Input = ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))
Expected Output = ABCDEF,GHIJKLMN,OPQRSTUVW
Tried so far
Output = Input.replace(/ *\([^)]*\)*/g, "");
Using a regex here probably won't work, or scale, because you expect nested parentheses in your input string. Regex works well when there is a known and fixed structure to the input. Instead, I would recommend that you approach this using a parser. In the code below, I iterate over the input string, one character at at time, and I use a counter to keep track of how many open parentheses there are. If we are inside a parenthesis term, then we don't record those characters. I also have one simple replacement at the end to remove whitespace, which is an additional step which your output implies, but you never explicitly mentioned.
var pCount = 0;
var Input = "ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))";
var Output = "";
for (var i=0; i < Input.length; i++) {
if (Input[i] === '(') {
pCount++;
}
else if (Input[i] === ')') {
pCount--;
}
else if (pCount == 0) {
Output += Input[i];
}
}
Output = Output.replace(/ /g,'');
console.log(Output);
If you need to remove nested parentheses, you may use a trick from Remove Nested Patterns with One Line of JavaScript.
var Input = "ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))";
var Output = Input;
while (Output != (Output = Output.replace(/\s*\([^()]*\)/g, "")));
console.log(Output);
Or, you could use a recursive function:
function remove_nested_parens(s) {
let new_s = s.replace(/\s*\([^()]*\)/g, "");
return new_s == s ? s : remove_nested_parens(new_s);
}
console.log(remove_nested_parens("ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))"));
Here, \s*\([^()]*\) matches 0+ whitespaces, (, 0+ chars other than ( and ) and then a ), and the replace operation is repeated until the string does not change.
Related
How do I remove a character from a string and remove the previous character as well?
Example:
"ABCXDEXFGHXIJK"
I want to split the string by "X" and remove the previous character which returns
"ABDFGIJK" // CX, EX, HX are removed
I found this thread but it removes everything before rather than a specific amount of characters: How to remove part of a string before a ":" in javascript?
I can run a for loop but I was wondering if there was a better/simpler way to achieve this
const remove = function(str){
for(let i = 0; i < str.length; i++){
if(str[i] === "X") str = str.slice(0, i - 1) + str.slice(i + 1);
}
return str
}
console.log(remove("ABCXDEXFGHXIJK")) // ABDFGIJK
You can use String.prototype.replace and regex.
"ABCXDEXFGHXIJK".replace(/.X/g, '')
The g at the end is to replace every occurrence of .X. You can use replaceAll as well, but it has less support.
"ABCXDEXFGHXIJK".replaceAll(/.X/g, '')
If you want it to be case insensitive, use the i flag as well.
"ABCXDEXFGHXIJK".replace(/.x/gi, '')
The simplest way is to use a regular expression inside replace.
"ABCXDEXFGHXIJK".replace(/.X/g, "")
.X means "match the combination of X and any single character before it, g flag after the expression body repeats the process globally (instead of doing it once).
While not the most computationally efficient, you could use the following one-liner that may meet your definition of "a better/simpler way to achieve this":
const remove = str => str.split("X").map((ele, idx) => idx !== str.split("X").length - 1 ? ele.slice(0, ele.length - 1) : ele).join("");
console.log(remove("ABCXDEXFGHXIJK"));
Maybe you can use recursion.
function removeChar(str, char){
const index = str.indexOf(char);
if(index < 0) return str;
// removes 2 characters from string
return removeChar(str.split('').splice(index - 2, index).join());
}
Try this way (Descriptive comments are added in the below code snippet itself) :
// Input string
const str = "ABCXDEXFGHXIJK";
// split the input string based on 'X' and then remove the last item from each element by using String.slice() method.
const splittedStrArr = str.split('X').map(item => item = item.slice(0, -1));
// Output by joining the modified array elements.
console.log(splittedStr.join(''))
By using RegEx :
// Input string
const str = "ABCXDEXFGHXIJK";
// Replace the input string by matching the 'X' and one character before that with an empty string.
const modifiedStr = str.replace(/.X/g, "")
// Output
console.log(modifiedStr)
So I am trying to create a code for palindrome this is how I tried. Is there another or better way of doing it?
But now it only shows that if the first value is equal or not and shows true or false?
var inpo= prompt("Please enter to check if palindrome")
var inp = parseFloat(inpo)
var a = inpo.split('')
var inpo2 = a.reverse()
var len= inpo.length
for (var i =0;i< len ;i++) {
if (inpo[i] == inpo2[i] )
alert("True")
else
alert("False")
}
A way to check if a word or an entire phrase is a palindrome or not:
function isPalindrome(str) {
// Escape the string: Eliminate punctuation and spaces, enforce lower case
let escaped = str.replace(/[^A-Za-z0-9_]/g,"").toLowerCase();
// Reverse the escaped string
let reversed = escaped.split('').reverse().join('');
//compare
return escaped == reversed;
}
console.log(isPalindrome('Level'));
console.log(isPalindrome('Red rum, sir, is murder'));
I hope the comments serve well as an explanation.
Also, you have a prompt example in THIS jsfiddle.
If you are creating a palindrome checker code, here is a simple way to do it. Split, reverse then join.
str1 = "xxaa";
str2 = str1.split('').reverse().join("");
if (str1 == str2) {
alert("good");
} else {
alert("not");
}
You can check the single character from the string consuming 1 character from right and another from left until you will find either the string is finished or there are 2 inequal character. I implemented with a classical for loop.
Note that bracket notation for strings [] is only recently supported, you can use charAt if memory serves me right
let inp = "otdto";
console.log(isPalindrome(inp));
function isPalindrome(inp) {
const len = inp.length;
for (let i = 0; i < Math.floor(len / 2); i++)
if (inp[i] != inp[len - i - 1])
return false;
return true;
}
This question already has answers here:
Strip HTML from Text JavaScript
(44 answers)
removing html tags from string
(3 answers)
Closed 7 years ago.
I need to get rid of any text inside < and >, including the two delimiters themselves.
So for example, from string
<brev-y>th</brev-y><sw-ex>a</sw-ex><sl>t</sl>
I would like to get this one
that
This is what i've tried so far:
var str = annotation.split(' ');
str.substring(str.lastIndexOf("<") + 1, str.lastIndexOf(">"))
But it doesn't work for every < and >.
I'd rather not use RegEx if possible, but I'm happy to hear if it's the only option.
You can simply use the replace method with /<[^>]*>/g.It matches < followed by [^>]* any amount of non> until > globally.
var str = '<brev-y>th</brev-y><sw-ex>a</sw-ex><sl>t</sl>';
str = str.replace(/<[^>]*>/g, "");
alert(str);
For string removal you can use RegExp, it is ok.
"<brev-y>th</brev-y><sw-ex>a</sw-ex><sl>t</sl>".replace(/<\/?[^>]+>/g, "")
Since the text you want is always after a > character, you could split it at that point, and then the first character in each String of the array would be the character you need. For example:
String[] strings = stringName.split("<");
String word = "";
for(int i = 0; i < strings.length; i++) {
word += strings[i].charAt(0);
}
This is probably glitchy right now, but I think this would work. You don't need to actually remove the text between the "<>"- just get the character right after a '>'
Using a regular expression is not the only option, but it's a pretty good option.
You can easily parse the string to remove the tags, for example by using a state machine where the < and > characters turns on and off a state of ignoring characters. There are other methods of course, some shorter, some more efficient, but they will all be a few lines of code, while a regular expression solution is just a single replace.
Example:
function removeHtml1(str) {
return str.replace(/<[^>]*>/g, '');
}
function removeHtml2(str) {
var result = '';
var ignore = false;
for (var i = 0; i < str.length; i++) {
var c = str.charAt(i);
switch (c) {
case '<': ignore = true; break;
case '>': ignore = false; break;
default: if (!ignore) result += c;
}
}
return result;
}
var s = "<brev-y>th</brev-y><sw-ex>a</sw-ex><sl>t</sl>";
console.log(removeHtml1(s));
console.log(removeHtml2(s));
There are several ways to do this. Some are better than others. I haven't done one lately for these two specific characters, so I took a minute and wrote some code that may work. I will describe how it works. Create a function with a loop that copies an incoming string, character by character, to an outgoing string. Make the function a string type so it will return your modified string. Create the loop to scan from incoming from string[0] and while less than string.length(). Within the loop, add an if statement. When the if statement sees a "<" character in the incoming string it stops copying, but continues to look at every character in the incoming string until it sees the ">" character. When the ">" is found, it starts copying again. It's that simple.
The following code may need some refinement, but it should get you started on the method described above. It's not the fastest and not the most elegant but the basic idea is there. This did compile, and it ran correctly, here, with no errors. In my test program it produced the correct output. However, you may need to test it further in the context of your program.
string filter_on_brackets(string str1)
{
string str2 = "";
int copy_flag = 1;
for (size_t i = 0 ; i < str1.length();i++)
{
if(str1[i] == '<')
{
copy_flag = 0;
}
if(str1[i] == '>')
{
copy_flag = 2;
}
if(copy_flag == 1)
{
str2 += str1[i];
}
if(copy_flag == 2)
{
copy_flag = 1;
}
}
return str2;
}
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
How would I split a javascript string such as foo\nbar\nbaz to an array of lines, while preserving the newlines? I'd like to get ['foo\n', 'bar\n', 'baz'] as output;
I'm aware there are numerous possible answers - I'm just curious to find a stylish one.
With perl I'd use a zero-width lookbehind assertion: split /(?<=\n)/, but they are not supported in javascript regexs.
PS. Extra points for handling different line endings (at least \r\n) and handling the missing last newline (as in my example).
You can perform a global match with this pattern: /[^\n]+(?:\r?\n|$)/g
It matches any non-newline character then matches an optional \r followed by \n, or the end of the string.
var input = "foo\r\n\nbar\nbaz";
var result = input.match(/[^\n]+(?:\r?\n|$)/g);
Result: ["foo\r\n", "bar\n", "baz"]
how about this?
"foo\nbar\nbaz".split(/^/m);
Result
["foo
", "bar
", "baz"]
The other answers and answers in comments are all flawed in different ways. I needed a function that works correctly on any string or file.
Here is a simple and correct answer:
function split_lines(s) {
return s.match(/[^\n]*\n|[^\n]+/g);
}
input = "foo\r\n\nbar\n\r\nba\rz\r\r\r";
a = split_lines(input);
Array(5) [ "foo\r\n", "\n", "bar\n", "\r\n", "ba\rz\r\r\r" ]
It effectively splits at each newline \n but includes the \n, and includes a final line without trailing \n if and only if it is not empty. It includes all input characters in the output. We don't need any special treatment for \r.
I've tested this on a large chunk of random data, it does preserve all input characters, and \n only occur at the end of the lines.
Here's a test script:
function split_lines(s) {
return s.match(/[^\n]*\n|[^\n]+/g);
}
function gen_random_string(n, ncharset=256, nlprob=0.05, crprob=0.05) {
var s = "";
for (let i = 0; i < n; ++i) {
var r = Math.random();
if (r < nlprob)
s += "\n";
else if (r < nlprob + crprob)
s += "\r";
else {
var cc = Math.floor(r / (1 - nlprob - crprob) * ncharset);
var c = String.fromCharCode(cc);
s += c;
}
}
return s;
}
function test(...args) {
var s = gen_random_string(...args);
console.log(`generated random string of length ${s.length} with args:`, ...args);
var ok = true, ok1;
var a = split_lines(s);
console.log(`split into ${a.length} lines`);
ok1 = s === a.join('');
ok = ok && ok1;
console.log("split lines combine to give the original string?", ok1 ? "OK" : "FAIL");
for (var i = 0; i < a.length; ++i) {
var s1 = a[i];
ok1 = s1.endsWith("\n") || i == a.length-1;
ok = ok && ok1;
ok1 = !s1.slice(0, -1).includes("\n");
ok = ok && ok1;
}
console.log("tested each line other than the last ends with \\n");
console.log("tested each line does not contain \\n before the last character");
console.log("Final result", ok ? "OK" : "FAIL");
}
test(10000, 256);
test(10000, 65536);
I'd stay away from split with regular expressions since IE has a failed implementation of it. Use match instead.
"foo\nbar\nbaz".match(/^.*(\r?\n|$)/mg)
Result: ["foo\n", "bar\n", "baz"]
One simple but crude method would be first to replace "\n"s with a 2 special characters. Split on the second one, and replace the first with "\n" after splitting. Not efficient and not elegant, but definitely works.