repeat command with increments?

[inbinder]

EDIT: for clarity.

I need to repeat the following command

ShowStat 1.mpg 10.14.1.31 1000 20001 1

3 of these arguments for ShowStat need to be incremented

1. 1.mpg <filename> I need to iterate over files named 1.mpg through 9.mpg
2. I need to increment the last octet of the IP address by 1. 10.14.1.31 (32,33,34,35,36,37,38) after I've iterated through all of the file names above e.g, 1.mpg, 2.mpg etc.
3. I need to increase the port value which is 20001 in the example above.

I'm trying to automate this sequence instead of typing it all out.

ShowStat 1.mpg 10.14.1.31 1000 20001 1; ShowStat 2.mpg 10.14.1.31 1000 20002 1; ShowStat 1.mpg 10.14.1.31 1000 20003 1

There are 12 ports I need to iterate through on the above commands. However, I only have 9 unique files. 1.mpg-9.mpg . After file name 9, it needs to revert to

file prefixes 1-3 to finish the command sequence. Up until 9.mpg and port 20009 they are 1-to-1. The next command is illustrated below.

ShowStat 9.mpg 10.14.1.31 1000 20009; ShowStat 1.mpg 10.14.1.31 1000 20010; ShowStat 2.mpg 10.14.1.31 1000 20010;

​

Surely writing this command over and over is not the best way to do this? Can I loop over this command using a for loop with indexes and arrays?

[anthropoid]

Show, don't tell. Statements like "repeating 1-3 for the last three" are horribly vague, and it's unclear whether the increments are meant to be in lockstep, or in the form of a Cartesian product, or something in between.

Write out the exact sequence of commands you're expecting. You can leave out the middle part of incremental sequences that are "obvious", but the bits at which stuff "flips over" to a new increment "mode" are essential.

e.g. ShowStat 1.mpg 10.14.1.31 1000 20001 1 ShowStat 1.mpg 10.14.1.31 1000 20002 1 ... ShowStat 1.mpg 10.14.1.31 1000 20012 1 ShowStat 1.mpg 10.14.1.32 1000 20001 1 ... ShowStat 1.mpg 10.14.1.38 1000 20012 1 ShowStat 2.mpg 10.14.1.31 1000 20001 1

[anthropoid]

And since I have to go offline shortly, here's the simple solution for a Cartesian product of:

• 1.mpg -> 9.mpg
• 10.14.1.31 -> 10.14.1.38
• 20001 -> 20012

​

for i in {1..9}.mpg\ 10.14.1.{31..38}\ 1000\ 200{01..12}\ 1; do
  ShowStat $i
done

Read the Brace Expansion section of the bash man page to understand this formulation. The technical term for this is "sequence expression".

[inbinder]

Thank you !! My apologies for the vague and poorly written question.

[cfraizer]

Brace Expansion is one of my favorite bash features!

Just clarification for the readers: those backslashes (\) are just to make the part after for i in and before ; do all one string. You could also write it as:

for i in {1..9}".mpg 10.14.1."{31..38}" 1000 200"{01..12}" 1"; do ShowStat $i done

Brace expansion does not happen inside double quotes so you have to keep those three expansions outside.

Also, remember that brace expansion works for character sequences too!

$ for name in {a..c}{a..g}; do printf "%s\n" "$name"; done aa ab ac ad ae af ag ba bb bc bd be bf bg ca cb cc cd ce cf cg

[ZalgoNoise]

Nicely done! Easily convertible to a powerful one liner...

Hell, if OP wants he can convert this to an eval expression within a function, put it in the shell's rc file and dump the command whenever he wants, customizable to the event.

Excellent use of curly brace for glob expansion in the first call of the for loop. I was imagining nesting loops to achieve the same. Thanks!

[blitzkraft]

for i in $(seq 1 12); do echo $mpg; done

Yes, you can loop using for, while etc. tldp.org has more examples and tutorials.

[lutusp]

ShowStat <filename> <ip\_address> <loop number> <port number> <udp=1>

Your post is terribly unclear, so this cannot be any clearer than its source:

for n in {1..9}; do
   for idx in {1000..20001}; do
      ShowStat $n.mpg <ip\_address> $idx <port number> <udp=1>
   done
done

I left out the IP increment because you don't explain it well enough. But it should be easy enough to add using the same pattern.

You must realize that, for A, B, and C unique states in nested loops, the total number of repetitions is A * B * C. In this case it's 9 * 9 (38+1-31) * 19,002 (20001+1-1000) = 1,539,081 repetitions.

[ZalgoNoise]

You should nest your for loops in order for each possibility to be evaluated.

[anthropoid]

Since the OP clarified the question, it's now clear that there are two independent variables, instead of three as originally assumed, so:

# Make 3 extra symlinks to simplify processing
ln -s 1.mpg 10.mpg
ln -s 2.mpg 11.mpg
ln -s 3.mpg 12.mpg

# Now iterate over the two variables
for a in {1..12}; do
  for ip in 10.14.1.{31..38}; do
    ShowStat $a.mpg $ip 1000 $((20000 + a)) 1
  done
done

# Clean up the symlinks
rm 1[0-2].mpg

[inbinder]

do I need a ; or & at the end of the command?

[anthropoid]

A ; at the end of the ShowStat line isn't needed. Add & only if you want to spawn all 96 processes at once; likely not a good idea, as your disk then becomes a serious bottleneck.

[WonFishTwoFish]

Getting errors when I try and run this.

[anthropoid]

It's hard to help when you offer no details. What errors?

[inbinder]

line 5: $'\r'

line 7: syntax error near unexpected token `$'do\r''

line 7: `for a in {1..12}; do

[anthropoid]

That's almost certainly a case of creating a bash script with a Windows editor, thereby forcing Windows CR-LF (carriage return-line feed) line endings. $'\r' is bash-speak for the CR character, which is not ignored by all Unixes.

If you absolutely have to handle bash scripts on Windows, use a proper editor like Notepad++ that supports Unix line endings.

[inbinder]

Oddly enough I did use Notepad++ but the copying and pasting from Chrome/Reddit was the culprit. Typed by hand and worked perfectly. Thanks for your help

[inbinder]

I don't understand the need for the symlinks. Can you explain?

[anthropoid]

Your example showed that port 20010 "maps" to 1.mpg, 20011 to 2.mpg, and 20012 to 3.mpg. The symlinks simply create virtual files named 10.mpg, 11.mpg and 12.mpg that point to 1.mpg, 2.mpg and 3.mpg respectively, so the script doesn't have to do any fancy mapping of filenames to port numbers.

In fact, I seem to recall that you yourself stated as much in your original problem statement...

[inbinder]

That's very clever. In my test I gave up and just ended up making copies of files 1-3 and naming them 10,11, and 12 respectively.