I have this function that finds whole words and should replace them. It identifies spaces but should not replace them, ie, not capture them.
function asd (sentence, word) {
str = sentence.replace(new RegExp('(?:^|\\s)' + word + '(?:$|\\s)'), "*****");
return str;
};
Then I have the following strings:
var sentence = "ich mag Äpfel";
var word = "Äpfel";
The result should be something like:
"ich mag *****"
and NOT:
"ich mag*****"
I'm getting the latter.
How can I make it so that it identifies the space but ignores it when replacing the word?
At first this may seem like a duplicate but I did not find an answer to this question, that's why I'm asking it.
Thank you
You should put back the matched whitespaces by using a capturing group (rather than a non-capturing one) with a replacement backreference in the replacement pattern, and you may also leverage a lookahead for the right whitespace boundary, which is handy in case of consecutive matches:
function asd (sentence, word) {
str = sentence.replace(new RegExp('(^|\\s)' + word + '(?=$|\\s)'), "$1*****");
return str;
};
var sentence = "ich mag Äpfel";
var word = "Äpfel";
console.log(asd(sentence, word));
See the regex demo.
Details
(^|\s) - Group 1 (later referred to with the help of a $1 placeholder in the replacement pattern): a capturing group that matches either start of string or a whitespace
Äpfel - a search word
(?=$|\s) - a positive lookahead that requires the end of string or whitespace immediately to the right of the current location.
NOTE: If the word can contain special regex metacharacters, escape them:
function asd (sentence, word) {
str = sentence.replace(new RegExp('(^|\\s)' + word.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&') + '(?=$|\\s)'), "$1*****");
return str;
};
Related
I would like to find all the matches of given strings (divided by spaces) in a string.
(The way for example, iTunes search box works).
That, for example, both "ab de" and "de ab" will return true on "abcde" (also "bc e a" or any order should return true)
If I replace the white space with a wild card, "ab*de" would return true on "abcde", but not "de*ab".
[I use * and not Regex syntax just for this explanation]
I could not find any pure Regex solution for that.
The only solution I could think of is spliting the search term and run multiple Regex.
Is it possible to find a pure Regex expression that will cover all these options ?
Returns true when all parts (divided by , or ' ') of a searchString occur in text. Otherwise false is returned.
filter(text, searchString) {
const regexStr = '(?=.*' + searchString.split(/\,|\s/).join(')(?=.*') + ')';
const searchRegEx = new RegExp(regexStr, 'gi');
return text.match(searchRegEx) !== null;
}
I'm pretty sure you could come up with a regex to do what you want, but it may not be the most efficient approach.
For example, the regex pattern (?=.*bc)(?=.*e)(?=.*a) will match any string that contains bc, e, and a.
var isMatch = 'abcde'.match(/(?=.*bc)(?=.*e)(?=.*a)/) != null; // equals true
var isMatch = 'bcde'.match(/(?=.*bc)(?=.*e)(?=.*a)/) != null; // equals false
You could write a function to dynamically create an expression based on your search terms, but whether it's the best way to accomplish what you are doing is another question.
Alternations are order insensitive:
"abcde".match(/(ab|de)/g); // => ['ab', 'de']
"abcde".match(/(de|ab)/g); // => ['ab', 'de']
So if you have a list of words to match you can build a regex with an alternation on the fly like so:
function regexForWordList(words) {
return new RegExp('(' + words.join('|') + ')', 'g');
}
'abcde'.match(['a', 'e']); // => ['a', 'e']
Try this:
var str = "your string";
str = str.split( " " );
for( var i = 0 ; i < str.length ; i++ ){
// your regexp match
}
This is script which I use - it works also with single word searchStrings
var what="test string with search cool word";
var searchString="search word";
var search = new RegExp(searchString, "gi"); // one-word searching
// multiple search words
if(searchString.indexOf(' ') != -1) {
search="";
var words=searchString.split(" ");
for(var i = 0; i < words.length; i++) {
search+="(?=.*" + words[i] + ")";
}
search = new RegExp(search + ".+", "gi");
}
if(search.test(what)) {
// found
} else {
// notfound
}
I assume you are matching words, or parts of words. You want space-separated search terms to limit search results, and it seems you intend to return only those entries which have all the words that the user supplies. And you intend a wildcard character * to stand for 0 or more characters in a matching word.
For example, if the user searches for the words term1 term2, you intend to return only those items which have both words term1 and term2. If the user searches for the word term*, it would match any word beginning with term.
There are suitable regular expressions which are equivalent to this search language and can be generated from it.
