Bash-替换并写入同一行

Question

I have a bash script which get details from many servers. It works good but i want that the lines get updated and not get written new.

while [ true ] ; do
for i in $(seq ${#details[@]}); do
    .... more code (irrelevant)

echo   ${server[$i]}
echo   $stat1
echo   $stat2
echo   $stat3
echo   $stat4


done
done

How can i do, that all lines get constantly updated into same line? I try with echo -ne but this makes that everything is in one long line.

I want that the line keep the place and just get updated with new value.

Would be great if somebody knows a trick.

Thank you!

UPDATE 1

@cbuckley:

Thanks for your answer, but its not working correctly. In this way with -ne i tryed it already. Result is (it always create new lines):

10.0.0.2
100310.0.0.1
72710.0.0.3
368310.0.0.2
100310.0.0.1
72710.0.0.3

Should be

10.0.0.1
17
1003
10.0.0.2
319
727
10.0.0.3
157
3683

values under IP should get updated constantly. (i think this should normaly work with -ne, but in my case it dont).

Answer 1

If you've already outputted across multiple lines, you can't remove those lines without clearing the screen. You have two options:

Using watch

You can write a script that outputs the stats once, and then use watch to repeatedly that script:

watch -n 10 ./script.sh # calls script every 10 seconds.

Clearing the screen

If that is not suitable, you'll need to clear the screen yourself:

while [ true ] ; do
    clear # clear the screen

    for i in $(seq ${#details[@]}); do
        # ...
    done

    sleep 10 # don't update the screen too often
done

However, at this point, you've pretty much implemented a basic version of watch anyway.

Answer 2

您可能想尝试使用带有-i选项的sed来“就地”编辑文件（即更改现有文件而不是编写新文件）