Develop Reference

check if any duplicate words exist in a string using regex

Using regex in javascript, how to check for the condition when duplicate words are present in a string?
The words could be located any where in the string:
Given: "but boys run fast boys are strong" or
"boys boys"
expected result: true, since there are 2 repeats of the word "boys"

\b matches word boundaries
\w+ will match 1 or more word characters
( ... ) creates a group for matches
\1 will match the contents of matching group #1.
Putting this together, you want a regex containing \b(\w+)\b.*\b\1\b
Plus appropriate backslash quoting, etc.
#guest27134 pointed out the above is not the complete solution, since the OP wanted true/false, not just the regex:
var result = a_string.match(/\b(\w+)\b.*\b\1\b/g) !== null
Or, even shorter, as suggested by O.P.:
var result = /\b(\w+)\b.*\b\1\b/g.test(myStr)

Taking into consideration that spaces are there after each word
var string = "but boys run fast boys are strong";
var strArray= string.split(" ");
var unique = [];
for(var i =0; i< strArray.length; i++)
{
eval(unique[strArray] = new Object());
}

If you know what word you're testing for duplicates you could use regular expressions like this str.match(/boys/g).length > 1 to test if that word shows up more than once (assuming your string is in the str variable).

Hello, here it is:
var temp = "but boys run fast boys are strong";
var count = (temp.match(/boys/g) || []).length;
console.log(count);
I hope I have helped!

You can count the occurrence of each matched word
var str = "but boys run fast boys are strong";
var matches = str.split(/\s/);
var res = matches.map(function(match) {
return str.match(new RegExp(match, "g")).length;
});
var bool = res.some(function(len) {return len > 1}));
console.log(bool);
for (var i = 0; i < matches.length; i++) {
if (res[i] > 1) console.log(matches[i], i);
}

Hello!
Here's an example with the exact answer you need:
var keyword = "boys"
var temp = "but boys run fast boys are strong";
var regex = new RegExp(keyword, "g");
var count = (temp.match(regex) || []).length;
if (count > 0) {
console.log("true, since there are " + count + " repeats of the word '" + keyword + "'");
} else {
console.log("false, not found.");
}
I hope I have helped!

Separate characters and numbers from a string

I have a string variable that contain character and numbers like this
var sampleString = "aaa1211"
Note that variable always start with a character/s and end with number/s. Character and number size is not fixed. It could be something like followings
var sampleString = "aaaaa12111"
var sampleString = "aaa12111"
I need to separate the characters and numbers and assign them into separate variables.
How could I do that ?
I try to use split and substring but for this scenario I couldn't apply those. I know this is a basic question but i'm search over the internet and I was unable to find an answer.
Thank you

Please use
[A-Za-z] - all letters (uppercase and lowercase)
[0-9] - all numbers
function myFunction() {
var str = "aaaaAZE12121212";
var patt1 = /[0-9]/g;
var patt2 = /[a-zA-Z]/g;
var letters = str.match(patt2);
var digits = str.match(patt1);
document.getElementById("alphabets").innerHTML = letters;
document.getElementById("numbers").innerHTML = digits;
}
Codepen-http://codepen.io/nagasai/pen/pbbGOB

A shorter solution if the string always starts with letters and ends with numbers as you say:
var str = 'aaaaa12111';
var chars = str.slice(0, str.search(/\d/));
var numbs = str.replace(chars, '');
console.log(chars, numbs);

You can use it in a single regex,
var st = 'red123';
var regex = new RegExp('([0-9]+)|([a-zA-Z]+)','g');
var splittedArray = st.match(regex);
var num= splittedArray[0];
var text = splittedArray[1];
this will give you both the text and number.

Using Match
const str = "ASDF1234";
const [word, digits] = str.match(/\D+|\d+/g);
console.log(word); // "ASDF"
console.log(digits); // "1234"
The above will work even if your string starts with digits.
Using Split
with Positive lookbehind (?<=) and Positive lookahead (?=):
const str = "ASDF1234";
const [word, digits] = str.split(/(?<=\D)(?=\d)/);
console.log(word); // "ASDF"
console.log(digits); // "1234"
where \D stands for not a digit and \d for digit.

Use isNaN() to differentiate
var sampleString = "aaa1211"
var newnum =""
var newstr =""
for(var i=0;i<sampleString.length;i++){
if(isNaN(sampleString[i])){
newstr = newstr+sampleString[i]
}else{
newnum= newstr+sampleString[i]
}
}
console.log(newnum) //1121
console.log(newstr) //aaa

If you're like me, you were looking to separate alphabets and numbers, no matter what their position is, Try this:
var separateTextAndNum = (x) => {
var textArray = x.split('')
var text = []
var num = []
textArray.forEach(t=>{
if (t>-1) {
num.push(t)
} else {
text.push(t)
}
})
return [text, num]
}
For ex - if you try this:
separateTextAndNum('abcd1234ava') // result [["a","b","c","d","a","v","a"],["1","2","3","4"]]

This isn't beautiful but it works.
function splitLettersAndNumbers(string) {
var numbers = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'];
var numbers, letters, splitIndex;
for (var i = 0; i < string.length; i++) {
if (numbers.indexOf(string[i]) > -1) {
letters = string.substring(0, i);
numbers = string.substring(i);
return [letters, numbers];
}
}
// in the chance that you don't find any numbers just return the initial string or array of the string of letters
return [string];
}
Essentially just looping through the string until you find a number and you split it at that index. Returning a tuple with your letters and numbers. So when you run it you can do something like this:
var results = splitLettersAndNumbers(string);
var letters = results[0];
var numbers = results[1];

