I am trying to detect whether a block of text (from a textarea) contains words that are prefixed with the #sign.
For example in the following text: Hey #John, i just saw #Smith
It will detect John and Smith respectively without the # symbol. I reckoned something like this would work:
#\w\w+
My question is how do i make javascript filter the text, assuming it is stored in a variable comment?
It should output only the names in the text that are prefixed with # without the # symbol.
Regards.
You use the g (global) flag, a capture group, and a loop calling RegExp#exec, like this:
var str = "Hi there #john, it's #mary, my email is mary#example.com.";
var re = /\B#(\w+)/g;
var m;
for (m = re.exec(str); m; m = re.exec(str)) {
console.log("Found: " + m[1]);
}
Output:
Found: john
Found: mary
Live example | source
With thanks to #Alex K for the boundary recommendation!
comment.match(/#\w+/g) will give you an array of the matches (["#John", "#Smith"]).
I added a check to the regex so that it won't match email addresses, in case you're interested.
var comment = "Hey #John, I just saw #Smith."
+ " (john#example.com)";
// Parse tags using ye olde regex.
var tags = comment.match(/\B#\w+/g);
// If no tags were found, turn "null" into
// an empty array.
if (!tags) {
tags = [];
}
// Remove leading space and "#" manually.
// Can't incorporate this into regex as
// lookbehind not always supported.
for (var i = 0; i < tags.length; i++) {
tags[i] = tags[i].substr(1);
}
var re = /#(\w+)/g; //set the g flag to match globally
var match;
while (match = re.exec(text)) {
//match is an array representing how the regex matched the text.
//match.index the position where it matches.
//it returns null if there are no matches, ending the loop.
//match[0] is the text matched by the entire regex,
//match[1] is the text between the first capturing group.
//each set of matching parenthesis is a capturing group.
}
Related
Having a text input, if there is a specific character it must convert it to a tag. For example, the special character is *, the text between 2 special characters must appear in italic.
For example:
This is *my* wonderful *text*
must be converted to:
This is <i>my</i> wonderful <i>text</i>
So I've tried like:
const arr = "This is *my* wonderful *text*";
if (arr.includes('*')) {
arr[index] = arr.replace('*', '<i>');
}
it is replacing the star character with <i> but doesn't work if there are more special characters.
Any ideas?
You can simply create wrapper and thereafter use regular expression to detect if there is any word that is surrounded by * and simply replace it with any tag, in your example is <i> tag so just see the following
Example
let str = "This is *my* wonderful *text*";
let regex = /(?<=\*)(.*?)(?=\*)/;
while (str.includes('*')) {
let matched = regex.exec(str);
let wrap = "<i>" + matched[1] + "</i>";
str = str.replace(`*${matched[1]}*`, wrap);
}
console.log(str);
here you go my friend:
var arr = "This is *my* wonderful *text*";
const matched = arr.match(/\*(?:.*?)\*/g);
for (let i = 0; i < matched.length; i++) {
arr = arr.replace(matched[i], `<i>${matched[i].replaceAll("*", "")}</i>`);
}
console.log(arr);
an explanation first of all we're matching the regex globaly by setting /g NOTE: that match with global flag returns an array.
secondly we're looking for any character that lies between two astrisks and we're escaping them because both are meta characters.
.*? match everything in greedy way so we don't get something like this my*.
?: for non capturing groups, then we're replacing every element we've matched with itself but without astrisk.
Here's an example string:
++++#foo+bar+baz++#yikes
I need to extract foo and only foo from there or a similar scenario.
The + and the # are the only characters I need to worry about.
However, regardless of what precedes foo, it needs to be stripped or ignored. Everything else after it needs to as well.
try this:
/\++#(\w+)/
and catch the capturing group one.
