '{ total=0; m=1; } { for (i=0; i < NF; i++) {total += $(NF-i)*m; m *= i >= 2 ? 24 : 60 }} {print total}'
Explanation:
解释：
- replace
-to:so that string becomes1:2:3:4instead of '1-2:3:4', set total to 0 and multiplier to 1 - split by :, start with the last field (seconds), multiply it by m = 1, add to total second number, m becomes 60 (seconds in a minute)
- add minutes fields multiplied by 60, m becomes 3600
- add hours * 3600
- add days * 3600 * 24
- 替换
-为:字符串，而1:2:3:4不是“1-2:3:4”，将总数设置为 0，乘数设置为 1 - 分割：，从最后一个字段（秒）开始，乘以 m = 1，加上总秒数，m 变为 60（一分钟内的秒数）
- 添加分钟字段乘以 60，m 变为 3600
- 添加小时 * 3600
- 添加天数 * 3600 * 24
回答by Andor
Here's mine Perl one liner:
这是我的 Perl 一个班轮：
ps -eo pid,comm,etime | perl -ane '@t=reverse split(/[:-]/,$F[2]); $s=$t[0]+$t[1]*60+$t[2]*3600+$t[3]*86400; print "$F[0]\t$F[1]\t$F[2]\t$s\n"'
Undefined values are rendering to zero, so they won't have effect on the sum of seconds.
未定义的值呈现为零，因此它们不会对秒数总和产生影响。
回答by James Thomas
Think I might be missing the point here but the simplest way of doing this is:
认为我可能在这里遗漏了一点，但最简单的方法是：
ps h -eo etimes
Note the 's' on the end of etime.
请注意 etime 末尾的“s”。
回答by bolD
Try to use my solution with sed+awk:
尝试将我的解决方案与 sed+awk 结合使用：
ps --pid $YOUR_PID -o etime= | sed 's/:\|-/ /g;' |\
awk '{print " "" "" "}' |\
awk '{print +*60+*3600+*86400}'
it splits the string with sed, then inverts the numbers backwards ("DD hh mm ss" -> "ss mm hh DD") and calculates them with awk.
它用 sed 拆分字符串，然后将数字向后反转（“DD hh mm ss”->“ss mm hh DD”）并用 awk 计算它们。
$ echo 21-18:26:30 | sed 's/:\|-/ /g;' | awk '{print " "" "" "}' | awk '{print +*60+*3600+*86400}'
1880790
$ echo 15:28:37 | sed 's/:\|-/ /g;' | awk '{print " "" "" "}' | awk '{print +*60+*3600+*86400}'
55717
$ echo 48:14 | sed 's/:\|-/ /g;' | awk '{print " "" "" "}' | awk '{print +*60+*3600+*86400}'
2894
$ echo 00:01 | sed 's/:\|-/ /g;' | awk '{print " "" "" "}' | awk '{print +*60+*3600+*86400}'
1
Also you can play with sed and remove all non-numeric characters from input string:
您也可以使用 sed 并从输入字符串中删除所有非数字字符：
sed 's/[^0-9]/ /g;' | awk '{print " "" "" "}' | awk '{print +*60+*3600+*86400}'
回答by markeissler
I've implemented a 100% bash solution as follows:
我已经实现了一个 100% bash 解决方案，如下所示：
#!/usr/bin/env bash
etime_to_seconds() {
local time_string=""
local time_string_array=()
local time_seconds=0
local return_status=0
[[ -z "${time_string}" ]] && return 255
# etime string returned by ps(1) consists one of three formats:
# 31:24 (less than 1 hour)
# 23:22:38 (less than 1 day)
# 01-00:54:47 (more than 1 day)
#
# convert days component into just another element
time_string="${time_string//-/:}"
# split time_string into components separated by ':'
time_string_array=( ${time_string//:/ } )
# parse the array in reverse (smallest unit to largest)
local _elem=""
local _indx=1
for(( i=${#time_string_array[@]}; i>0; i-- )); do
_elem="${time_string_array[$i-1]}"
# convert to base 10
_elem=$(( 10#${_elem} ))
case ${_indx} in
1 )
(( time_seconds+=${_elem} ))
;;
2 )
(( time_seconds+=${_elem}*60 ))
;;
3 )
(( time_seconds+=${_elem}*3600 ))
;;
4 )
(( time_seconds+=${_elem}*86400 ))
;;
esac
(( _indx++ ))
done
unset _indx
unset _elem
echo -n "$time_seconds"; return $return_status
}
main() {
local time_string_array=( "31:24" "23:22:38" "06-00:15:30" "09:10" )
for timeStr in "${time_string_array[@]}"; do
local _secs="$(etime_to_seconds "$timeStr")"
echo " timeStr: "$timeStr""
echo " etime_to_seconds: ${_secs}"
done
}
main
回答by Larry
[[ $(ps -o etime= REPLACE_ME_WITH_PID) =~ ((.*)-)?((.*):)?(.*):(.*) ]]
printf "%d\n" $((10#${BASH_REMATCH[2]} * 60 * 60 * 24 + 10#${BASH_REMATCH[4]} * 60 * 60 + 10#${BASH_REMATCH[5]} * 60 + 10#${BASH_REMATCH[6]}))
PureBASH. Requires BASH 2+ (?) for the BASH_REMATCH variable. The regex matches any of the inputs and places the matched strings into the array BASH_REMATCH, which parts of are used to compute number of seconds.
纯BASH。BASH_REMATCH 变量需要 BASH 2+ (?)。正则表达式匹配任何输入并将匹配的字符串放入数组 BASH_REMATCH，其中的部分用于计算秒数。
回答by Delyan Tashev
#!/bin/bash
echo | sed 's/-/:/g' | awk -F $':' -f <(cat - <<-'EOF'
{
if (NF == 1) {
print
}
if (NF == 2) {
print *60 +
}
if (NF == 3) {
print *60*60 + *60 + ;
}
if (NF == 4) {
print *24*60*60 + *60*60 + *60 + ;
}
if (NF > 4 ) {
print "Cannot convert datatime to seconds"
exit 2
}
}
EOF
) < /dev/stdin
Then to run use:
然后运行使用：
ps -eo etime | ./script.sh
回答by rinogo
Here's a PHP alternative, readable and fully unit-tested:
这是一个 PHP 替代方案，可读且完全经过单元测试：
//Convert the etime string $s (as returned by the `ps` command) into seconds
function parse_etime($s) {
$m = array();
preg_match("/^(([\d]+)-)?(([\d]+):)?([\d]+):([\d]+)$/", trim($s), $m); //Man page for `ps` says that the format for etime is [[dd-]hh:]mm:ss
return
$m[2]*86400+ //Days
$m[4]*3600+ //Hours
$m[5]*60+ //Minutes
$m[6]; //Seconds
}
回答by JonB
Ruby version:
红宝石版本：
def psETime2Seconds(etime)
running_secs = 0
if etime.match(/^(([\d]+)-)?(([\d]+):)?([\d]+):([\d]+)$/)
running_secs += .to_i * 86400 # days
running_secs += .to_i * 3600 # hours
running_secs += .to_i * 60 # minutes
running_secs += .to_i # seconds
end
return running_secs
end
