Regular expression for parsing string inside ""_问答_开发者

<A  "SystemTemperatureOutOfSpec" >

What should be the regular expression for parsing the string inside "". In t开发者_JAVA百科he above sample it is 'SystemTemperatureOutOfSpec'

In JavaScript, this regexp:

/"([^"]*)"/

ex.

> /"([^"]*)"/.exec('<A  "SystemTemperatureOutOfSpec" >')[1]
  "SystemTemperatureOutOfSpec"

Similar patterns should work in a bunch of other programming languages.

try this

string Exp = "\"!\"";

I am not sure I understand your question well but if you need to match everything between double quotes, here it is: /(?<=").*?(?=")/s

(?<=<A\s")(?<content>.*)(?="\s>)

Regular expressions don't get much easier than this, so you should be able to solve it by yourself. Here's how you go about doing that:

The first step is to try to define as precisely as possible what you want to find. Let's start with this: you want to find a quote, followed by some number of characters other than a quote, followed by a quote. Is that correct? If so, our pattern has three parts: "a quote", "some characters other than a quote", and "a quote".

Now all we need to do is figure out what the regular expressions for those patterns are.

A quote

For "a quote", the pattern is literally ". Regular expressions have special characters which you have to be aware of (*, ., etc). Anything that's not a special character matches itself, and " is one of those characters. For a complete list of special characters for your language, see the documentation.

Characters other than a quote

So now the question is, how do we match "characters other than a quote"? That sounds like a range. A range is square brackets with a list of allowable characters. If the list begins with ^ it means it is a list of not-allowed characters. We want any characters other than a quote, so that means [^"].

"Some"

That range just means any one of the characters in the range, but we want "some". "Some" usually means either zero-or-more, or one-or-more. You can place * after a part of an expression to mean zero-or-more of that part. Likewise, use + to mean one-or-more (and ? means zero-or-one). There are a few other variations, but that's enough for this problem.

So, "some characters other than a quote" is the range [^"] (any character other than a quote) followed by * (zero-or-more). Thus, [^"]*

Putting it all together

This is the easy part: just combine all the pieces. A quote, followed by some characters other than a quote, followed by a quote, is "[^"]*".

Capturing the interesting part

The pattern we have will now match your string. What you want, however, is just the part inside the quotes. For that you need a "capturing group", which is denoted by parenthesis. To capture a part of a regular expression, put it in parenthesis. So, if we want to capture everything but the beginning and ending quote, the pattern becomes "([^"]*)".

And that's how you learn regular expressions. Break your problem down into a precise statement composed of short sequences of characters, figure out the regular expression for each sequence, then put it all together.

The pattern in this answer may not actually be the perfect answer for you. There are some edge cases to worry about. For example, you may only want to match a quote following a non-word character, or only quotes at the beginning or end of a word. That's all possible, but is highly dependent on your exact problem. Figuring out how to do that is just as easy though -- decide what you want, then look at the documentation to see how to accomplish that.

Spend one day practicing on regular expressions and you'll never have to ask anyone for help with regular expressions for the rest of your career. They aren't hard, but they do require concentrated study.

Are you sure you need regular expression matching here? Looking at your "string" you might be better off using a Xml parser?