string 来自第一个 indexof 子字符串的 Shell 脚本子字符串
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Shell script substring from first indexof substring
提问by pathikrit
I want to accomplish the equivalent of the following pseudo-code in bash (both a and b are inputs to my script) :
我想在 bash 中完成以下伪代码的等效操作（a 和 b 都是我的脚本的输入）：
String a = "some long string";
String b = "ri";
print (a.substring(a.firstIndexOf(b), a.length()); //prints 'ring'
How can I do this in shell script?
我怎样才能在shell脚本中做到这一点？
采纳答案by codaddict
You can do:
你可以做：
$ a="some long string"
$ b="ri"
$ echo $a | grep -o "$b.*"
ring
回答by Josshad
Try:
尝试：
$ a="some long string"
$ b="ri"
$ echo ${a/*$b/$b}
ring
$ echo ${a/$b*/$b}
some long stri
回答by lessyv
grep, sedand so on can be used but it is not pure-bash.
grep,sed等等都可以使用，但它不是纯 bash。
expris a good choice but indexparameter is not, because it matches character not the whole string, try with a = "some wrong string"it matches the first r.
expr是一个不错的选择，但index参数不是，因为它匹配字符而不是整个字符串，请尝试使用a = "some wrong string"它匹配第一个r.
Instead use expr matchwith its regular expression parameter :
而是expr match与其正则表达式参数一起使用：
a="some long string";
b="ri";
echo ${a:$(expr match "$a" ".*${b}") - $(expr length "$b")}
It also works with a = "some wrong string"
它也适用于 a = "some wrong string"
回答by Fritz G. Mehner
Try this:
尝试这个：
a="some long string"
b="ri"
echo ${b}${a#*${b}}