You can simply use the match() method.
var str = "++++#foo+bar+baz++#yikes";
var res = str.match(/\w+/g);
console.log(res[0]); // foo
console.log(res); // foo,bar,baz,yikes
Or use exec
var str = "++++#foo+bar+baz++#yikes";
var match = /(\w+)/.exec(str);
alert(match[1]); // foo
Using exec with a g modifier (global) is meant to be used in a loop getting all sub matches.
var str = "++++#foo+bar+baz++#yikes";
var re = /\w+/g;
var match;
while (match = re.exec(str)) {
// In array form, match is now your next match..
}
How exactly do + and # play a role in identifying foo? If you just want any string that follows # and is terminated by + that's as simple as:
var foostring = '++++#foo+bar+baz++#yikes';
var matches = (/\#([^+]+)\+/g).exec(foostring);
if (matches.length > 1) {
// all the matches are found in elements 1 .. length - 1 of the matches array
alert('found ' + matches[1] + '!'); // alerts 'found foo!'
}
To help you more specifically, please provide information about the possible variations of your data and how you would go about identifying the token you want to extract even in cases of differing lengths and characters.
If you are just looking for the first segment of text preceded and followed by any combination of + and #, then use:
var foostring = '++++#foo+bar+baz++#yikes';
var result = foostring.match(/[^+#]+/);
// will be the single-element array, ['foo'], or null.
Depending on your data, using \w may be too restrictive as it is equivalent to [a-zA-z0-9_]. Does your data have anything else such as punctuation, dashes, parentheses, or any other characters that you do want to include in the match? Using the negated character class I suggest will catch every token that does not contain a + or a #.
base on the following string
...here..
..there...
.their.here.
How can i remove the . on the beginning and end of string like the trim that removes all spaces, using javascript
the output should be
here
there
their.here
These are the reasons why the RegEx for this task is /(^\.+|\.+$)/mg:
Inside /()/ is where you write the pattern of the substring you want to find in the string:
/(ol)/ This will find the substring ol in the string.
var x = "colt".replace(/(ol)/, 'a'); will give you x == "cat";
The ^\.+|\.+$ in /()/ is separated into 2 parts by the symbol | [means or]
^\.+ and \.+$
^\.+ means to find as many . as possible at the start.
^ means at the start; \ is to escape the character; adding + behind a character means to match any string containing one or more that character
\.+$ means to find as many . as possible at the end.
$ means at the end.
The m behind /()/ is used to specify that if the string has newline or carriage return characters, the ^ and $ operators will now match against a newline boundary, instead of a string boundary.
The g behind /()/ is used to perform a global match: so it find all matches rather than stopping after the first match.
To learn more about RegEx you can check out this guide.
Try to use the following regex
var text = '...here..\n..there...\n.their.here.';
var replaced = text.replace(/(^\.+|\.+$)/mg, '');
Here is working Demo
Use Regex /(^\.+|\.+$)/mg
^ represent at start
\.+ one or many full stops
$ represents at end
so:
var text = '...here..\n..there...\n.their.here.';
alert(text.replace(/(^\.+|\.+$)/mg, ''));
Here is an non regular expression answer which utilizes String.prototype
String.prototype.strim = function(needle){
var first_pos = 0;
var last_pos = this.length-1;
//find first non needle char position
for(var i = 0; i<this.length;i++){
if(this.charAt(i) !== needle){
first_pos = (i == 0? 0:i);
break;
}
}
//find last non needle char position
for(var i = this.length-1; i>0;i--){
if(this.charAt(i) !== needle){
last_pos = (i == this.length? this.length:i+1);
break;
}
}
return this.substring(first_pos,last_pos);
}
alert("...here..".strim('.'));
alert("..there...".strim('.'))
alert(".their.here.".strim('.'))
alert("hereagain..".strim('.'))
