匹配和替换函数表达式 [英] Matching and replacing function expressions

问题描述

我需要做的C＃（实际上的 transpiled剃刀代码）取代的功能列表，调用与文本替换。

I need to do some very light parsing of C# (actually transpiled Razor code) to replace a list of function calls with textual replacements.

如果给包含一组 {Foo.myFunc：\def\} 应该替换这个代码：

If given a set containing {"Foo.myFunc" : "\"def\"" } it should replace this code:

var res = "abc" + Foo.myFunc(foo, Bar.otherFunc( Baz.funk()));

与此：

with this:

var res = "abc" + "def"

我不在乎如果您对本嵌套表达式

I don't care about the nested expressions.

这似乎相当琐碎，我想我应该能避免使用像这样的映射集的每一个成员建立一个整个C＃编译器：

This seems fairly trivial and I think I should be able to avoid building an entire C# parser using something like this for every member of the mapping set:

• 找到表达启动（例如 Foo.myFunc ）


• 按（） / POP（）括号在的Stack 直到计数== 0 。


• 标记这是表达停止


• 取代一切从表达式开始到停止的表达

• find expression start (e.g. Foo.myFunc)
• Push()/Pop() parentheses on a Stack until Count == 0.
• Mark this as expression stop
• replace everything from expression start until expression stop

但也许我并不需要...有一个（可能是内置在）.NET库，能为我做到这一点？计数不可能在语言的家庭，稀土是，但也许是的扩展的在C＃中的正则表达式语法，可以处理这个不知何故使用反向引用？

But maybe I don't need to ... Is there a (possibly built-in) .NET library that can do this for me? Counting is not possible in the family of languages that RE is in, but maybe the extended regex syntax in C# can handle this somehow using back references?

编辑：
至于这个答案的评论演示仅仅统计括号会没有足够的一般，因为像 trollMe（（）将甩开这些算法，只有真正的解析会话足够了，我猜（？）。

edit: As the comments to this answer demonstrates simply counting brackets will not be sufficient generally, as something like trollMe("(") will throw off those algorithms. Only true parsing would then suffice, I guess (?).

推荐答案

对于普通的字符串的诀窍是：

The trick for a normal string will be:

(?>"(\\"|[^"])*")

（大于（\\[^]）*？）

一个逐字字符串：

A verbatim string:

(?>@"(""|[^"])*")

也许这可以帮助，但我不知道，这将适用于所有情况>

Maybe this can help, but I'm not sure that this will work in all cases:

<func>(?=\()((?>/\*.*?\*/)|(?>@"(""|[^"])*")|(?>"(\\"|[^"])*")|\r?\n|[^()"]|(?<open>\()|(?<-open>\)))+?(?(open)(?!))

替换< FUNC方式> 与您的函数名

没用的，说 trollMe（\（，（（@ABC德（（F）作品如预期。

Useless to say that trollMe("\"(", "((", @"abc""de((f") works as expected.

演示

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