public list[str] deleteBlockComments(list[str] fileLines)
{
bool blockComment = false;
list[str] sourceFile = [];
for(fileLine <- fileLines)
{
fileLine = trim(fileLine);
println(fileLine);
if (/^[\t]*[\/*].*$/ := fileLine)
{
blockComment = true;
}
if (/^[\t]*[*\/].*$/ := fileLine)
{
blockComment = false;
}
println(blockComment);
if(!blockComment)
{
sourceFile = sourceFile + fileLine;
}
}
return sourceFile;
}
For some reason, I am not able to detect /* at the beginning of a string. If I execute this on the command line, it seems to work fine.
Can someone tell me what I am doing wrong? In the picture below you can see the string to be compared above the comparison result (false).
[\/*] is a character set that matches forward slash or star, not both one after the other. Simply remove the square brackets and your pattern should start behaving as you expect.
While we're at it, let's also get rid of the superfluous square brackets around \t
^\t*\/*.*$
I am making a simple dialogue system, and would like to "dynamise" some of the sentences.
For exemple, I have a Sentence
Hey Adventurer {{PlayerName}} !
Welcome in the world !
Now In code I am trying to replace that by the real value of the string in my game. I am doing something like this. But it doesn't work. I do have a string PlayerName in my component where the function is situated
Regex regex = new Regex("(?<={{)(.*?)(?=}})");
MatchCollection matches = regex.Matches(sentence);
for(int i = 0; i < matches.Count; i++)
{
Debug.Log(matches[i]);
sentence.Replace("{{"+matches[i]+"}}", this.GetType().GetField(matches[i].ToString()).GetValue(this) as string);
}
return sentence;
But this return me an error, even tho the match is correct.
Any idea of a way to do fix, or do it better?
Here's how I would solve this.
Create a dictionary with keys as the values you wish to replace and values as what you will be replacing them to.
Dictionary<string, string> valuesToReplace;
valuesToReplace = new Dictionary<string, string>();
valuesToReplace.Add("[playerName]", "Max");
valuesToReplace.Add("[day]", "Thursday");
Then check the text for the values in your dictionary.
If you make sure all of your keys start with "[" and end with "]" this will be quick and easy.
List<string> replacements = new List<string>();
//We will save all of the replacements we are about to perform here.
//This is done so we won't be modifying the original string while working on it, which will create problems.
//We will save them in the following format: originalText}newText
for(int i = 0; i < text.Length; i++) //Let's loop through the entire text
{
int startOfVar = 9999;
if(text[i] == '[') //We have found the beginning of a variable
{
startOfVar = i;
}
if(text[i] == ']') //We have found the ending of a variable
{
string replacement = text.Substring(startOfVar, i - startOfVar); //We have found the section we wish to replace
if (valuesToReplace.ContainsKey(replacement))
replacements.Add(replacement + "}" + valuesToReplace[replacement]); //Add the replacement we are about to perform to our dictionary
}
}
//Now let's perform the replacements:
foreach(string replacement in replacements)
{
text = text.Replace(replacement.Split('}')[0], replacement.Split('}')[1]); //We split our line. Remember the old value was on the left of the } and the new value was on the right
}
This will also work much faster, since it allows you to add as many variables as you wish without making the code slower.
Using Regex.Replace method, and a MatchEvaluator delegate (untested):
Dictionary<string, string> Replacements = new Dictionary<string, string>();
Regex DialogVariableRegex = new Regex("(?<={{)(.*?)(?=}})");
string Replace(string sentence) {
DialogVariableRegex.Replace(sentence, EvaluateMatch);
return sentence;
}
string EvaluateMatch(Match match) {
var matchedKey = match.Value;
if (Replacements.ContainsKey(matchedKey))
return Replacements[matchedKey];
else
return ">>MISSING KEY<<";
}
This is kind of old now, but I figured I'd update the accepted code above. It won't work since the start index is reset every time the loop iterates, so setting startOfVar = i gets completely reset by the time it hits the closing character. Plus there are problems if there's an open bracket '[' and no closing one. You can also no longer use those brackets in your text.
There's also setting the splitter to a single character. It tests fine, but if I set my player name to "Rob}ert", that will cause problems when it performs the replacements.
