trying to echo a line that contains echo in a file

aristosv · 12-30-2015, 07:34 AM

I want to add this line to the file named "playing"

Code:

/usr/bin/mpg123 -Z /root/media/* 2>&1 | while read i; do echo "$(date): $i"; done | grep --line-buffered Playing

To do that I use the command echo

Code:

echo "/usr/bin/mpg123 -Z /root/media/* 2>&1 | while read i; do echo "$(date): $i"; done | grep --line-buffered Playing"

The problem is that the word "date" becomes the actual date when echoed into the file "playing". Like this:

Code:

/usr/bin/mpg123 -Z /root/media/* 2>&1 | while read i; do echo Wed Dec 30 15:31:48 EET 2015: ; done | grep --line-buffered Playing

How can I avoid this, and actually put the word "date" in the file?

Thanks

MensaWater · 12-30-2015, 07:45 AM

I fear I'm missing something in your question.

$() says to use the returned output of the command within the parentheses. You are doing: $(date) so you are getting the actual output of the command named "date".

If you want the literal word "date" don't put $() around date but instead put quotes around it.

i.e.
DO:
echo "date"
NOT:
echo $(date)

aristosv · 12-30-2015, 07:51 AM

The file "playing" will become an executable which will later run, to show which song is playing.
The command "date" has to run, to show the date and time the song started to play. That's why I need $(date).
"date" is not working

aristosv · 12-30-2015, 07:53 AM

To put it simply, I need to put the following command in a file. Using echo, sed, awk or whatever's available.

Code:

/usr/bin/mpg123 -Z /root/media/* 2>&1 | while read i; do echo "$(date): $i"; done | grep --line-buffered Playing

MensaWater · 12-30-2015, 08:22 AM

ah - that makes it clearer.

1) Encapsulate the whole string in double quotes.
2) Escape the dollar signs and parentheses by putting the \ character in front of them

e.g.
echo "/usr/bin/mpg123 -Z /root/media/* 2>&1 | while read i; do echo "\$$date$: \$i"; done | grep --line-buffered Playing" >test2

That would add your original line to the file named test2.

cat test2 would show:
/usr/bin/mpg123 -Z /root/media/* 2>&1 | while read i; do echo $(date): $i; done | grep --line-buffered Playing

aristosv · 12-30-2015, 08:33 AM

That did it. Thanks for your help.

chrism01 · 12-30-2015, 04:32 PM

Use of single quotes is less messy; specifically no need for '\'s

Code:

echo '/usr/bin/mpg123 -Z /root/media/* 2>&1 | while read i; do echo "$(date): $i"; done | grep --line-buffered Playing' >t.t

cat t.t

/usr/bin/mpg123 -Z /root/media/* 2>&1 | while read i; do echo "$(date): $i"; done | grep --line-buffered Playing

MensaWater · 01-05-2016, 12:39 PM

Quote:

Originally Posted by chrism01

Use of single quotes is less messy; specifically no need for '\'s

Good point and good for what the OP was doing.

I used the single quotes to day in doing a bunch of repetitive output that had things like slashes, semi-colons and brackets that would have been a bit of a pain to escape singly.

One note on it though is that the single quotes can't be used if you do want a variable expanded in one of the echo lines as I did today. The single quotes make it take ${var} literally rather than giving the value of ${var}. Not an issue in what the OP wrote but was an issue for what I was doing so for that one line I did have to do double quotes. Luckily that one line didn't also require me to escape anything else on the line but I could see situations where it would.

Just posting to note there are times when both are appropriate.