如何在bash中读取字符串

Question

我的代码

FOO="dsad; dsadasddas; dadsad; dsdsad asdsdd"
echo "${FOO%%;*}"   
result: dsad

FOO="dsad; dsadasddas; dadsad; dsdsad asdsdd"
echo "${FOO##*;}"
result: dadsad; dsdsad asdsdd

如何打印第一个分号的所有字符？

FOO="dsad; dsadasddas; dadsad; dsdsad asdsdd"
echo "${FOO???}"
result: dsadasddas; dadsad; dsdsad asdsdd

如何在最后一个分号前打印和所有字符？

FOO="dsad; dsadasddas; dadsad; dsdsad asdsdd"
echo "${FOO???}"
result: dsadasddas; dadsad; dsdsad asdsdd

帮助任何人？谢谢;）

Answer 1

使用#和%代替##和%%：

foo='one; two; three; four'

# All characters after first semicolon
echo "${foo#*;}"  # two; three; four

# All characters before last semicolon
echo "${foo%;*}"  # one; two; three

${var##glob}和${var%%glob}会尽可能地带走（贪婪匹配）。 ${var#glob}和${var%glob}将尽可能少地带走（不情愿的匹配）。