Here is my updated take on the code which I've tested works in Unity:
public string EvaluateVariables(string str)
{
Dictionary<string, string> varDict = GetVariableDictionary();
List<string> varReplacements = new List<string>();
string matchGuid = Guid.NewGuid().ToString();
bool matched = false;
int start = int.MaxValue;
for (int i = 0; i < str.Length; i++)
{
if (str[i] == '{')
{
if (str[i + 1] == '$')
{
start = i;
matched = true;
}
}
else if (str[i] == '}' && matched)
{
string replacement = str.Substring(start, (i - start) + 1);
if (varDict.ContainsKey(replacement))
{
varReplacements.Add(replacement + matchGuid + varDict[replacement]);
}
start = int.MaxValue;
matched = false;
}
}
foreach (string replacement in varReplacements)
{
str = str.Replace(replacement.Split(new string[] { matchGuid }, StringSplitOptions.None)[0], replacement.Split(new string[] { matchGuid }, StringSplitOptions.None)[1]);
}
return str;
}
private Dictionary<string, string> GetVariableDictionary()
{
Dictionary<string, string> varDict = new Dictionary<string, string>();
varDict.Add("{$playerName}", playerName);
varDict.Add("{$npcName}", npcName);
return varDict;
}
I can't find a way to match the text before opening curly bracket (i.e. p) using regex and Qt. My input file reads :
solvers
{
p
{
solver PCG;
preconditioner DIC;
tolerance 1e-06;
relTol 0.05;
}
q
{
solver PCG;
relTol 0.03;
}
}
and corresponding code from .cpp is :
rule.pattern = QRegularExpression("\\b(\\w+)(?=[\\s+\n]?\\{)",
QRegularExpression::MultilineOption);
Is anyone with better knowledge of Qt and regex can explain to me a way to achieve that?
EDIT #1
Thanks for the reply and comment. Two things :
I mistype my input file had no ">" symbol so I edited it in the above completed input.
I was trying to match the "p" of p-block and the "q" of q-block. A more extended version of my input is now edited above.
I found \}\s*(\w+)(?=\s*\{) to matched the "q" q-block but does not work in the code.
It seems to struggle with the return to line between "p" and the bracket "{".
EDIT #2 : show the code
in highlighter.cpp
#include "highlighter.h"
Highlighter::Highlighter(QTextDocument *parent)
: QSyntaxHighlighter(parent)
{
HighlightingRule rule;
(...)
varFormat.setFontWeight(QFont::Bold);
varFormat.setForeground(Qt::darkMagenta);
rule.pattern = QRegularExpression("^\\s+(\\w+)\\s*$",QRegularExpression::MultilineOption);
rule.format = varFormat;
highlightingRules.append(rule);
(...) }
void Highlighter::highlightBlock(const QString &text)
{
foreach (const HighlightingRule &rule, highlightingRules) {
QRegularExpressionMatchIterator matchIterator = rule.pattern.globalMatch(text);
while (matchIterator.hasNext()) {
QRegularExpressionMatch match = matchIterator.next();
setFormat(match.capturedStart(), match.capturedLength(), rule.format);
}
}
setCurrentBlockState(0);
int startIndex = 0;
if (previousBlockState() != 1)
startIndex = text.indexOf(commentStartExpression);
while (startIndex >= 0) {
QRegularExpressionMatch match = commentEndExpression.match(text, startIndex);
int endIndex = match.capturedStart();
int commentLength = 0;
if (endIndex == -1) {
setCurrentBlockState(1);
commentLength = text.length() - startIndex;
} else {
commentLength = endIndex - startIndex
+ match.capturedLength();
}
setFormat(startIndex, commentLength, multiLineCommentFormat);
startIndex = text.indexOf(commentStartExpression, startIndex + commentLength);
}
}
Have a look at [\\s+\n]?, it matches 1 or 0 occurrences of any whitespace or + characters. But there are more than 1 whitespace betwee solvers and {.
Replacing (?=[\\s+\n]?\\{) with (?=\\s*{) will already fix the issue. But you may also use
QRegularExpression("^\\s*(\\w+)\\s*\\{", QRegularExpression::MultilineOption)
to match the
^ - start of a line
\\s* - 0+ whitespaces
(\\w+) - Group 1 (you can get it via match.captured(1)): one or more word chars
\\s* - 0+ whitespaces followed with
\{ - a literal {.
See the regex demo.
Because p is not after {, but is after }
You can go this way:
[\{\}]\s*(\w+)(?=\s*\{) see https://regex101.com/r/wA1vu2/3
Or this this one:
(?P<tagname>[^{}\s]*)(?P<postspace>\s*)(?P<json_item>\{[^{}]*\})
?P<tagname> name of the match
?P<json_item>\{[^{}]*\} - leaf level item
(?P<postspace>\s*) - space between leaf item and leaf name
(?P<tagname>[^{}\s]*) - leaf name
https://regex101.com/r/wA1vu2/1/
I need to define a bunch of vector sequences, which are all a series of L,D,R,U for left, down, right, up or x for break. There are optional parts, and either/or parts. I have been using my own invented system for noting it down, but I want to document this for other, potentially non-programmers to read.
