sass - 在 if 语句中声明的变量导致 "undefined variable"

Question

我希望在 if 语句中定义变量可以在 Sass 中工作，但不幸的是我收到错误消息，说变量未定义。这是我尝试过的:

@for !i from 1 through 9
    !foo = #000
    @if !i == 1
        !bg_color = #009832
    @if !i == 2
        !bg_color = #195889
    ...

    #bar#{!i} 
        color: #{!foo}
        background-color: #{!bg_color}

使用此代码，我会收到以下错误:

Undefined variable: "!bg_color".

Answer 1

Sass 变量仅对声明它们的缩进级别以及嵌套在其下的缩进级别可见。所以你只需要在你的 for 循环之外声明 !bg_color :

!bg_color = #FFF
@for !i from 1 through 9
    !foo = #000
    @if !i == 1
        !bg_color = #009832
    @if !i == 2
        !bg_color = #195889

    #bar#{!i} 
        color: #{!foo}
        background-color: #{!bg_color}

你会得到以下 css:

#bar1 {
  color: black;
  background-color: #009832; }

#bar2 {
  color: black;
  background-color: #195889; }

#bar3 {
  color: black;
  background-color: #195889; }

#bar4 {
  color: black;
  background-color: #195889; }

#bar5 {
  color: black;
  background-color: #195889; }

#bar6 {
  color: black;
  background-color: #195889; }

#bar7 {
  color: black;
  background-color: #195889; }

#bar8 {
  color: black;
  background-color: #195889; }

#bar9 {
  color: black;
  background-color: #195889; }