Using Regex to split string with quoted substrings

Question

I am trying to match the below text using Regex,

"1bbl" 2bbl "is as" 0.22 "3"

When I match I am not getting the output as expected in my list.

string val = "\"1bbl\" 2bbl \"is as\" 0.22 \"3\"";
var reg = new Regex(@"([a-z0-9\s]+)");
Match match = reg.Match(val);
List<string> list = new List<string>();
while (match.Success)
{
    list.Add(match.ToString());
    match = match.NextMatch();
}

Current Output:

"1bbl"
" 2bbl "
"is as"
" 0"
"22 "
"3"

Expected Output:

"1bbl"
"2bbl"
"is as"
"0.22"
"3"

I've edited out unrelated references to XML - feel free to revert... In either case please explain why `"is as"` should be selected as a string. — Alexei Levenkov, Mar 20 '15 at 06:11

Avinash Raj · Answer 1 · 2015-03-20T06:06:55.463

0

You need to include the pattern for matching decimal numbers also.

System.Text.RegularExpressions.Regex(@"\b(?:\d+\.\d+|[a-z\s0-9]+)\b");

\b at the start and at the end helps to avoid matching of unnecessary leading and trailing spaces.

DEMO

edited Mar 20 '15 at 06:06

answered Mar 20 '15 at 06:01

Avinash Raj

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Thanks..This almost works. But when I pass two strings like 2bbl 3bbl I don't get it right. The list should have 2 strings but 1 – Edwin Vivek N Mar 20 '15 at 06:22
see https://regex101.com/r/pT4tM5/22 it matches correctly the both. – Avinash Raj Mar 20 '15 at 06:27

score 0 · Answer 2 · answered Mar 20 '15 at 06:17

Your regex ([a-z0-9\s]+) does not capture special characters. Hence, it doesn't capture the decimal point.

You might want to add the special characters that you need to capture inside the regex. Say, you need to allow'@', '#' and '.' to be captured, then you regex will be:

System.Text.RegularExpressions.Regex(@"([a-z0-9\s@#\.]+)")

Since the point '.' has a special meaning in regex, we need to use a backlash, hence '\.' .

Jerry · Accepted Answer · 2015-03-20T06:30:49.337

0

Since it looks like you are matching a space delimited string with quotes used as text qualifiers, you might want to use a regex like the below instead:

@"""[^""]+""|[^""\s]+"

That is, try to match any text field, grab everything inside the text qualifiers (quotes) ""([^""]+)"", else, grab everything using space as delimiter ([^""\s]+).

ideone demo

Though if you don't want to get the quotes in the match, you can use named capture groups, for example:

@"""(?<1>[^""]+)""|(?<1>[^""\s]+)"

ideone demo

edited Mar 20 '15 at 06:30

answered Mar 20 '15 at 06:25

Jerry

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very precise and what I was looking for. Thanks – Edwin Vivek N Mar 20 '15 at 06:39

Using Regex to split string with quoted substrings

3 Answers3