bash bash中的while循环使用pgrep来检查服务是否存在
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/38666191/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
while loop in bash that uses pgrep to check if service exists
提问by ewcz
Can someone explain to me why this doesn't work?
有人可以向我解释为什么这不起作用吗？
/usr/bin/mysqld_safe
STATUS=$(/usr/bin/pgrep mysql | wc -l)
while $STATUS -eq 0; do
echo "$STATUS"
sleep 1
done
It defies any logic. Yes I'm not versed in bash :)
它无视任何逻辑。是的，我不精通 bash :)
p.s. I'm trying to wait till service is up, but after running this even echo "$STATUS" is not printed! so loop is not running
ps 我正在尝试等到服务启动，但是在运行此命令后，甚至没有打印 echo "$STATUS" ！所以循环没有运行
回答by John1024
This runs mysqld_safe:
这运行mysqld_safe：
/usr/bin/mysqld_safe
This runs pgrepand stores the result in STATUS:
这将运行pgrep并将结果存储在 STATUS 中：
STATUS=$(/usr/bin/pgrep mysql | wc -l)
STATUS is never updated again. This loops using a constant fixed value of STATUS:
STATUS 永远不会再次更新。此循环使用 STATUS 的恒定固定值：
while $STATUS -eq 0; do
echo "$STATUS"
sleep 1
done
Note that the test in the while loop is malformed. It should read:
请注意，while 循环中的测试格式错误。它应该是：
while [ "$STATUS" -eq 0 ]; do
To get live updates, pgrepshould be run within the loop. Further, because pgrepsets an exit code, the test command, [...], is superfluous:
要获得实时更新，pgrep应在循环内运行。此外，因为pgrep设置了退出代码，所以测试命令[...], 是多余的：
To keep the loop running while there are no instances of mysql:
要在没有 mysql 实例时保持循环运行：
while ! /usr/bin/pgrep mysql >/dev/null; do
pgrepreturns success (exit code=0) when it finds a matching process. Since you seem to want the loop to repeat when there is no matching process, we invert the exit code using !.
pgrep找到匹配的进程时返回成功（退出代码=0）。由于您似乎希望在没有匹配进程时重复循环，因此我们使用!.
Or, putting it all back together:
或者，将它们全部重新组合在一起：
/usr/bin/mysqld_safe
while ! /usr/bin/pgrep mysql >/dev/null; do
echo "No such process"
sleep 1
done
Assuming that mysqld_safesuccessfully starts, the whileloop will never run. You should only see output from the while loop is mysqld_safefails to start.
假设mysqld_safe成功启动，while循环将永远不会运行。您应该只看到 while 循环的输出mysqld_safe无法启动。
If you instead want a continuous status update:
如果您想要持续的状态更新：
/usr/bin/mysqld_safe
while true; do
/usr/bin/pgrep mysql >/dev/null
echo "Current status: $?"
sleep 1
done
回答by ewcz
perhaps a solution could be something like this
也许解决方案可能是这样的
STATUS=0
while [ $STATUS -eq 0 ]; do
echo $STATUS
sleep 1
STATUS=$(/usr/bin/pgrep mysql | wc -l)
done
one needs to update the STATUSvariable within the loop to reflect the current status.
需要更新STATUS循环内的变量以反映当前状态。
