My tag struct looks like this:
<sml8/>
combination of < , sml , digits (one or two) and />
Is there anyway to capture number inside tag?
for example in above I want capture 8 inside
I've defined regular expression and I tried to capture it by digit position but it's not working for me.
QRegExp rxlen("<sml(.*)/>");
int index = rxlen.pos(3);
I guess it's not correct way and it gives me position of digit although I want value of digit (or digits).
You need to use capturedTexts() together with <sml(\\d{1,2})/> regex (it matches <sml literally, then 1 or 2 digits capturing them into Captured group 1, then />:
QString str = "<sml8/>";
QRegExp rxlen("<sml(\\d{1,2})/>");
int pos = rxlen.indexIn(str);
QStringList list = rxlen.capturedTexts();
QString my_number = list[1];
Related
I have a QString like this:
QString fileData = "SOFT_PACKAGES.ABC=MY_DISPLAY_OS:MY-Display-OS.2022-3.10.25.10086-1.myApplication"
What I need to do is to create substrings as follow:
SoftwareName = MY_DISPLAY_OS //text after ':'
Version = 10.25.10086-1
Release = 2022-3
I tried using QString QString::sliced(qsizetype pos, qsizetype n) const but didn't worked as I'm using 5.9 and this is supported on 6.0.
QString fileData = "SOFT_PACKAGES.ABC=MY_DISPLAY_OS:MY-Display-OS.2022-3.10.25.10086-1.myApplication";
QString SoftwareName = fileData.sliced(fileData.lastIndexOf(':'), fileData.indexOf('.'));
Please help me to code this in Qt.
Use QString::split 3 times:
Split by QLatin1Char('=') to two parts:
SOFT_PACKAGES.ABC
MY_DISPLAY_OS:MY-Display-OS.2022-3.10.25.10086-1.myApplication
Next, split 2nd part by QLatin1Char(':'), probably again to just 2 parts if there can never be more than 2 parts, so the 2nd part can contain colons:
MY_DISPLAY_OS
MY-Display-OS.2022-3.10.25.10086-1.myApplication
Finally, split 2nd part of previous step by QLatin1Char('.'):
MY-Display-OS
2022-3
10
25
10086-1
myApplication
Now just assemble your required output strings from these parts. If exact number of parts is unknown, you can get Version = 10.25.10086-1 by removing two first elements and last element from the final list above, and then joining the rest by QLatin1Char('.'). If indexes are known and fixed, you can just use QStringLiteral("%1.%2.%3").arg(....
One way is using
QString::mid(int startIndex, int howManyChar);
so you probably want something like this:
QString fileData = "SOFT_PACKAGES.ABC=MY_DISPLAY_OS:MY-Display-OS.2022-3.10.25.10086-1.myApplication";
QString SoftwareName = fileData.mid(fileData.indexOf('.')+1, (fileData.lastIndexOf(':') - fileData.indexOf('.')-1));
To extract the other part you requested and if the number of '.' characters remains constant along all strings you want to check you can use the second argument IndexOf to find shift the starting location to skip known many occurences of '.', so for example
int StartIndex = 0;
int firstIndex = fileData.indexOf('.');
for (int i=0; i<=6; i++) {
StartIndex += fileData.indexOf('.', firstIndex+StartIndex);
}
int EndIndex = fileData.indexOf('.', StartIndex+8);
should give the right indices to be cut out with
QString SoftwareVersion = fileData.mid(StartIndex, EndIndex - StartIndex);
If the strings to be parsed stay less consistent in this way, try switching to regular expressions, they are the more flexible approach.
In my experience, using regular expressions for these types of tasks is generally simpler and more robust. You can do this with a regular expressions with the following:
// Create the regular expression.
// Using C++ raw string literal to reduce use of escape characters.
QRegularExpression re(R"(.+=([\w_]+):[\w-]+\.(\d+-\d+)\.(\d+\.\d+\.\d+-?\d+))");
// Match against your string
auto match = re.match("SOFT_PACKAGES.ABC=MY_DISPLAY_OS:MY-Display-OS.2022-3.10.25.10086-1.myApplication");
// Now extract the portions you are interested in
// match.captured(0) is always the full string that matched the entire expression
const auto softwareName = match.captured(1);
const auto version = match.captured(3);
const auto release = match.captured(2);
Of course for this to make sense, you have to understand regex, so here is my explanation of the regex used here:
.+=([\w_]+):[\w-]+\.(\d+-\d+)\.(\d+\.\d+\.\d+-?\d+)
.+=
get all characters up to and including the first equals sign
([\w_]+)
capture one or more word characters (alphanumeric characters) or underscores
:
a colon
[\w-]+\.
one or more alphanumeric or dash characters followed by a single period
(\d+-\d+)
capture one or more of digits followed by a dash followed by one or more digits
\.
a single period
(\d+\.\d+\.\d+-?\d*)
capture three sets of digits with periods in between, then an optional dash, and any number of digits (could be zero digits)
I think it is generally easier to make a regex that handles changes to the input - lets say version becomes 10.25.10087 - more easily than manually parsing things by index.