A simple example, the word term, can be asserted in regex by converting to \bterm\b. But two or more words which must match in any order require lookahead assertions. Using extended syntax, the equivalent regex is:
(?= .* \b term1 \b )
(?= .* \b term2 \b )
The asterisk wildcard can be asserted in regex with a character class followed by asterisk. The character class identifies which letters you consider to be part of word. For example, you might find that [A-Za-z0-9]* fits the bill.
In short, you might be satisfied if you convert an expression such as:
foo ba* quux
to:
(?= .* \b foo \b )
(?= .* \b ba[A-Za-z0-9]* \b )
(?= .* \b quux \b )
That is a simple matter of search and replace. But do be careful to sanitize the input string to avoid injection attacks by removing punctuation, etc.
I think you may be barking up the wrong tree with RegEx. What you might want to look at is the Levenshtein distance of two input strings.
There's a Javascript implementation here and a usage example here.
I'm trying to write a regex that finds all punctuation marks [.!?] in order to capitalize the next word, however if the period is part of the string 'D.C.' it should be ignored, so far I have the first part working, but not sure about how to ignore 'D.C.'
const punctuationCaps = /(^|[.!?]\s+)([a-z])/g;
You can match the D.C. part and use an alternation using the 2 capturing groups that you already have.
In the replacement check for one of the groups. If it is present, concatenate them making group 2 toUpperCase(), else return the match keeping D.C. in the string.
const regex = /D\.C\.|(^|[.!?]\s+)([a-z])/g;
let s = "this it D.C. test. and? another test! it is.";
s = s.replace(regex, (m, g1, g2) => g2 ? g1 + g2.toUpperCase() : m);
console.log(s);
Use a negative lookahead:
var str = 'is D.C. a capital? i don\'t know about X.Y. stuff.';
var result = str.replace(/(^|[.!?](?<![A-Z]\.[A-Z]\.)\s+)([a-z])/g, (m, c1, c2) => { return c1 + c2.toUpperCase(); });
console.log('in: '+str);
console.log('out: '+result);
Console output:
in: is D.C. a capital? i don't know about X.Y. stuff.
out: Is D.C. a capital? I don't know about X.Y. stuff.
Explanation:
(^|[.!?]) - expect start of string, or a punctuation char
(?<![A-Z]\.[A-Z]\.) - negative lookahead: but not a sequence of upper char and dot, repeated twice
\s+ - expect one or more whitespace chars
all of the above is captured because of the parenthesis
([a-z]) - expect a lower case char, in parenthesis for second capture group
In the javascript code below I need to find in a text exact words, but excluding the words that are between quotes. This is my attempt, what's wrong with the regex? It should find all the words excluding word22 and "word3". If I use only \b in the regex it selects exact words but it doesn't exclude the words between quotes.
var text = 'word1, word2, word22, "word3" and word4';
var words = [ 'word1', 'word2', 'word3' , 'word4' ];
words.forEach(function(word){
var re = new RegExp('\\b^"' + word + '^"\\b', 'i');
var pos = text.search(re);
if (pos > -1)
alert(word + " found in position " + pos);
});
First, we'll use a function to escape the characters of the word, just in case there's some that have special meaning for regexp.
// from https://stackoverflow.com/a/30851002/240443
function regExpEscape(literal_string) {
return literal_string.replace(/[-[\]{}()*+!<=:?.\/\\^$|#\s,]/g, '\\$&');
}
Then, we construct a regular expression as an alternation between individual word regexps. For each word, we assert that it starts with a word boundary, ends with a word boundary, and has an even number of quote characters between its end, and the end of string. (Note that from the end of word3, there is only one quote till the end of string, which is odd.)
let text = 'word1, word2, word22, "word3" and word4';
let words = [ 'word1', 'word2', 'word3' , 'word4' ];
let regexp = new RegExp(words.map(word =>
'\\b' + regExpEscape(word) + '\\b(?=(?:[^"]*"[^"]*")*[^"]*$)').join('|'), 'g')
text.match(regexp)
// => word1, word2, word4
while ((m = regexp.exec(text))) {
console.log(m[0], m.index);
}
// word1 0
// word2 7
// word4 34
EDIT: Actually, we can speed the regexp up a bit if we factor out the surrounding conditions:
let regexp = new RegExp(
'\\b(?:' +
words.map(regExpEscape).join('|') +
')\\b(?=(?:[^"]*"[^"]*")*[^"]*$)', 'g')
Your excluding of the quote character is wrong, that's actually matching the beginning of the string followed by a quote. Trying this instead
var re = new RegExp('\\b[^"]' + word + '[^"]\\b', 'i');
Also, this site is amazing to help you debug regex : https://regexpal.com
Edit: Because \b will match on quotation marks, this needs to be tweaked further. Unfortunately javascript doesn't support lookbehinds, so we have to get a little tricky.
