I have:
var text = 'LTE CSSR (East xr) (301-LT_King_St_PC)'
and I want to split from (East xr) (301-LT_King_St_PC) only as follow :
var result = text.split('(')
result = (East xr) (301-LT_King_St_PC)
You can use a regular expression with the match function to get this done. Depending on how you want the result try one of the following:
var text = 'LTE CSSR (East xr) (301-LT_King_St_PC)'
console.log('one string:', text.match(/\(.*\)/)[0])
console.log('array of strings:', text.match(/\([^\)]*\)/g))
The first does what you seem to be asking for - a single output of everything between the first ( and the second ). It does this by searching for /\(.*\)/ which is a regex that says "everything in between the parentheses".
The second match splits it into all parenthesised strings using /\([^\)]*\)/g which is a regex that says "each parenthesised string" but the g at the end says "all of those" so an array of each is given.
You can do it using substring() and indexOf() :
var text = 'LTE CSSR (East xr) (301-LT_King_St_PC)';
var result = text.substring(text.indexOf('('));
console.log(result);
You're quite close with your current method. One way you could try this is to use multiple characters within your split
var result = text.split(") (")
# Output -> result = ["(East xr", "301-LT_King_St_PC)"]
From here, some string manipulation could get rid of the brackets.
Alternatively you can also use String.match and join its result:
const text = 'LTE CSSR (East xr) (301-LT_King_St_PC)';
const cleaned = text.match(/\(.+[^\)]\)/).join(` `);
console.log(cleaned);
Related
I have string like below
BANKNIFTY-13-FEB-2020-31200-ce
I want to convert the string to 13-FEB-31200-ce
so I tried below code
str.match(/(.*)-(?:.*)-(?:.*)-(.*)-(?:.*)-(?:.*)/g)
But its returning whole string
Two capture groups is probably the way to go. Now you have two options to use it. One is match which requires you to put the two pieces together
var str = 'BANKNIFTY-13-FEB-2020-31200-ce'
var match = str.match(/[^-]+-(\d{2}-[A-Z]{3}-)\d{4}-(.*)/)
// just reference the two groups
console.log(`${match[1]}${match[2]}`)
// or you can remove the match and join the remaining
match.shift()
console.log(match.join(''))
Or just string replace which you do the concatenation of the two capture groups in one line.
var str = 'BANKNIFTY-13-FEB-2020-31200-ce'
var match = str.replace(/[^-]+-(\d{2}-[A-Z]{3}-)\d{4}-(.*)/, '$1$2')
console.log(match)
Regex doesn't seem to be the most appropriate tool here. Why not use simple .split?
let str = 'BANKNIFTY-13-FEB-2020-31200-ce';
let splits = str.split('-');
let out = [splits[1], splits[2], splits[4], splits[5]].join('-');
console.log(out);
If you really want to use regexp,
let str = 'BANKNIFTY-13-FEB-2020-31200-ce';
let splits = str.match(/[^-]+/g);
let out = [splits[1], splits[2], splits[4], splits[5]].join('-');
console.log(out);
I would not use Regex at all if you know exact positions. Using regex is expensive and should be done differently if there is way. (https://blog.codinghorror.com/regular-expressions-now-you-have-two-problems/)
const strArr = "BANKNIFTY-13-FEB-2020-31200-ce".split("-"); // creates array
strArr.splice(0,1); // remove first item
strArr.splice(2,1); // remove 2020
const finalStr = strArr.join("-");
If the pattern doesn't need to be too specific.
Then just keep it simple and only capture what's needed.
Then glue the captured groups together.
let str = 'BANKNIFTY-13-FEB-2020-31200-ce';
let m = str.match(/^\w+-(\d{1,2}-[A-Z]{3})-\d+-(.*)$/)
let result = m ? m[1]+'-'+m[2] : undefined;
console.log(result);
In this regex, ^ is the start of the string and $ the end of the string.
You can have something like this by capturing groups with regex:
const regex = /(\d{2}\-\w{3})(\-\d{4})(\-\d{5}\-\w{2})/
const text = "BANKNIFTY-13-FEB-2020-31200-ce"
const [, a, b, c] = text.match(regex);
console.log(`${a}${c}`)
I have the following string
"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,
I need to get string after "ssu":" the Result should be 89c4eef0-3a0d-47ae-a97f-42adafa7cf8f. How do I do it in Javascript but very simple? I am thinking to collect 36 character after "ssu":".
