output on the same line

ygloo · 11-10-2006, 11:39 AM

hello,
bash script outputs data continiously in terminal...
new data is printed on a new line
nothing unusual so far....

how to print new output in the place of previous (on the same line)?

druuna · 11-10-2006, 12:31 PM

Hi,

Haven't used it that much myself, but take a look at tput.

It enables you to put string/text anywhere on the screen (and overwrite if something is already there), nice for building 'simple', but reasonably good looking menu's.

Hope this helps.

demon_vox · 11-10-2006, 12:38 PM

if you want to print on the same line, you should never go to the second one

For example:

Code:

echo -n "This is the first line"
sleep 1
echo -e "\rsecond line on top of the first line!"

In the echo command the -n specifies that the echo should not add a newline at the end. And the -e specifies that there might be special escaped characters in the string (in this case there is a \r at the beginning which is the carriage return).

Cheers!

druuna · 11-10-2006, 01:34 PM

Hi,

@demon_vox: It does not overwrite the text that's already there, it puts the new text after the text already there. This makes the chance of line wrapping (by the terminal) greater.

demon_vox · 11-10-2006, 02:04 PM

Hi druuna, I test that code and it worked. The second echo starts writting at the begining of the line. Perhaps you missed the -e, in which case it would write after the text like you said.
The only problem that you might get is that the second echo might not cover up completely the output of the first one if the second string is shorter. But that can be fixed with spaces.

Be sure you copied the example exactly.

druuna · 11-10-2006, 02:25 PM

Hi,

I did copy it exactly, there's another issue I did not mention. The ! at the end. Here's my output:

Code:

$ echo -n "This is the first line"
This is the first line$ sleep 1
$ echo -e "\rsecond line on top of the first line!"
-bash: !": event not found
$

It just hits me why it doesn't work (my way): Im doing this from the command line (which fails), not from within a script (which works nicely!).

I understand it not being on the same line if done from the command line, but the ! error puzzles me:

Code:

$ echo "second line on top of the first line!"
-bash: !": event not found

Ok, ! is a special character:
But escaping the ! has not the effect I expected:

$ echo "second line on top of the first line\!"
second line on top of the first line\!

ygloo · 11-10-2006, 02:36 PM

i try in this script, it won't work:

Code:

#!/bin/bash
# a clock

seconds=0
minutes=0
hours=0

sec ()
{
	while [ "$seconds" -le 59 ]; do
###
		echo -n "$hours " :"$minutes " ."$seconds "
		sleep 1
###
		echo -e "\r$hours " :"$minutes " ."$seconds "
		let "seconds += 1"
	done
	seconds=0 
	let "minutes += 1"
}

min ()
{
	while [ "$minutes" -le 59 ]; do
		sec
	done
	minutes=0 
	let "hours += 1"
}

hr ()
{
	while [ "$hours" -lt 24 ]; do
		min
	done
	hours=0 
}

while true; do
	hr	
done

druuna · 11-10-2006, 02:46 PM

Hi,

I can't get it to work either with demon_vox's solution. It will work with tput (changes are in bold):

Code:

#!/bin/bash
# a clock
clear
seconds=0
minutes=0
hours=0

sec ()
{
     while [ "$seconds" -le 59 ]; do
          # place cursor at row 5 column 10
          tput cup 5 10
          echo "$hours " :"$minutes " ."$seconds "
          let "seconds += 1"
     done
     seconds=0
     let "minutes += 1"
}

min ()
{
     while [ "$minutes" -le 59 ]; do
          sec
     done
     minutes=0
     let "hours += 1"
}

hr ()
{
     while [ "$hours" -lt 24 ]; do
          min
     done
     hours=0
}

while true; do
     hr
done

ygloo · 11-10-2006, 03:10 PM

found how to do it!!

echo -en "\r$hours :$minutes .$seconds "

demon_vox · 11-10-2006, 03:14 PM

Hi all,
the error with the ! is simple: just dont put that character in the string

Its not the same runnning it separately from a command line than to execute them all togheter from a script (and that's why it didnt worked for you at first).

In regards with the ygloo script, its not working because you are using my example too literal

The second echo always add a newline so you get your clock to write the time one line below the other. You should have used just one echo:

Code:

#!/bin/bash
# a clock

seconds=0
minutes=0
hours=0

sec ()
{
	while [ "$seconds" -le 59 ]; do
###
		echo -en "$hours " :"$minutes " ."$seconds "
		sleep 1
###
		let "seconds += 1"
	done
	seconds=0 
	let "minutes += 1"
}

min ()
{
	while [ "$minutes" -le 59 ]; do
		sec
	done
	minutes=0 
	let "hours += 1"
}

hr ()
{
	while [ "$hours" -lt 24 ]; do
		min
	done
	hours=0 
}

while true; do
	hr	
done

NOTE: I copied druuna idea of the modifications in bold! (good idea druuna!

)

I think both ideas are good, each one has its strengths and disadvantages: my idea is simpler and it can be used easily in any language. But, it is quite limited since you cant do much more than that (and if you happen to output a newline somewhere else in your code, your output would look messy).
druuna idea's of using tput is probably more flexible, since tput will let you do more things (I learned today about this command, but having a flash look at the man page, it seems that it has more things you can do with the screen). On the other side, it is a bit more complex to use, and its not that easy to use in other language (not that it matters for this case, but its something I like to take into consideration

)

In conclusion: we all learned new things today!

Cheers!

druuna · 11-10-2006, 03:33 PM

Hi,

And I found the solution to the echo "foo!" problem (because not using it is a workaround, not a solution

):

Code:

$ eval echo "foo\!"
foo!

If anybody knows why this echo "foo\!" results in foo\!, please let me know.