var re = new RegExp('(?:^|[^"\\w])' + word + '(?:$|[^"\\w])','i')
So what this is doing is saying
(?: Don't capture this group
^ | [^"\w]) either match the start of the line, or any non word (alphanumeric and underscore) character that isn't a quote
word capture and match your word here
(?: Don't capture this group either
$|[^"\w) either match the end of the line, or any non word character that isn't a quote again
How to capitalize each words on starting of a string and after dot(.) sign?
I made a research on google and stackoverflow, below are the codes that I achieved but this will only capitalize starting of a string. Example as belows;
var str = 'this is a text. hello world!';
str = str.replace(/^(.)/g, str[0].toUpperCase());
document.write(str);
I want the string to be This is a text. Hello world!.
I have tried to use css, text-transform: capitalize; but this will result in each word to be capitalize.
I use a function like this, that takes an optional second parameter that will convert the entire string to lowercase initially. The reason is that sometimes you have a series of Title Case Items. That You Wish to turn into a series of Title case items. That you wish to have as sentence case.
function sentenceCase(input, lowercaseBefore) {
input = ( input === undefined || input === null ) ? '' : input;
if (lowercaseBefore) { input = input.toLowerCase(); }
return input.toString().replace( /(^|\. *)([a-z])/g, function(match, separator, char) {
return separator + char.toUpperCase();
});
}
The regex works as follows
1st Capturing Group (^|\. *)
1st Alternative ^
^ asserts position at start of the string
2nd Alternative \. *
\. matches the character `.` literally (case sensitive)
* matches the character ` ` literally (case sensitive)
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
2nd Capturing Group ([a-z])
Match a single character present in the list below [a-z]
a-z a single character in the range between a (ASCII 97) and z (ASCII 122) (case sensitive)
You would implement it in your example like so:
var str = 'this is a text. hello world!';
str = sentenceCase(str);
document.write(str); // This is a text. Hello world!
Example jsfiddle
PS. in future, i find regex101 a hugely helpful tool for understanding and testing regex's
If you are using jquery, use this code
function capitalize(str) {
return str.charAt(0).toUpperCase() + str.slice(1);
}
var str = "my name is Jhon. are you good. is it";
var str1 = str.split('.');
var str2 = "";
$.each(str1,function(i){
str2 += capitalize($.trim(str1[i]))+'. ';
});
console.log(str2);
catch the out put in str2.
in my case its like the following.
My name is Jhon. Are you good. Is it.
I'm trying to match all the words starting with # and words between 2 # (see example)
var str = "#The test# rain in #SPAIN stays mainly in the #plain";
var res = str.match(/(#)[^\s]+/gi);
The result will be ["#The", "#SPAIN", "#plain"] but it should be ["#The test#", "#SPAIN", "#plain"]
Extra: would be nice if the result would be without the #.
Does anyone has a solution for this?
You can use
/#\w+(?:(?: +\w+)*#)?/g
See the demo here
The regex matches:
# - a hash symbol
\w+ - one or more alphanumeric and underscore characters
(?:(?: +\w+)*#)? - one or zero occurrence of:
(?: +\w+)* - zero or more occurrences of one or more spaces followed with one or more word characters followed with
# - a hash symbol
NOTE: If there can be characters other than word characters (those in the [A-Za-z0-9_] range), you can replace \w with [^ #]:
/#[^ #]+(?:(?: +[^ #]+)*#)?/g
See another demo
var re = /#[^ #]+(?:(?: +[^ #]+)*#)?/g;
var str = '#The test-mode# rain in #SPAIN stays mainly in the #plain #SPAIN has #the test# and more #here';
var m = str.match(re);
if (m) {
// Using ES6 Arrow functions
m = m.map(s => s.replace(/#$/g, ''));
// ES5 Equivalent
/*m = m.map(function(s) {
return s.replace(/#$/g, '');
});*/ // getting rid of the trailing #
document.body.innerHTML = "<pre>" + JSON.stringify(m, 0, 4) + "</pre>";
}
You can also try this regex.
#(?:\b[\s\S]*?\b#|\w+)
(?: opens a non capture group for alternation
\b matches a word boundary
\w matches a word character
[\s\S] matches any character
See demo at regex101 (use with g global flag)