I now want to use a subset (I don't plan on using any wildcards, or infinite repetition for example) of regex to define the vector sequence and a script to produce all possible matching strings...
/LDR/ produces ['LDR']
/LDU?R/ produces ['LDR','LDUR']
/R(LD|DR)U/ produces ['RLDU','RDRU']
/DxR[DL]U?RDRU?/ produces ['DxRDRDR','DxRDRDRU','DxRDURDR','DxRDURDRU','DxRLRDR','DxRLRDRU','DxRLURDR','DxRLURDRU']
Is there an existing library I can use to generate all matches?
EDIT
I realised I will only be needing or statements, as optional things can be specified by thing or nothing maybe a, or b, both optional could be (a|b|). Is there another language I could use to define what I am trying to do?
By translating the java code form the link provided by #Dukeling into javascript, I think I have solved my problem...
var Node = function(str){
this.bracket = false;
this.children = [];
this.s = str;
this.next = null;
this.addChild = function(child){
this.children.push(child);
}
}
var printTree = function(root,prefix){
prefix = prefix.replace(/\./g, "");
for(i in root.children){
var child = root.children[i]
printTree(child, prefix + root.s);
}
if(root.children.length < 1){
console.log(prefix + root.s);
}
}
var Stack = function(){
this.arr = []
this.push = function(item){
this.arr.push(item)
}
this.pop = function(){
return this.arr.pop()
}
this.peek = function(){
return this.arr[this.arr.length-1]
}
}
var createTree = function(s){
// this line was causing errors for `a(((b|c)d)e)f` because the `(((` was only
// replacing the forst two brackets.
// var s = s.replace(/(\(|\||\))(\(|\||\))/g, "$1.$2");
// this line fixes it
var s = s.replace(/[(|)]+/g, function(x){ return x.split('').join('.') });
var str = s.split('');
var stack = new Stack();
var root = new Node("");
stack.push(root); // start node
var justFinishedBrackets = false;
for(i in str){
var c = str[i]
if(c == '('){
stack.peek().next = new Node("Y"); // node after brackets
stack.peek().bracket = true; // node before brackets
} else if (c == '|' || c == ')'){
var last = stack.peek(); // for (ab|cd)e, remember b / d so we can add child e to it
while (!stack.peek().bracket){ // while not node before brackets
stack.pop();
}
last.addChild(stack.peek().next); // for (b|c)d, add d as child to b / c
} else {
if (justFinishedBrackets){
var next = stack.pop().next;
next.s = "" + c;
stack.push(next);
} else {
var n = new Node(""+c);
stack.peek().addChild(n);
stack.push(n);
}
}
justFinishedBrackets = (c == ')');
}
return root;
}
// Test it out
var str = "a(c|mo(r|l))e";
var root = createTree(str);
printTree(root, "");
// Prints: ace / amore / amole
I only changed one line, to allow more than two consecutive brackets to be handled, and left the original translation in the comments
I also added a function to return an array of results, instead of printing them...
var getTree = function(root,prefix){
this.out = this.out || []
prefix = prefix.replace(/\./g, "");
for(i in root.children){
var child = root.children[i]
getTree(child, prefix + root.s, out);
}
if(root.children.length < 1){
this.out.push(prefix + root.s);
}
if(!prefix && !root.s){
var out = this.out;
this.out = null
return out;
}
}
// Test it
var str = "a(b|c)d";
var root = createTree(str);
console.log(getTree(root, ""));
// logs ["abd","acd"]
The last part, to allow for empty strings too, so... (ab|c|) means ab or c or nothing, and a convenience shortcut so that ab?c is translated into a(b|)c.
var getMatches = function(str){
str = str.replace(/(.)\?/g,"($1|)")
// replace all instances of `(???|)` with `(???|µ)`
// the µ will be stripped out later
str = str.replace(/\|\)/g,"|µ)")
// fix issues where last character is `)` by inserting token `µ`
// which will be stripped out later
str = str+"µ"
var root = createTree(str);
var res = getTree(root, "");
// strip out token µ
for(i in res){
res[i] = res[i].replace(/µ/g,"")
}
// return the array of results
return res
}
getMatches("a(bc|de?)?f");
// Returns: ["abcf","adef","adf","af"]
The last part is a little hack-ish as it relies on µ not being in the string (not an issue for me) and solves one bug, where a ) at the end on the input string was causing incorrect output, by inserting a µ at the end of each string, and then stripping it from the results. I would be happy for someone to suggest a better way to handle these issues, so it can work as a more general solution.
