等到服务在 bash-script 中启动

Question

我有一个在后台启动一些服务的bash脚本。此服务成功启动后，它会将"服务器处于活动状态"打印到标准输出。我需要等到这个字符串出现，然后继续执行我的脚本。我怎样才能做到这一点？

Answer 1

我会这样做。

./server > /tmp/server-log.txt &
sleep 1
while ! grep -m1 'Server is active' < /tmp/server-log.txt; do
sleep 1
done
echo Continue

在这里，-m1告诉grep(1)在第一场比赛时退出。

我在下面用我的玩具"服务"验证了我的答案：

#! /bin/bash
trap "echo 'YOU killed me with SIGPIPE!' 1>&2 " SIGPIPE
rm -f /tmp/server-output.txt
for (( i=0; i<5; ++i )); do
echo "i==$i"
sleep 1;
done
echo "Server is active"
for (( ; i<10; ++i )); do
echo "i==$i"
sleep 1;
done
echo "Server is shutting down..." > /tmp/server-output.txt

如果将echo Continue替换为echo Continue; sleep 1; ls /tmp/server-msg.txt，您将看到ls: cannot access /tmp/server-output.txt: No such file or directory证明"继续"操作是在Server is active输出后立即触发的。

Answer 2

使用grep -q.-q选项使grep安静，当文本出现时，它将立即退出。

下面的命令在后台启动./some-service，并阻止直到 stdout 上出现"服务器处于活动状态"。

(./some-service &) | grep -q "Server is active"

Answer 3

让我读取有关service应用程序的服务状态：

$ /sbin/service network status
network.service - Network Connectivity
Loaded: loaded (/lib/systemd/system/network.service; enabled)
Active: active (exited) since Ср 2014-01-29 22:00:06 MSK; 1 day 15h ago
Process: 15491 ExecStart=/etc/rc.d/init.d/network start (code=exited, status=0/SUCCESS)
$ /sbin/service httpd status 
httpd.service - SYSV: Apache is a World Wide Web server.  It is used to serve HTML files and CGI.
Loaded: loaded (/etc/rc.d/init.d/httpd)
Active: activating (start) since Пт 2014-01-31 13:59:06 MSK; 930ms ago

它可以通过代码完成：

function is_in_activation {
activation=$(/sbin/service "$1" status | grep "Active: activation" )
if [ -z "$activation" ]; then
true;
else
false;
fi
return $?;
}
while is_in_activation network ; do true; done

Answer 4

您是否要求将标准输出重定向到标准输出？

./yourscript.sh  2>&1 |grep "Server is active" && echo "continue executing my script"