Problems calling a SASS built-in function [duplicate]

问题

This question already has an answer here:

SASS: randomly pick background-image from a list (1 answer)

Closed 4 years ago.

I have this fragment of SASS code:

@font-face {
  $rndnum: random(1000);
  font-family: myfont;
  src:url('fonts/myfont#{$rndnum}.svg')
  font-weight: bold;
  font-style: normal;
}

I need to put a random number after the name of the font, I know it seems stupid operation, but I need it. However the problem is that I get the following result:

@font-face {
  font-family: myfont;
  src:url('fonts/myfontrandom(1000).svg') /* <== HERE */
  font-weight: bold;
  font-style: normal;
}

The function does not get called for some reason... Where am I doing wrong?

Something more...

I tried calling another function like percentage:

@font-face {
  $rndnum: percentage(0.2);
  ...

And it worked in that case... looks like it is not recognizing random as a valid function...

回答1:

Have you tried just concatenating the string?

src:url('fonts/myfont'+$rndnum+'.svg');

And do you have the code for what $rndnum is being set as since this could also be a problem with how that variable is set.

Edit 1:

Your code works for me. Do you just need the semi-colon after the src line?

What version of sass are you using?

I tried it here and it seemed to work perfectly fine on Sass version 3.3.3 but not on version 3.2.15 (the two versions offered on that site).

来源：https://stackoverflow.com/questions/22538430/problems-calling-a-sass-built-in-function