Regex - If contains '%', can only contain '%20'

Question

Hi All,

I am wanting to create a regular expression for the following scenario:

If a string contains the percentage character (%) then it can only contain the following: %20, and cannot be preceded by another '%'.

So if there was for instance, %25 it would be rejected. For instance, the following string would be valid:

http://www.test.com/?&Name=My%20Name%20Is%20Vader

But these would fail:

http://www.test.com/?&Name=My%20Name%20Is%20VadersAccountant%25

%%%25

Any help would be greatly appreciated,

Kyle

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EDIT:

The scenario in a nutshell is that a link is written to an encoded state and then launched via JavaScript. No decoding works. I tried .net decoding and JS decoding, each having the same result - The results stay encoded when executed.

Answer 1

Doesn't require a %:

/^[^%]*(%20[^%]*)*$/

Answer 2

I think that would find what you need

/^([^%]|%%|%20)+$/

Edit: Added case where %% is valid string inside URI
Edit2: And fixed it for case where it should fail :-)
Edit3:

In case you need to use it in editor (which would explain why you can't use more programmatic way), then you have to correctly escape all special characters, for example in Vim that regex should lool:

/^\([^%]\|%%\|%20\)\+$/

Answer 3

Another solution if look-arounds are not available:

^([^%]|%([013-9a-fA-F][0-9a-fA-F]|2[1-9a-fA-F]))*$

Answer 4

Maybe a better approach is to deal with that validation after you decode that string:

string name = HttpUtility.UrlDecode(Request.QueryString["Name"]);

Answer 5

Which language are you using?

Most languages have a Uri Encoder / Decoder function or class. I would suggest you decode the string first and than check for valid (or invalid) characters.

i.e. something like /[\w ]/ (empty is a space)

With a regex in the first place you need to respect that www.example.com/index.html?user=admin&pass=%%250 means that the pass really is "%250".

Answer 6

I agree with dominic's comment on the question. Don't use Regex.

If you want to avoid scanning the string twice, you can just iteratively search for % and then check that it is being followed by 20 and nothing else. (Update: allow a % after to be interpreted as a literal %nnn sequence)

// pseudo code
pos = 0
while (pos = mystring.find(pos, '%'))
{
     if mystring[pos+1] = "%" then
         pos = pos + 2 // ok, this is a literal, skip ahead
     else if mystring.substring(pos,2) != "20" 
          return false; // string is invalid
     end if
}
return true;

Answer 7

/^([^%]|%20)*$/

Answer 8

Reject the string if it matches %[^2][^0]

Answer 9

This requires a test against the "bad" patterns. If we're allowing %20 - we don't need to make sure it exists.

As others have said before, %% is valid too... and %%25would be %25

The below regex matches anything that doesn't fit into the above rules

/(?<![^%]%)%(?!(20|%))/

The first brackets check whether there is a % before the character (meaning that it's %%) and also checks that it's not %%%. it then checks for a %, and checks whether the item after doesn't match 20

This means that if anything is identified by the regex, then you should probably reject it.