Regex is a powerful tool once you get used to it, but it can certainly seem daunting at first.
Example of this regex on regex101.com: https://regex101.com/r/dj3Z4U/1
I tried the following code. However, the result is not what I want.
$strLine = "100.11 Q9"
$sortString = StringRegExp ($strLine,'([0-9\.]{1,7})', $STR_REGEXPARRAYMATCH)
MsgBox(0, "", $sortString[0],2)
The output shows 100.11, but I want 100.11 9. How could I display it this way using a regular expression?
$sPattern = "([0-9\.]+)\sQ(\d+)"
$strLine = "100.11 Q9"
$sortString = StringRegExpReplace($strLine, $sPattern, '\1 \2')
MsgBox(0, "$sortString", $sortString, 2)
$strLine = "100.11 Q9"
$sortString = StringRegExp($strLine, $sPattern, 3); array of global matches.
For $i1 = 0 To UBound($sortString) -1
MsgBox(0, "$sortString[" & $i1 & "]", $sortString[$i1], 2)
Next
The pattern is to get the 2 groups being 100.11 and 9.
The pattern will 1st match the group with any digit and dot until it reach
/s which will match the space. It will then match the Q. The 2nd group
matches any remaining digits.
StringRegExpReplace replaces the whole string with 1st and 2nd groups
separated with a space.
StringRegExp get the 2 groups as 2 array elements.
Choose 1 from the 2 types regexp above of which you prefer.
Context. I'm developing a Lexer/Tokenizing engine, which would use regex as a backend. The lexer accepts rules, which define the token types/IDs, e.g.
<identifier> = "\\b\\w+\\b".
As I envision, to do the regex match-based tokenizing, all of the rules defined by regexes are enclosed in capturing groups, and all groups are separated by ORs.
When the matching is being executed, every match we produce must have an index of the capturing group it was matched to. We use these IDs to map the matches to token types.
So the problem of this question arises - how to get the ID of the group?
Similar question here, but it does not provide the solution to my specific problem.
Exactly my problem here, but it's in JS, and I need a C/C++ solution.
So let's say I've got a regex, made up of capturing groups separated by an OR:
(\\b[a-zA-Z]+\\b)|(\\b\\d+\\b)
which matches the the whole numbers or alpha-words.
My problem requires that the index of the capture group the regex submatch matched to could be known, e.g. when matching the string
foo bar 123
3 iterations will be done. The group indexes of the matches of every iteration would be 0 0 1, because the first two matches matched the first capturing group, and the last match matched the second capturing group.
I know that in standard std::regex library it's not entirely possible (regex_token_iterator is not a solution, because I don't need to skip any matches).
I don't have much knowledge about boost::regex or PCRE regex library.
What is the best way to accomplish this task? Which is the library and method to use?
You may use the sregex_iterator to get all matches, and once there is a match you may analyze the std::match_results structure and only grab the ID-1 value of the group that participated in the match (note only one group here will match, either the first one, or the second), which can be conveniently checked with the m[index].matched:
std::regex r(R"((\b[[:alpha:]]+\b)|(\b\d+\b))");
std::string s = "foo bar 123";
for(std::sregex_iterator i = std::sregex_iterator(s.begin(), s.end(), r);
i != std::sregex_iterator();
++i)
{
std::smatch m = *i;
std::cout << "Match value: " << m.str() << " at Position " << m.position() << '\n';
for(auto index = 1; index < m.size(); ++index ){
if (m[index].matched) {
std::cout << "Capture group ID: " << index-1 << std::endl;
break;
}
}
}
See the C++ demo. Output:
Match value: foo at Position 0
Capture group ID: 0
Match value: bar at Position 4
Capture group ID: 0
Match value: 123 at Position 8
Capture group ID: 1
Note that R"(...)" is a raw string literal, no need to double backslashes inside it.
Also, index is set to 1 at the start of the for loop because the 0th group is the whole match, but you want group IDs to be zero-based, that is why 1 is subtracted later.
I want to accomplish the following requirements using Regex only (no C# code can be used )
• BTN length is 12 and BTN starts with 0[123456789] then it should remove one digit from left and one digit from right.
WORKING CORRECTLY
• BTN length is 12 and it’s not the case stated above then it should always return 10 right digits by removing 2 from the start. (e.g. 491234567891 should be changed to 1234567891)
NOT WORKING CORRECTLY
• BTN length is 11 and it should remove one digit from left. WORKING CORRECTLY
for length <=10 BTNs , nothing is required to be done , they would remain as it is or Regex may get failed too on them , thats acceptable .
USING SQL this can be achieved like this
case when len(BTN) = 12 and BTN like '0[123456789]%' then SUBSTRING(BTN,2,10) else RIGHT(BTN,10) end
but how to do this using Regex .
