Shell脚本将字符串从一种形式转换为另一种形式 [英] Shell script to convert a string from one form to another

问题描述

我有两个字符串

internal func add(a: Int, b: Int) -> Int
add(a:b:)

如何使第一个字符串转换为第二个字符串.我可以使用编程语言来做到这一点，但无法找到我所需的bash脚本解决方案.

How can I make the first string convert into the second string. I can do this in programming languages, but unable to figure out a solution for bash script that I need.

基本上，所需的是func(包括)func之前的所有内容，并且->(包括)->之后的所有内容都应该删除，变量的类型也应该删除.所有空格也应该消失.

Essentially whats needed is anything before (and including) func should be trimmed and anything following (and including) -> should be removed and the type of variables should be removed. All spaces should also go.

此外，我尝试使用正则表达式.我做了(?:.*func\s(.*)\s->.*)，它从第一个字符串中取出了add(a: Int, b: Int)，但是，我不确定如何进一步消除该字符串.

Also, I tried using regex. I made (?:.*func\s(.*)\s->.*) that takes out the add(a: Int, b: Int) from the 1st string, but, I am not sure, how to further eliminate this string.

我最好在使用awk或sed的bash方面需要一些帮助，但是regex对我也很​​好.

I preferably need some help in bash using awk or sed but regex would also do well for me.

推荐答案

使用SED:

string='internal func add(a: Int, b: Int) -> Int'

sed -e '
s/^.*func \+//
s/ *\->.*$//
s/:[^,)]\+/:/g
s/[, ]//g
' <<< "$string"

• s/^.*func \+//删除func，func之前的前(^)部分，并删除func之后的空格；
• s/ *\->.*$//删除空格，后跟->以及字符串末尾的任何内容；
• s/:[^,)]\+/:/g删除冒号后的类型名称；
• s/[, ]//g通过删除逗号和空格来执行最终清除.
• s/^.*func \+// removes the front (^) part before func, func, and spaces after func;
• s/ *\->.*$// removes spaces followed by -> and anything after it at the end of the string;
• s/:[^,)]\+/:/g removes the type names after colons;
• s/[, ]//g performs final cleanup by removing commas and spaces.

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