A functional approach...
var sampleString = "aaaaa12111";
var seperate = sampleString.split('').reduce(function(start , item){
Number(item) ? start[0] += item : start[1] += item;
return start
},['',''])
console.log(seperate) //["12111", "aaaaa"]

You can loop through the string length, check it & add to the variable.
It is not clear if you want to assign each of the character to a variable or all alphabets to one variable & integers to another.
var sampleString = "aaa12111"
var _num = "";
var _alp = "";
for (var i = 0; i < sampleString.length; i++) {
if (isNaN(sampleString[i])) {
_num += sampleString[i];
} else {
_alp += sampleString[i];
}
}
console.log(_num, _alp)

How can i match and replace a string and its before one letter in javascript?

var str="itss[BACK][BACK][BACK][BACK][BACK][BACK] it's a test stringgg[BACK][BACK]";
var word = '[BACK]';
var substrings = str.split(word);
var cnt= substrings.length - 1;
for(var i = 0;i<cnt;i++){
str = str.replace(/.{1}\[BACK\]{1}/i,""); //remove backspace and one character before it.
}
The above script returns something like "[BACK it's a test string" I need to get this result as "it's a test string" please help me....

It's easier to do this without a regex actually.
String.prototype.replaceFromIndex=function(index, length, replace) {
return this.substr(0, index) + replace + this.substr(index+length);
}
var search = '[BACK]';
var str="itss[BACK][BACK][BACK][BACK][BACK][BACK] it's a test stringgg[BACK][BACK]";
while((index = str.indexOf(search)) >= 0){
str = str.replaceFromIndex(index-1, search.length+1, '');
}
alert(str);
Check http://jsfiddle.net/fRThH/2/ for a working example.
Wrap it in a function and you are ready to go!
Courtesy to Cem Kalyoncu ( https://stackoverflow.com/a/1431113/187018 ) for a slightly modified version of String.prototype.replaceAt

My idea is to count all the backspaces [BACK] and then replace them with an empty string one by one:
var str="itss[BACK][BACK][BACK][BACK][BACK][BACK] it's a test stringgg[BACK][BACK]";
var backspaces = str.match(/\[BACK\]/g).length;
for(i=0; i<backspaces; i++)
{
str = str.replace(/.?\[BACK\]/, '');
}
document.write( str );
working example: jsFiddle

If I understood correctly
var dat = str.split('[BACK]').filter(function(e){return e})[1];
here is the working demo.

One of the problems that I found out was that you didn't set a condition in which you would not have to remove the first character when the string '[BACK]' is in position zero.
Well, the solution I am posting here first search for the position of the first '[BACK]' string, and then creates a substring of the characters that we want to remove, so, if there is a character before the string '[BACK]', it is included in the substring. Then, the substring is removed from the main string, and it continues looping until all the '[BACK]' s are removed.
var str = "itss[BACK][BACK][BACK][BACK][BACK][BACK] it's a test stringgg[BACK][BACK]";
var word = '[BACK]';
var substrings = str.split(word);
var cnt = substrings.length - 1;
for (i = 0; i < cnt; i++) {
pos = str.search("[BACK]");
if (pos - 1 > 0) {
str = str.replace(str.substring(pos - 2, pos + 5), '');
} else {
str = str.replace(str.substring(pos - 1, pos + 5), '');
}
}
Here is the code in jsfiddle:

How do i get numbers from this string?

i have this string:
var s = 'http://xxxxxxx.xxx/abcd123456789?abc=1';
how do i get digits 123456789 (between "d" and "?") ?
these digits may vary. the number of digits may vary as well.
How do i get them?? Regex? Which one?

try
'http://xxxxxxx.xxx/abcd123456789?abc=1'.match(/\d+(?=\?)/)[0];
// ^1 or more digits followed by '?'

Try
var regexp = /\/abcd(\d+)\?/;
var match = regexp.exec(input);
var number = +match[1];

Are the numbers always between "abcd" and "?"?
If so, then you can use substring():
s.substring(s.indexOf('abcd'), s.indexOf('?'))
If not, then you can just loop through character by character and check if it's numeric:
var num = '';
for (var i = 0; i < s.length; i++) {
var char = s.charAt(i);
if (!isNaN(char)) {
num += char;
}
}

Yes, regex is the right answer. You'll have something like this:
var s = 'http://xxxxxxx.xxx/abcd123456789?abc=1';
var re = new RegExp('http\:\/\/[^\/]+\/[^\d]*(\d+)\?');
re.exec(s);
var digits = $1;

Stripping some characters out of a string

I try to remove some non-safe characters from a string but i believe i have a problem on my RegExp object.
What i try to do below is if there are chars whose encoded length is greater than 3 chars they should be replaced with a space.
So if encoded value is %3D which is = sign, it is ok to have in my string. But if it is an ’ apostrophe %E2%80%99 it should be replaced with a space.
val = "Angelina’s";
valEnc = encodeURIComponent(val);
for(var i = 0; i < val.length; i++){
var a = val.substr(i,1);
if(encodeURIComponent(a).length > 3){
console.log(a, encodeURIComponent(a));
var re = new RegExp(encodeURIComponent(a),"ig");
valEnc.replace(re," ");
};
};
console.log(decodeURIComponent(valEnc));
This code works and logs me the unwanted chars but it can not replace them with spaces, what am i doing wrong? Thanks.

You seem to be using regular expressions unnecessarily here. One way to do this is to build up the result string one character at a time:
val = "Angelina’s";
valEnc = "";
for(var i = 0; i < val.length; i++){
var a = val.substr(i,1);
var e = encodeURIComponent(a);
if(e.length <= 3){
valEnc += e;
}
}
console.log(decodeURIComponent(valEnc));