如何让一个字符串减除掉字符串的最后2个字符？

reyleon

1. echo aaabb | awk 'gsub(/..$/,"")'
   

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bikong0411

echo ${VAR//b/}

li0924

1. echo aaabb | awk -F"" '{NF-=2}1'

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Buring__

echo 'aaabb'|sed 's/..$//'

o枫叶o飘零

python
print VAR[:-2]

stupid_lee

1. echo ${a:0:$((${#a}-2))}

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rulebook

回复 2# Looiml


    学习了！

qz357

膜拜2楼，还有用awk的，学习了。
echo $var | sed 's/\(...\).*/\1/g'

聆雨淋夜

回复 8# Looiml
能否解释下这里的"?"是什么用法，好像不是正则也不是通配符


   

Looiml

回复 19# 聆雨淋夜

${parameter%word}
       ${parameter%%word}
              Remove matching suffix pattern.  The word is expanded to produce
              a pattern just as in pathname expansion.  If the pattern matches
              a  trailing portion of the expanded value of parameter, then the
              result of the expansion is the expanded value of parameter  with
              the  shortest  matching  pattern (the ``%'' case) or the longest
              matching pattern (the ``%%'' case) deleted.  If parameter  is  @
              or  *,  the  pattern  removal operation is applied to each posi‐
              tional parameter in turn, and the  expansion  is  the  resultant
              list.   If  parameter is an array variable subscripted with @ or
              *, the pattern removal operation is applied to  each  member  of
              the array in turn, and the expansion is the resultant list.

看到了吗？和pathname expansion用法一致，活学活用而已。