and see it working here : http://jsfiddle.net/cettox/VQPbp/
Slightly more code-golfy, if not readable, non-regexp prototype extension:
String.prototype.strim = function(needle) {
var out = this;
while (0 === out.indexOf(needle))
out = out.substr(needle.length);
while (out.length === out.lastIndexOf(needle) + needle.length)
out = out.slice(0,out.length-needle.length);
return out;
}
var spam = "this is a string that ends with thisthis";
alert("#" + spam.strim("this") + "#");
Fiddle-ige
Use RegEx with javaScript Replace
var res = s.replace(/(^\.+|\.+$)/mg, '');
We can use replace() method to remove the unwanted string in a string
Example:
var str = '<pre>I'm big fan of Stackoverflow</pre>'
str.replace(/<pre>/g, '').replace(/<\/pre>/g, '')
console.log(str)
output:
Check rules on RULES blotter
I'm having trouble with removing all characters up to and including the 3 third slash in JavaScript. This is my string:
http://blablab/test
The result should be:
test
Does anybody know the correct solution?
To get the last item in a path, you can split the string on / and then pop():
var url = "http://blablab/test";
alert(url.split("/").pop());
//-> "test"
To specify an individual part of a path, split on / and use bracket notation to access the item:
var url = "http://blablab/test/page.php";
alert(url.split("/")[3]);
//-> "test"
Or, if you want everything after the third slash, split(), slice() and join():
var url = "http://blablab/test/page.php";
alert(url.split("/").slice(3).join("/"));
//-> "test/page.php"
var string = 'http://blablab/test'
string = string.replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'')
alert(string)
This is a regular expression. I will explain below
The regex is /[\s\S]*\//
/ is the start of the regex
Where [\s\S] means whitespace or non whitespace (anything), not to be confused with . which does not match line breaks (. is the same as [^\r\n]).
* means that we match anywhere from zero to unlimited number of [\s\S]
\/ Means match a slash character
The last / is the end of the regex
var str = "http://blablab/test";
var index = 0;
for(var i = 0; i < 3; i++){
index = str.indexOf("/",index)+1;
}
str = str.substr(index);
To make it a one liner you could make the following:
str = str.substr(str.indexOf("/",str.indexOf("/",str.indexOf("/")+1)+1)+1);
You can use split to split the string in parts and use slice to return all parts after the third slice.
var str = "http://blablab/test",
arr = str.split("/");
arr = arr.slice(3);
console.log(arr.join("/")); // "test"
// A longer string:
var str = "http://blablab/test/test"; // "test/test";
You could use a regular expression like this one:
'http://blablab/test'.match(/^(?:[^/]*\/){3}(.*)$/);
// -> ['http://blablab/test', 'test]
A string’s match method gives you either an array (of the whole match, in this case the whole input, and of any capture groups (and we want the first capture group)), or null. So, for general use you need to pull out the 1th element of the array, or null if a match wasn’t found:
var input = 'http://blablab/test',
re = /^(?:[^/]*\/){3}(.*)$/,
match = input.match(re),
result = match && match[1]; // With this input, result contains "test"
let str = "http://blablab/test";
let data = new URL(str).pathname.split("/").pop();
console.log(data);
text = 'ticket number #1234 and #8976 ';
r = /#(\d+)/g;
var match = r.exec(text);
log(match); // ["#1234", "1234"]
In the above case I would like to capture both 1234 and 8976. How do I do that. Also the sentence can have any number of '#' followed by integers. So the solution should not hard not be hard coded assuming that there will be at max two occurrences.
Update:
Just curious . Checkout the following two cases.
var match = r.exec(text); // ["#1234", "1234"]
var match = text.match(r); //["#1234", "#8976"]
Why in the second case I am getting # even though I am not capturing it. Looks like string.match does not obey capturing rules.
exec it multiple times to get the rest.
while((match = r.exec(text)))
log(match);
Use String.prototype.match instead of RegExp.prototype.exec:
var match = text.match(r);
That will give you all matches at once (requires g flag) instead of one match at a time.
Here's another way
var text = 'ticket number #1234 and #8976 ';
var r = /#(\d+)/g;
var matches = [];
text.replace( r, function( all, first ) {
matches.push( first )
});
log(matches);
// ["1234", "8976"]