Capture variable string in a regular expression?

Question

In C#, I'm attempting to use the following capture pattern with a variable - I wonder if I'm going about this wrong. info_name is a string variable I pass to the method.

    Regex g = new Regex(@"""" + info_name + """>.+</span>");
    // capture "info">Capture pattern</span>

But it gives me an error, ')' expected about halfway through. This gives no error:

    Regex g = new Regex(@"""" + info_name +">.+</span>");
                                         //^ 1 quote, not 3

I can't use this as a solution, I need to capture the " just before the close of the tag.

But be wary of what you pass in in info_name - if it happens to contain any special Regex characters, it may mess up your Regex. — Nikki Locke
– Nikki Locke, Commented Aug 7, 2011 at 21:44
@Nikki: Good point. That can be handled using Regex.Escape if necessary. I didn't add that to my own answer because it's possible that info_name has already been pre-processed or suitably sanitised. — LukeH
– LukeH, Commented Aug 7, 2011 at 21:47

LukeH · Accepted Answer · 2011-08-07 21:39:58Z

4

You're using two string literals there, so you need to apply the @ both times:

Regex g = new Regex(@"""" + info_name + @""">.+</span>");

// or alternatively
Regex g = new Regex("\"" + info_name + "\">.+</span>");

answered Aug 7, 2011 at 21:39

LukeH

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3 Comments

N0ug4t Over a year ago

Thanks! For some reason escaping a quote " with \" doesn't work, it stops the expression. I have to use "" to escape it. Your first answer seemed to work! Thanks a bunch! Also, I read why parsing HTML with Regex is an unholy sin like the work of satan, so I might try another way to do this.

Nikki Locke Over a year ago

Or use new Regex("\"" + info_name + "\">.+</span>"). @ is often used in Regex strings, because Regexes often contain backslash characters, and you have to escape each one with another backslash in regular strings (without the @).

N0ug4t Over a year ago

Ah, that's why the \ was cancelling it out. I have a lot to learn. =\

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