I'm writing a bash script to make DVD authoring more automated (but, mainly, so that I can learn some more bash scripting) and I'm trying to find out if it's possible to control how an exterrnal command presents its output.
For instance, the output from ffmpeg is a load of (to me) irrelevant cruft about options, libraries, streams, progress and so on.
What I really want is to be able to select for display only the lines with the input and output filenames and then to display the progress on the same line each time. Similarly for mkisofs and wodim.
I've tried Googling for this and am beginning to suspect that either it's not possible or nobody's thought of it before (or, possibly, that it's so obvious that nobody thinks it necessary to say how :-) ).
Many thanks, in advance,
David Shaw
You want to use grep and pipes. They are your friends. You want to pipe the output of the ffmpeg into grep and have it output only lines containing the text you want.
Assuming you have the input and output file names as command lines arguments $1 and $2 to your shell script, you might try something like
ffmpeg .... | grep "$1\|$2"
^ ^
| +-- escape and OR character
+--pipe character
The '\|' is an escape and an OR character for regular expressions. The OR '|' is also the pipe character so you have to escape that.
This will output only output lines that contain the files you are looking for.
This assumes all output is via stdout. If ffmpeg is outputting text via stderr then you will need to add some redirects at the end of ffmpeg line to redirect those back to stdout.
EDIT: I used the wrong quotes in the first example. Use double quotes or it won't expand the parameters $1 and $2
Related
I have following snippet:
host="https://example.com"
port="80"
url="${host}:${port}"
echo $url
the output is:
:80ps://example.com
How can I escape the colon here. I also tried:
url="${host}\:${port}"
but it did not work.
Expected output is:
https://example.com:80
You've most likely run into what I call the Linefeed-Limbo.
If I copy the code you provided from StackOverflow and run it on my machine (bash version 4.4.19(1)), then it outputs correctly
user#host:~$ cat script.sh
host="https://example.com"
port="80"
url="${host}:${port}"
echo $url
user#host:~$ bash script.sh
https://example.com:80
What is Linefeed-Limbo?
Different operating systems use different ASCII symbols to represent when a new line occurs in a text, such as a script. This Wikipedia article gives a good introduction.
As you can see, Unix and Unix-like systems use the single character \n, also called a "Line Feed". Windows, as well as other systems, use \r\n, so a "carriage return" followed by a "line feed".
What happens now is when you write a script on Windows on an editor such as notepad, what you write is host="example.com"\r\n. When you copy this file into Linux, Linux interprets the \r as if it were part of the script, since only \n is considered a new line. And indeed, when I change my newline style to DOS-style, I get the exact output you get.
How can I fix this?
You have several options to fix this issue.
Converting the script (with dos2unix)
Since all you need to do is replacing every instance of \r\n with \n, you could use any text-editing software you want. However, if you like simple solutions, then dos2unix (and its sister unix2dos) might be what you looking for:
user#host:~$ dos2unix script.sh
dos2unix: converting file script.sh to Unix format...
That's it. Run your file now and you will see it behaves well.
Encoding the source-file correctly
By using a more advanced text editor such as Notepad++, you can define which style of newline you would like to use.
By changing the newline-type to whichever system you intend to run your script on, you will not run into any problems like this anymore.
Bonus round: Why does it output :80ps://example.com?
To understand why your output is like this, you have to look at what your script is doing, and what \r means.
Try thinking of your terminal as an old-fashioned typewriter. Returning the carriage means you start writing on the left again. Making a "new line" means sliding the paper. These two things are seperate, and I think that's why some systems decided to use these two characters as a logical "new line".
But I digress. Let's look at the first line, host="https://example.com"\r.
What this means when printed is "Print https://example.com, then put the carriage back at the start". When you then print :80\r, it doesn't start after ".com", it starts at the beginning of the line, because that's where you (unknowingly) told the cursor to go. it then overwites the first few characters, resulting in ":80ps://example.com" to be written. Keep in mind that after 80, you again placed a carriage return symbol, so any new text you would have written ends up overwriting the beginning again.
It works for me, try to remove carriage returns in variables and then try.
new_host=$(echo "$host" | tr -d '\r')
new_port=$(echo "$port" | tr -d '\r')
new_url="${new_host}:${new_port}"
Examples:
Create an ISO image and burn it directly to a CD.
mkisofs -V Photos -r /home/vivek/photos | cdrecord -v dev=/dev/dvdrw -
Change to the previous directory.
cd -
Listen on port 12345 and untar data sent to it.
nc -l -p 12345 | tar xvzf -
What is the purpose of the dash and how do I use it?
If you mean the naked - at the end of the tar command, that's common on many commands that want to use a file.
It allows you to specify standard input or output rather than an actual file name.
That's the case for your first and third example. For example, the cdrecord command is taking standard input (the ISO image stream produced by mkisofs) and writing it directly to /dev/dvdrw.
With the cd command, every time you change directory, it stores the directory you came from. If you do cd with the special - "directory name", it uses that remembered directory instead of a real one. You can easily switch between two directories quite quickly by using that.
Other commands may treat - as a different special value.
It's not magic. Some commands interpret - as the user wanting to read from stdin or write to stdout; there is nothing special about it to the shell.
- means exactly what each command wants it to mean. There are several common conventions, and you've seen examples of most of them in other answers, but none of them are 100% universal.
