Bash 如何从 PID 获取完整的脚本名称
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Bash how to get full script name from PID
提问by user1105430
I have this php function that checks the script's name from the given PID, and compares it to itself.
我有这个 php 函数，它从给定的 PID 中检查脚本的名称，并将其与自身进行比较。
function isRunning($pid) {
$filename = exec('ps -p '.$pid.' -o "%c"');
$self = basename($_SERVER['SCRIPT_NAME']);
return ($filename == $self) ? TRUE : FALSE;
}
From what I know, I usually use this command to get the script name from the PID:
据我所知，我通常使用此命令从 PID 中获取脚本名称：
ps -o PID -o "%c"
It returns me the filename, but only the first 15 characters. Since my script's name is
它返回文件名，但只返回前 15 个字符。由于我的脚本名称是
daily_system_check.php
Daily_system_check.php
the function always returns FALSE, because it's comparing itself with
该函数总是返回 FALSE，因为它正在将自己与
daily_system_ch
Daily_system_ch
Is there another bash command for Centos 6 that will return me script's full name?
Centos 6 是否还有另一个 bash 命令可以返回脚本的全名？
回答by julumme
You didn't specify what is your OS, but in Ubuntu Linux I can see full name of the script with adding --contextto the ps call:
您没有指定您的操作系统是什么，但在 Ubuntu Linux 中，我可以通过添加--context到 ps 调用来查看脚本的全名 ：
# ps -p 17165 --context
PID CONTEXT COMMAND
17165 unconfined /bin/bash ./testing_long_script_name.sh
#
回答by amc
read the the proc cmdlinefile:
读取proccmdline文件：
cat /proc/$pid/cmdline | awk 'BEGIN {FS="function isRunning($pid) {
$filename = basename(exec('ps -o cmd= '.$pid));
$self = basename($_SERVER['SCRIPT_NAME']);
return ($filename == $self) ? TRUE : FALSE;
}
"} {print }'
回答by user1105430
There seems to be no flag or collumn in "ps" command to show the whole filename without the filepath or it being cutoff. PHP's basename()gets the job done.
“ps”命令中似乎没有标志或列来显示没有文件路径或被截断的整个文件名。PHP 的basename()完成了工作。
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