how to write a while loop in a bash file

Question

I am trying to write a while loop in bash as below:

while [!(netstat -pnlt | grep ":9393" | wc -l)];
do
    echo "server not ready... waiting..."
    sleep 2
done

I get

syntax error near unexpected token `netstat'

from console

what I want to do is:

to grep the service that using port 9393, count its lines. If lines number is 0, it means the service is not running, so I will keep it wait.

OS: debian:jessie

please help me correct it... I looked up a lot of documents

ghoti · Accepted Answer · 2017-01-17 09:03:54Z

2

Your test syntax is incorrect. Have a look at the man page for usage, or at the examples all over the Internet.

while [[ $(netstat -pnlt | grep -c ":9393 .*LISTEN") -eq '1' ]]; do
    echo "server not ready... waiting..."
    sleep 2
done

But with the edit you made to your question, the following would be better:

while ! netstat -pnlt | grep -q ":9393 .*LISTEN"; do
    echo "server not ready... waiting..."
    sleep 2
done

answered Jan 17, 2017 at 8:53

ghoti

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Comments

Ipor Sircer · Accepted Answer · 2017-01-17 08:59:46Z

2

It's better to use lsof

while ! lsof -iTCP:9393 -sTCP:LISTEN >/dev/null; do
    echo "server not ready... waiting..."
    sleep 2
done

answered Jan 17, 2017 at 8:59

Ipor Sircer

3,1693 gold badges12 silver badges18 bronze badges

One command instead of 3. Lower usage of resources, faster run time, easier to read by humans.