There is nothing magic about the - character as far as the shell is concerned (except that the shell itself, and some of its built-in commands like cd and echo, use it in conventional ways). Some characters, like \, ', and ", are "magical", having special meanings wherever they appear. These are "shell metacharacters". - is not like that.
To see how a given command uses -, read the documentation for that command.
It means to use the program's standard input stream.
In the case of cd, it means something different: change to the prior working directory.
The magic is in the convention. For millennia, people have used '-' to distinguish options from arguments, and have used '-' in a filename to mean either stdin or stdout, as appropriate. Do not underestimate the power of convention!
I have an output file (namely a log from screen) containing several control characters. Inside the screen, I have programs running that use control characters to refresh certain lines (examples would be top or anything printing progress bars).
I would like to output a tail of this file using PHP. If I simply read in that file and echo its contents (either using PHP functions or through calling tail, the output is messy and much more than these last lines as it also includes things that have been overwritten. If I instead run tail in the command line, it returns just what I want because the terminal evaluates the control characters.
So my question is: Is there a way to evaluate the control characters, getting the output that a terminal would show me, in a way that I could then use elsewhere (e.g., write to a file)?
#5gon12eder's answer got rid of some control characters (thanks for that!) but it did not handle the carriage return part that was even more important to me.
I figured out that I could just delete anything from the beginning of a line to the last carriage return inside that line and simply keep everything after that, so here is my sed command accomplishing that:
sed 's/^.*\r\([^\r]\+\)\r\?$/\1\r/g'
The output can then be further cleaned using #5gon12eder's answer:
cat screenlog.0 | sed 's/^.*\r\([^\r]\+\)\r\?$/\1\r/g' | sed 's,\x1B\[[0-9?;]*[a-zA-Z],,g'
Combined, this looks exactly like I wanted.
I'm not sure what you mean by “evaluating” the control characters but you could remove them easily.
Here is an example using sed but if you are already using PHP, its internal regex processing functionality seems more appropriate. The command
$ sed 's,\x1B\[[0-9?;]*[a-zA-Z],,g' file.dat
will dump the contents of file.dat to standard output with all ANSI escape sequences removed. (And I'm pretty sure that nothing else is removed except if your file contains invalid escape sequences in which case the operation is ill-defined anyway.)
Here is a little demo:
$ echo -e "This is\033[31m a \033[umessy \033[46mstring.\033[0m" > file.dat
$ cat file.dat
# The output of the above command is not shown to protect small children
# that might be browsing this site.
$ reset # your terminal
$ sed 's,\x1B\[[0-9?;]*[a-zA-Z],,g' file.dat
This is a messy string.
The less program has some more advanced logic built in to selectively replace some escape sequences. Read the man page for the relevant options.
I'm trying to write a quick batch file. It will take the result of a command, put some extra text and quotes around it, and put that into a new file.The problem is that the result of the command I'm running includes a new line. Here's the command:
p4 changelists -m 1 -t //depot/...> %FILENAME%
The output of that p4 command has a newline at the end of it. The file I'm putting it into needs to have quotes surrounding the output of that command, but the fact that the command contains a newline in it means that the "closing quote" appears on a new line in the file, which doesn't work for what I'm doing.
I've tried writing the output of that command into a file and reading it back in, and also trying to run FINDSTR on a file containing the output, but I always seem to get back the stupid trailing whitespace. I've even tried inserting backspaces into the file, but that just put a backspace character into the file instead of actually executing a backspace...
Is there anything to be done about this?
I'm no perl wizard, but the following seems to work:
p4 changelists -m 1 -t //depot/...| perl -p -e "s/^/\042/;s/$/\042/"
Check out Strawberry Perl, which provides a Windows version of Perl.
I'm always looking at my Unix tools when solving problems like this, even under Windows. sed and gawk will also get you there, check out msysgit for a nice bundle of Unix tools that will run on Windows.
I'm combining sox and lame to generate a new music file, but in order to do everything on one line using pipes, it seems necessary to 'mark' the output and input boundaries with a - character. I've inherited this code, so let me show.
sox $DIRNAME/$BASENAME -e signed-integer -r 8000 -c 2 -t wav - trim $POSITIONS | lame -v -V4 --resample 8 - $DIRNAME/${NOEXT}.mp3
The - between wav and trim is the output file, and the - between --resample 8 and $DIRNAME/${NOEXT}.mp3 is the input file.
I'm trying to find further information on this, like whether any character can be used, or if - is special in this way. What is this called, and what makes it work?
Many Unix command-line utilities use "-" as a shorthand to mean "don't use a real file here, use stdin (or stdout) instead". Sox is one of these programs:
This is from the sox manpage
SoX can be used in simple pipeline operations by using the special
filename '-' which, if used in place of an input filename, will cause
SoX will read audio data from 'standard input' (stdin), and which, if
used in place of the output filename, will cause SoX will send audio
data to 'standard output' (stdout). Note that when using this option,
the file-type (see -t below) must also be given.
By convention, unix and friends use - to represent stdin and stdout.
It's not 100% universal, but it's a pretty widely used.
In your example, it's the same thing as
/dev/stdin
Try to replace your - with it, you will see.
"-" is often use as a convention instead of a file name to say "use the standard input (instead or reading for the file) or standard output (instead of writing to the file)". This is not a feature of the command shell (i.e. bash), so in that sense, it is not a special character. It is a feature of some commands (like in your case "sox" and "lame") and is very useful to chain these commands through pipes.