ksh shell脚本查找字符串中_的第一个匹配项并删除所有内容，直到

Question

Im New To Shell Scripting.Using KSH Shell. Could you please help me in this.

My string is like errorfile101_ApplicationData_2_333.txt. I want to remove everything until the first occurence of _.

My output should be ApplicationData_2_333.txt

Answer 1

This is an easy one, assuming you can assign your string to a variable, ie

str="errorfile101_ApplicationData_2_333.txt"
echo ${str#*_}

output

ApplicationData_2_333.txt

The # operator in ${str#*_} means remove the following pattern from the left of the variable's value.

There is also ## , which removes the longest match from the left, which would give you

333.txt

There are also similar removal operators for working from the right side of the string, % and a longest match (from right) with %% .

All versions of ksh (and bash, and other shells) support these operators. (sorry if this is the wrong term).

Versions of ksh93 and greater ( bash , zsh and probably others) also support a sed like pattern match/sub value like

echo ${str/*_/xx}
#----------|--|>replacement
#----------> pattern to match

output

xx333.txt

which means that / works like sed matching the longest possible string.

IHTH

Answer 2

您可以使用cut命令：

echo "errorfile101_ApplicationData_2_333.txt" | cut -d"_" -f2-