So far I have used and able to get some result correct using this regex
[0*|\d\d]*(.{10}) but by this regex I am not able to correctly remove 1st and last character of a BTN like this 015732888810 to 1573288881 as this regex returns me this 5732888810 which is wrong
code is
string s = "111112573288881,0573288881000,057328888105,005732888810,15732888815,344956345335,004171511326,01777203102,1772576210,015732888810,494956345335";
string[] arr = s.Split(',');
foreach (string ss in arr)
{
// Match mm = Regex.Match(ss, #"\b(?:00(\d{10})|0(\d{10})\d?|(\d{10}))\b");
// Match mm = Regex.Match(ss, "0*(.{10})");
// ([0*|\\d\\d]*(.{10}))|
Match mm = Regex.Match(ss, "[0*|\\d\\d]*(.{10})");
// Match mm = Regex.Match(ss, "(?(^\\d{12}$)(.^{12}$)|(.^{10}$))");
// Match mm = Regex.Match(ss, "(info)[0*|\\d\\d]*(.{10}) (?(1)[0*|\\d\\d]*(.{10})|[0*|\\d\\d]*(.{10}))");
string m = mm.Groups[1].Value;
Console.WriteLine("Original BTN :"+ ss + "\t\tModified::" + m);
}
This should work:
(0(\d{10})0|\d\d(\d{10}))
UPDATE:
(0(\d{10})0|\d{1,2}(\d{10}))
1st alternate will match 12-digits with 0 on left and 0 on right and give you only 10 in between.
2nd alternate will match 11 or 12 digits and give you the right 10.
EDIT:
The regex matches the spec, but your code doesn't read the results correctly. Try this:
Match mm = Regex.Match(ss, "(0(\\d{10})0|\\d{1,2}(\\d{10}))");
string m = mm.Groups[2].Value;
if (string.IsNullOrEmpty(m))
m = mm.Groups[3].Value;
Groups are as follows:
index 0: returns full string
index 1: returns everything inside the outer closure
index 2: returns only what matches in the closure inside the first alternate
index 3: returns only what matches in the closure inside the second alternate
NOTE: This does not deal with anything greater than 12 digits or less than 11. Those entries will either fail or return 10 digits from somewhere. If you want results for those use this:
"(0(\\d{10})0|\\d*(\\d{10}))"
You'll get rightmost 10 digits for more than 12 digits, 10 digits for 10 digits, nothing for less than 10 digits.
EDIT:
This one should cover your additional requirements from the comments:
"^(?:0|\\d*)(\\d{10})0?$"
The (?:) makes a grouping excluded from the Groups returned.
EDIT:
This one might work:
"^(?:0?|\\d*)(\\d{10})\\d?$"
(?(^\d{12}$)(?(^0[1-9])0?(?<digit>.{10})|\d*(?<digit>.{10}))|\d*(?<digit>.{10}))
which does the exact same thing as sql query + giving result in Group[1] all the time so i didn't had to change the code a bit :)
Can regex extract the values embedded within a string, as identified by a variable template defined earlier within the same string? Or is this better handled in Java?
For example: "2012 Ferrari [F12] - Ostrich Leather interior [F12#OL] - Candy Red Metallic [F12#3]" The variable template is the first string encountered with square brackets, e.g. [F12], and the desired variables are found within subsequent instances of that template, e.g. 'OL' and '3'.
Since you are mentioning Java, I'll assume you are using the Java implementation, Pattern.
Java's Pattern supports so called back references, which can be used to match the same value a previous capturing group matched.
Unfortunately you cannot extract multiple values from a single capturing group, so you'll have to hardcode the number of templates you want to match, if you want to do this with a single pattern.
For one variable, it could look like this:
\[(.*?)\].*?\[\1#(.*?)\]
^^^^^ ^^^^^ variable
template ^^ back reference to whatever template matched
You can add more optional matches by wrapping them in optional non-capturing groups like this:
\[(.*?)\].*?\[\1#(.*?)\](?:.*?\[\1#(.*?)\])?(?:.*?\[\1#(.*?)\])?
^ optional group ^ another one
This would match up to three variables:
String s = "2012 Ferrari [F12] - Ostrich Leather interior [F12#OL] - Candy Red Metallic [F12#3]";
String pattern = "\\[(.*?)\\].*?\\[\\1#(.*?)\\](?:.*?\\[\\1#(.*?)\\])?(?:.*?\\[\\1#(.*?)\\])?";
Matcher matcher = Pattern.compile(pattern).matcher(s);
if (matcher.find()) {
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println(matcher.group(i));
}
}
// prints F12, OL, 3, null
If you'll need to match any number of variables, however, you'll have to resort to extracting the template in a first pass and then embedding it in a second pattern:
// compile once and store in a static variable
Pattern templatePattern = Pattern.compile("\\[(.*?)\\]");
String s = "2012 Ferrari [F12] - Ostrich Leather interior [F12#OL] - Candy Red Metallic [F12#3]";
Matcher templateMatcher = templatePattern.matcher(s);
if (!templateMatcher.find()) {
return;
}
String template = templateMatcher.group(1);
Pattern variablePattern = Pattern.compile("\\[" + Pattern.quote(template) + "#(.*?)\\]");
Matcher variableMatcher = variablePattern.matcher(s);
while (variableMatcher.find()) {
System.out.println(variableMatcher.group(1));
}