Linux 庆典 : timeout a program if the program hangs in there and echo 'timeout'

Question

我有一个程序可以打印出通过或失败。我想检测卡在那里的程序并回显“超时”

我写了这样一个脚本:

#!/bin/bash

echo -n 'test' && timeout 5 ./mytest | grep -q -i 'passed' && echo ', passed'|| echo ', failed'

if [ $? -eq 124 ]; then
    echo 'timeout'
fi

但它会将卡在那里的程序视为“失败”并终止该程序。任何建议将不胜感激。谢谢!

Answer 1

人工超时:

NAME
       timeout - run a command with a time limit

SYNOPSIS
       timeout [OPTION] DURATION COMMAND [ARG]...
       timeout [OPTION]

DESCRIPTION
   Start COMMAND, and kill it if still running after DURATION.
...

这是一个例子:

cmd_output=`timeout 5 ./mytest`

if [ $? -eq 124 ]; then
    echo 'timeout'
else
    echo $cmd_output
fi