This code as it stands does everything I need. Thanks for all your help!
I'd imagine what you're trying is quite easy with a tree (as long as it's only or-statements).
Parse a(b|c)d (or any or-statement) into a tree as follows: a has children b and c, b and c have a mutual child d. b and c can both consist of 0 or more nodes (as in c could be g(e|f)h in which case (part of) the tree would be a -> g -> e/f (2 nodes) -> h -> d or c could be empty, in which case (part of) the tree would be a -> d, but an actual physical empty node may simplify things, which you should see when trying to write the code).
Generation of the tree shouldn't be too difficult with either recursion or a stack.
Once you have a tree, it's trivial to recursively iterate through the whole thing and generate all strings.
Also, here is a link to a similar question, providing a library or two.
EDIT:
"shouldn't be too difficult" - okay, maybe not
Here is a somewhat complicated example (Java) that may require some advanced knowledge about stacks.
Here is a slightly simpler version (Java) thanks to inserting a special character between each ((, )), |(, etc.
Note that neither of these are particularly efficient, the point is just to get the idea across.
Here is a JavaScript example that addresses parsing the (a|b) and (a|b|) possibilities, creates an array of possible substrings, and composes the matches based on this answer.
var regex = /\([RLUD]*\|[RLUD]*\|?\)/,
str = "R(LD|DR)U(R|L|)",
substrings = [], matches = [], str_tmp = str, find
while (find = regex.exec(str_tmp)){
var index = find.index
finds = find[0].split(/\|/)
substrings.push(str_tmp.substr(0, index))
if (find[0].match(/\|/g).length == 1)
substrings.push([finds[0].substr(1), finds[1].replace(/.$/, '')])
else if (find[0].match(/\|/g).length == 2){
substrings.push([finds[0].substr(1), ""])
substrings.push([finds[1], ""])
}
str_tmp = str_tmp.substr(index + find[0].length)
}
if (str_tmp) substrings.push([str_tmp])
console.log(substrings) //>>["R", ["LD", "DR"], "U", ["R", ""], ["L", ""]]
//compose matches
function printBin(tree, soFar, iterations) {
if (iterations == tree.length) matches.push(soFar)
else if (tree[iterations].length == 2){
printBin(tree, soFar + tree[iterations][0], iterations + 1)
printBin(tree, soFar + tree[iterations][1], iterations + 1)
}
else printBin(tree, soFar + tree[iterations], iterations + 1)
}
printBin(substrings, "", 0)
console.log(matches) //>>["RLDURL", "RLDUR", "RLDUL", "RLDU", "RDRURL", "RDRUR", "RDRUL", "RDRU"]
I am writing a text editor using the wxWidgets framework. I need to get the word under caret from the text control. Here is what I came up with.
static bool IsWordBoundary(wxString& text)
{
return (text.Cmp(wxT(" ")) == 0 ||
text.Cmp(wxT('\n')) == 0 ||
text.Cmp(wxT('\t')) == 0 ||
text.Cmp(wxT('\r')) == 0);
}
static wxString GetWordUnderCaret(wxTextCtrl* control)
{
int insertion_point = control->GetInsertionPoint();
wxTextPos last_position = control->GetLastPosition();
int start_at, ends_at = 0;
// Finding starting position:
// from the current caret position, move back each character until
// we hit a word boundary.
int caret_pos = insertion_point;
start_at = caret_pos;
while (caret_pos)
{
wxString text = control->GetRange (caret_pos - 1, caret_pos);
if (IsWordBoundary (text)) {
break;
}
start_at = --caret_pos;
}
// Finding ending position:
// from the current caret position, move forward each character until
// we hit a word boundary.
caret_pos = ends_at = insertion_point;
while (caret_pos < last_position)
{
wxString text = control->GetRange (caret_pos, caret_pos + 1);
if (IsWordBoundary (text)) {
break;
}
ends_at = ++caret_pos;
}
return (control->GetRange (start_at, ends_at));
}
This code works as expected. But I am wondering is this the best way to approach the problem? Do you see any possible fixes on the above code?
Any help would be great!
Is punctuation part of a word? It is in your code -- is that what you want?
Here is how I would do it:
wxString word_boundary_marks = " \n\t\r";
wxString text_in_control = control->GetValue();
int ends_at = text_in_control.find_first_of( word_boundary_marks, insertion_point) - 1;
int start_at = text_in_control.Mid(0,insertion_point).find_last_of(word_boundary_marks) + 1;
I haven't tested this, so there likely are one or two "off-by-one" errors and you should add checks for "not found", end of string, and any other word markers. My code should give you the basis for what you need.