Regular Expression working in regex tester, but not in c#

Question

I've searched around a bit, but it's hard to tell if this exact question has been answered before. I know you'll let me know if this is a duplicate.

I have a regular expression that matches a series of one or more positive integers preceded by a backslash. For example: \12345 would match, but \1234f or 12345 would not match.

The regex I'm using is ^\\(\d+)$

When I test the expression using various testers it works. For example, see: http://regex101.com/r/cY2bI1/1

However, when I implement it in the following c# code, I fail to get a match.

The implementation:

public string ParseRawUrlAsAssetNumber(string rawUrl) {
    var result = string.Empty;
    const string expression = @"^\\([0-9]+)$";
    var rx = new Regex(expression);
    var matches = rx.Matches(rawUrl);
    if (matches.Count > 0)
    {
        result = matches[0].Value;
    }
    return result;
}

The failing test (NUnit):

[Test]
public void ParseRawUrlAsAssetNumber_Normally_ParsesTheUrl() {
    var f = new Forwarder();
    var validRawUrl = @"\12345";
    var actualResult = f.ParseRawUrlAsAssetNumber(validRawUrl);
    var expectedResult = "12345";
    Assert.AreEqual(expectedResult, actualResult);
}

The test's output:

Expected string length 5 but was 6. Strings differ at index 0.
Expected: "12345"
But was:  "\\12345"
-----------^

Any ideas?

Resolution:

Thanks everyone for the input. In the end I took the following route based on your recommendations, and it is passing tests now.

public string ParseRawUrlAsAssetNumber(string rawUrl)
{
    var result = string.Empty;
    const string expression = @"^\\([0-9]+)$";
    var rx = new Regex(expression);
    var matches = rx.Matches(rawUrl);
    if (matches.Count > 0)
    {
        result = matches[0].Groups[1].Value;
    }
    return result;
}

p.s.w.g · Accepted Answer · 2014-07-10 22:13:35Z

7

The problem is this line:

var rx = new Regex(Regex.Escape(expression));

By escaping your expression you're turning all your special regex characters into literals. Calling Regex.Escape(@"^\$\d+)$") will return "\^\\\\\(\\d\+$\$", which will only match the literal string "^\\(\d+)$"

Try just this:

var rx = new Regex(expression);

See MSDN: Regex.Escape for a full explanation and examples of how this method is intended to be used.

Given your updated question, it seems like you also have an issue here:

result = matches[0].Value;

This will return the entire matched substring, not just the first capture group. For that you'll have to use:

result = matches[0].Groups[1].Value;

edited Jul 10, 2014 at 22:13

answered Jul 10, 2014 at 22:01

p.s.w.g

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6 Comments

Joe Over a year ago

Oops, that was an old version of the code. I'll make an edit and show the current version (the problem still exists).

p.s.w.g Over a year ago

See my updated answer. Also, while it's fine to edit the question to provide more information, please try not to edit it in a way that largely invalidates the answers. If you must, post the updated code under the original code and clearly mark it as an edit.

Joe Over a year ago

Ah, very good. I wasn't aware that matches[0] would return the full substring. Thanks.

Joe Over a year ago

Good point about editing the post in that way. I'll avoid that in the future.

p.s.w.g Over a year ago

@SergeyBerezovskiy Yes, I noticed the error after posting my initial update. Please check again.

|

Sergey Berezovskiy · Accepted Answer · 2014-07-10 22:10:40Z

3

Don't escape pattern. Also simply use Regex.Match thus you are going to have single match here. Use Match.Success to check if input matched your pattern. And return group value - digits are in group of your matched expression:

public string ParseRawUrlAsAssetNumber(string rawUrl)
{            
    const string pattern = @"^\\(\d+)$";

    var match = Regex.Match(rawUrl, pattern);
    if (!match.Success)
        return String.Empty;

    return match.Groups[1].Value;
}

edited Jul 10, 2014 at 22:10

answered Jul 10, 2014 at 22:04

Sergey Berezovskiy

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2 Comments

p.s.w.g Over a year ago

+1 For the clean re-write, but the if is not necessary either. Even if it fails to match .Groups[1].Value will be an empty string.

Sergey Berezovskiy Over a year ago

@p.s.w.g thanks! Didn't know about this trick with empty group returned by default. Very cool. Whole method can be written as return Regex.Match(rawUrl, @"^\\(\d+)$").Groups[1].Value

Kyle Gobel · Accepted Answer · 2014-07-10 22:05:17Z

1

What if you try get the group result instead?

match.Groups[1].Value

When I get to a real computer I'll test, but seems like it should work

answered Jul 10, 2014 at 22:05

Kyle Gobel

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1 Comment

sarh Over a year ago

Yes, it should be the solution stackoverflow.com/questions/6375873/… match.Groups[0] is always the same as match.Value, which is the entire match. match.Groups[1] is the first capturing group in your regular expression.

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