Question

我的印象是设置IFS会更改read在将一行文字分成字段时使用的分隔符，但显然我遗漏了一些内容：

# OK: 'read' sees 3 items separated by spaces
$ (IFS=' '; x="aa bb cc"; echo "'$x'"; read a b c <<< $x;\
  echo "'$a' '$b' '$c'")
'aa bb cc'
'aa' 'bb' 'cc'

# OK: 'read' sees a single item after IFS is changed
$ (IFS=','; x="aa bb cc"; echo "'$x'"; read a b c <<< $x;\
  echo "'$a' '$b' '$c'")
'aa bb cc'
'aa bb cc' '' ''

# Why doesn't 'read' see 3 items?
$ (IFS=','; x="dd,ee,ff"; echo "'$x'"; read a b c <<< $x;\
  echo "'$a' '$b' '$c'")
'dd,ee,ff'
'dd ee ff' '' ''

# OK: 'read' sees a single item when IFS is restored.
$ (IFS=' '; x="dd,ee,ff"; echo "'$x'"; read a b c <<< $x;\
  echo "'$a' '$b' '$c'")
'dd,ee,ff'
'dd,ee,ff' '' ''

# OK: 'read' again sees 3 items separated by spaces.
$ (IFS=' '; x="gg hh ii"; echo "'$x'"; read a b c <<< $x;\
  echo "'$a' '$b' '$c'")
'gg hh ii'
'gg' 'hh' 'ii'

为什么IFS=','不{/ 1}}将read解析为dd,ee,ff三个字段？

Answer 1

# Why doesn't 'read' see 3 items?
$ (IFS=','; x="dd,ee,ff"; echo "'$x'"; read a b c <<< $x;\
  echo "'$a' '$b' '$c'")
'dd,ee,ff'
'dd ee ff' '' ''

因为你没有引用变量。

$ ( IFS=','; x="dd,ee,ff"; echo "'$x'"; read a b c <<< "$x";\
  echo "'$a' '$b' '$c'")
'dd,ee,ff'
'dd' 'ee' 'ff'

编辑：当未引用变量时，扩展会导致Word Splitting：

shell扫描参数扩展命令的结果替换和未发生的算术扩展用于分词的双引号。

bash：`read`似乎不尊重IFS

1 个答案: