Is there a way to make this string:
foo
bar
Appear like this:
foo\r\n\tbar
It would really help with debugging a lexer.
The key is to escape \n in a string replace.
let specialCharacters = [
{regex: /\n/g, replacement: '\\n'},
{regex: /\t/g, replacement: '\\t'}
];
function escapeSpecialCharacters(str){
specialCharacters.forEach(c => {
str = str.replace(c.regex, c.replacement);
});
return str;
}
console.log(escapeSpecialCharacters(`test
test
test
1234`));
If debugging is all you want to do, you can display all escaped characters it in the browser console, but putting the string into an array:
let string = "test\ntest";
let arr = [];
arr.push(string);
console.log(arr);
But that will not change your string in any way, it still contains new line, instead of "\n".
Related
I parse the data from website and try to change to a json object.
Here is my function:
function outPutJSON() {
for (var i = 0; i < movieTitle.length; i++) {
var handleN = movieContent[i];
console.log('===\n');
console.log(handleN);
data.movie.push({
mpvieTitle: movieTitle[i],
movieEnTitle: movieEnTitle[i],
theDate: theDate[i],
theLength: theLength[i],
movieVersion: movieVersion[i],
youtubeId: twoId[i],
content: movieContent[i]
});
};
return JSON.stringify(data);
}
console.log will print movieContent[0] like:
but i return JSON.stringfy(data);
it will become:
There are so many /n i want to remove it.
I try to change return JSON.stringfy(data); to this:
var allMovieData = JSON.stringify(data);
allMovieData = allMovieData.replace(/\n/g, '');
return allMovieData;
It's not working the result is the same.
How to remove /n when i use JSON.stringfy() ?
Any help would be appreciated . Thanks in advance.
In your data screenshots, you literally see "\n".
This probably means that the actual string doesn't contain a newline character (\n), but a escaped newline character (\\n).
A newline character would have been rendered as a linebreak. You wouldn't see the \n.
To remove those, use .replace(/\\n/g, '') instead of .replace(/\n/g, '')
just :=>
JSON.stringify(JSON.parse(<json object>))
JSON.stringify converts new lines (\n) and tab (\t) chars into string, so, when you will try to parse it, the string will contain those again.
So, you need to search the string \n, you can do that with something like that.
const stringWithNewLine = {
x: `this will conatin
new lines`
};
const json = JSON.stringify(stringWithNewLine);
console.log(json.replace(/\\n/g, ''))
I know this is an old question but it is always good to have options. You can use a string literal as a simple wrapper. The string literal will honor the string's line breaks, empty spaces etc. Like:
let jsonAsPrettyString = `${JSON.stringify(jsonObject, null, 2)}`;
I am writing js code to get array of elements after splitting using regular expression.
var data = "ABCXYZ88";
var regexp = "([A-Z]{3})([A-Z]{3}d{2})";
console.log(data.split(regexp));
It returns
[ 'ABCXYZ88' ]
But I am expecting something like
['ABC','XYZ','88']
Any thoughts?
I fixed your regex, then matched it against your string and extracted the relevant capturing groups:
var regex = /([A-Z]{3})([A-Z]{3})(\d{2})/g;
var str = 'ABCXYZ88';
let m = regex.exec(str);
if (m !== null) {
console.log(m.slice(1)); // prints ["ABC", "XYZ", "88"]
}
In your case, I don't think you can split using a regex as you were trying, as there don't seem to be any delimiting characters to match against. For this to work, you'd have to have a string like 'ABC|XYZ|88'; then you could do 'ABC|XYZ|88'.split(/\|/g). (Of course, you wouldn't use a regex for such a simple case.)
Your regexp is not a RegExp object but a string.
Your capturing groups are not correct.
String.prototype.split() is not the function you need. What split() does:
var myString = 'Hello World. How are you doing?';
var splits = myString.split(' ', 3);
console.log(splits); // ["Hello", "World.", "How"]
What you need:
var data = 'ABCXYZ88';
var regexp = /^([A-Z]{3})([A-Z]{3})(\d{2})$/;
var match = data.match(regexp);
console.log(match.slice(1)); // ["ABC", "XYZ", "88"]
Try this. I hope this is what you are looking for.
var reg = data.match(/^([A-Z]{3})([A-Z]{3})(\d{2})$/).slice(1);
https://jsfiddle.net/m5pgpkje/1/
Given an input field, I'm trying to use a regex to find all the URLs in the text fields and make them links. I want all the information to be retained, however.
So for example, I have an input of "http://google.com hello this is my content" -> I want to split that by the white space AFTER this regex pattern from another stack overflow question (regexp = /(ftp|http|https)://(\w+:{0,1}\w*#)?(\S+)(:[0-9]+)?(/|/([\w#!:.?+=&%#!-/]))?/) so that I end up with an array of ['http://google.com', 'hello this is my content'].
Another ex: "hello this is my content http://yahoo.com testing testing http://google.com" -> arr of ['hello this is my content', 'http://yahoo.com', 'testing testing', 'http://google.com']
How can this be done? Any help is much appreciated!
First transform all the groups in your regular expression into non-capturing groups ((?:...)) and then wrap the whole regular expression inside a group, then use it to split the string like this:
var regex = /((?:ftp|http|https):\/\/(?:\w+:{0,1}\w*#)?(?:\S+)(?::[0-9]+)?(?:\/|\/(?:[\w#!:.?+=&%#!-/]))?)/;
var result = str.split(regex);
Example:
var str = "hello this is my content http://yahoo.com testing testing http://google.com";
var regex = /((?:ftp|http|https):\/\/(?:\w+:{0,1}\w*#)?(?:\S+)(?::[0-9]+)?(?:\/|\/(?:[\w#!:.?+=&%#!-/]))?)/;
var result = str.split(regex);
console.log(result);
You had few unescaped backslashes in your RegExp.
var str = "hello this is my content http://yahoo.com testing testing http://google.com";
var captured = str.match(/(ftp|http|https):\/\/(\w+:{0,1}\w*#)?(\S+)(:[0-9]+)?(\/|\/([\w#!:.?+=&%#!-/]))?/g);
var nonCaptured = [];
str.split(' ').map((v,i) => captured.indexOf(v) == -1 ? nonCaptured.push(v) : null);
console.log(nonCaptured, captured);
Remove strings with numbers and special characters using regular expression.Here is my code
var string = "[Account0].&[1]+[Account1].&[2]+[Account2].&[3]+[Account3].&[4]";
var numbers = string.match(/(\d+)/gi);
alert(numbers.join(','));
here output is : 0,1,1,2,2,3,3,4
But i want the following output 1,2,3,4
Can any one please help me.
Thanks,
Seemd what you want is [\d+], use exec like this,
var myRe = /\[(\d+)\]/gi;
var myArray, numbers = [];
while ((myArray = myRe.exec(string)) !== null) {
numbers.push(myArray[1]);
}
http://jsfiddle.net/xE265/
You can do:
string = "[Account0].&[1]+[Account1].&[2]+[Account2].&[3]+[Account3].&[4]";
repl = string.replace(/.*?\[(\d+)\][^\[]*/g, function($0, $1) { return $1 });
//=> "1234"
I guess the simplest solution in this case would be:
\[(\d+)\]
simply saying that you only want the digits enclosed by brackets.
Regards
I have the following string:
",'first string','more','even more'"
I want to transform this into an Array but obviously this is not valid due to the first comma. How can I remove the first comma from my string and make it a valid Array?
I’d like to end up with something like this:
myArray = ['first string','more','even more']
To remove the first character you would use:
var myOriginalString = ",'first string','more','even more'";
var myString = myOriginalString.substring(1);
I'm not sure this will be the result you're looking for though because you will still need to split it to create an array with it. Maybe something like:
var myString = myOriginalString.substring(1);
var myArray = myString.split(',');
Keep in mind, the ' character will be a part of each string in the split here.
In this specific case (there is always a single character at the start you want to remove) you'll want:
str.substring(1)
However, if you want to be able to detect if the comma is there and remove it if it is, then something like:
if (str[0] == ',') {
str = str.substring(1);
}
One-liner
str = str.replace(/^,/, '');
I'll be back.
var s = ",'first string','more','even more'";
var array = s.split(',').slice(1);
That's assuming the string you begin with is in fact a String, like you said, and not an Array of strings.
Assuming the string is called myStr:
// Strip start and end quotation mark and possible initial comma
myStr=myStr.replace(/^,?'/,'').replace(/'$/,'');
// Split stripping quotations
myArray=myStr.split("','");
Note that if a string can be missing in the list without even having its quotation marks present and you want an empty spot in the corresponding location in the array, you'll need to write the splitting manually for a robust solution.
var s = ",'first string','more','even more'";
s.split(/'?,'?/).filter(function(v) { return v; });
Results in:
["first string", "more", "even more'"]
First split with commas possibly surrounded by single quotes,
then filter the non-truthy (empty) parts out.
To turn a string into an array I usually use split()
> var s = ",'first string','more','even more'"
> s.split("','")
[",'first string", "more", "even more'"]
This is almost what you want. Now you just have to strip the first two and the last character:
> s.slice(2, s.length-1)
"first string','more','even more"
> s.slice(2, s.length-2).split("','");
["first string", "more", "even more"]
To extract a substring from a string I usually use slice() but substr() and substring() also do the job.
s=s.substring(1);
I like to keep stuff simple.
You can use directly replace function on javascript with regex or define a help function as in php ltrim(left) and rtrim(right):
1) With replace:
var myArray = ",'first string','more','even more'".replace(/^\s+/, '').split(/'?,?'/);
2) Help functions:
if (!String.prototype.ltrim) String.prototype.ltrim = function() {
return this.replace(/^\s+/, '');
};
if (!String.prototype.rtrim) String.prototype.rtrim = function() {
return this.replace(/\s+$/, '');
};
var myArray = ",'first string','more','even more'".ltrim().split(/'?,?'/).filter(function(el) {return el.length != 0});;
You can do and other things to add parameter to the help function with what you want to replace the char, etc.
this will remove the trailing commas and spaces
var str = ",'first string','more','even more'";
var trim = str.replace(/(^\s*,)|(,\s*$)/g, '');
remove leading or trailing characters:
function trimLeadingTrailing(inputStr, toRemove) {
// use a regex to match toRemove at the start (^)
// and at the end ($) of inputStr
const re = new Regex(`/^${toRemove}|{toRemove}$/`);
return inputStr.replace(re, '');
}