I know that /play -r Origins.txt plays a random line from file?
But how to read a line into a variable?
I need this for making random quit messages.
var %Var = $read(file.txt)
Nimue, /var needs no = - /var %var bleh will work as /var %var = bleh does.
Most of the time it will work without the = yes. Sometimes it will not.
It is good practice to use the correct syntax ALL the time to avoid errors when the incorrect syntax will not work.
Anyway i just needed $read(file.txt). Thanks again
That rand statement is completly useless. When you don't provide the line to read from, the alias does the same exact thing. Can you say unneccisary typing?
//var %b $remove(abc,%a) | echo -s %a
Type that.
/var %var = value is the documented way, nowhere in the help file does it say that you can omit the =. There are other things that aren't mentioned in the help file and do work, but this one (omitting the =) is an abuse imo. Mainly because there are cases where it breaks.
//var %b $remove(abc,%a) | echo -s %a
this doesnt work because you echo %a - but %a doesnt exist.
type //var %b $remove(abc,b) | echo -s %b
this works
You're missing the point. If you see more carefully, the example I mentioned does not generate an "/echo: insufficient parameters", but a "$remove(): invalid parameters": the script halts before it reaches the /echo command. Just replace "echo -s %a" with "echo -s something %a" if you're not convinced yet. Also note that your example works because it's a different case. I used a variable as a second parameter to an identifier (%a and $remove()) , you used a letter as a second parameter to the same identifier. Test stuff some more, you'll definitely get a better picture of how things (don't) work.
nope, you are missing the point - the question was if the = is necessary in the /var command.
I denied that because its not. Your example has nothing to do with this topic because //var %b $remove(abc,%a) | echo -s %a of course will give an * Invalid format: $remove - this is because %a doesnt exist (so it doesnt return anything), but u use it as a substring in $remove, it's the same as if you would use $remove(abc,)
so the point is that the error is not that the = is missing in the //var command but in your $remove, if you use it properly, e.g. //var %b $remove(abc,b) it works perfect.
Please don't keep it up, you're making a fool of yourself.
You want a variable that exists? Type this:
//var %b b | var %a $remove(abc,%b) | echo -s %a
Now type this:
//var %b b | var %a = $remove(abc,%b) | echo -s %a
Now read the thread again more carefully and let's hope you get it.