You could build a valid JSON string and parse it and get the wanted property ssu.
var string = '"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,',
object = JSON.parse(`{${string.slice(0, -1)}}`), // slice for removing the last comma
ssu = object.ssu;
console.log(ssu);
One solution would be to use the following regular expression:
/\"ssu\":\"([\w-]+)\"/
This pattern basically means:
\"ssu\":\" , start searching from the first instance of "ssu":"
([\w-]+) , collect a "group" of one or more alphanumeric characters \w and hypens -
\", look for a " at the end of the group
Using a group allows you to extract a portion of the matched pattern via the String#match method that is of interest to you which in your case is the guid that corresponds to ([\w-]+)
A working example of this would be:
const str = `"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,`
const value = str.match(/\"ssu\":\"([\w-]+)\"/)[1]
console.log(value);
Update: Extract multiple groupings that occour in string
To extract values for multiple occurances of the "ssu" key in your input string, you could use the String#matchAll() method to achieve that as shown:
const str = `"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,"ssu":"value-of-second-ssu","ssu":"value-of-third-ssu"`;
const values =
/* Obtain array of matches for pattern */
[...str.matchAll(/\"ssu\":\"([\w-]+)\"/g)]
/* Extract only the value from pattern group */
.map(([,value]) => value);
console.log(values);
Note that for this to work as expected, the /g flag must be added to the end of the original pattern. Hope that helps!
Use this regExp: /(?!"ssu":")(\w+-)+\w+/
const str = '"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,';
const re = /(?!"ssu":")(\w+-)+\w+/;
const res = str.match(re)[0];
console.log(res);
You can use regular expressions.
var str = '"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,'
var minhaRE = new RegExp("[a-z|0-9]*-[a-z|0-9|-]*");
minhaRE.exec(str)
OutPut: Array [ "89c4eef0-3a0d-47ae-a97f-42adafa7cf8f" ]
Looks almost like a JSON string.
So with a small change it can be parsed to an object.
var str = '"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049, ';
var obj = JSON.parse('{'+str.replace(/[, ]+$/,'')+'}');
console.log(obj.ssu)
I have a string that has the following format: <strong>FirstName LastName</strong>
How can I change this into an array with the first element firstName and second lastName?
I did this, but no luck, it won't produce the right result:
var data = [myString.split('<strong>')[1], myString.split('<strong>')[2]]
How can I produce ["firstName", "lastName"] for any string with that format?
In order to parse HTML, use the best HTML parser out there, the DOM itself!
// create a random element, it doesn't have to be 'strong' (e.g., it could be 'div')
var parser = document.createElement('strong');
// set the innerHTML to your string
parser.innerHTML = "<strong>FirstName LastName</strong>";
// get the text inside the element ("FirstName LastName")
var fullName = parser.textContent;
// split it into an array, separated by the space in between FirstName and LastName
var data = fullName.split(" ");
// voila!
console.log(data);
EDIT
As #RobG pointed out, you could also explicitly use a DOM parser rather than that of an element:
var parser = new DOMParser();
var doc = parser.parseFromString("<strong>FirstName LastName</strong>", "text/html");
console.log(doc.body.textContent.split(" "));
However, both methods work perfectly fine; it all comes down to preference.
Just match everything between <strong> and </strong>.
var matches = "<strong>FirstName LastName</strong>".match(/<strong>(.*)<\/strong>/);
console.log(matches[1].split(' '));
The preferred approach would be to use DOM methods; create an element and get the .textContent then match one or more word characters or split space character.
let str = '<strong>FirstName LastName</strong>';
let [,first, last] = str.split(/<[/\w\s-]+>|\s/g);
console.log(first, last);
/<[/\w\s-]+>|\s/g
Splits < followed by one or more word, space or dash characters characters followed by > character or space to match space between words in the string.
Comma operator , within destructuring assignment is used to omit that index from the result of .split() ["", "FirstName", "LastName", ""].
this is my approach of doing your problem. Hope it helps!
var str = "<strong>FirstName LastName</strong>";
var result = str.slice(0, -9).substr(8).split(" ");
Edit: it will only work for this specific example.
Another way to do this in case you had something other than an html
var string = "<strong>FirstName LastName</strong>";
string = string.slice(0, -9); // remove last 9 chars
string = string.substr(8); // remove first 8 chars
string = string.split(" "); // split into an array at space
console.log(string);
In my Javascript code, I get one very long line as a string.
This one line only has around 65'000 letters. Example:
config=123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...
What I have to do is replace all & with an break (\n) first and then pick only the line which starts with "path_of_code=". This line I have to write in a variable.
The part with replace & with an break (\n) I already get it, but the second task I didn't.
var obj = document.getElementById('div_content');
var contentJS= obj.value;
var splittedResult;
splittedResult = contentJS.replace(/&/g, '\n');
What is the fastest way to do it? Please note, the list is usually very long.
It sounds like you want to extract the text after &path_of_code= up until either the end of the string or the next &. That's easily done with a regular expression using a capture group, then using the value of that capture group:
var rex = /&path_of_code=([^&]+)/;
var match = rex.exec(theString);
if (match) {
var text = match[1];
}
Live Example:
var theString = "config=123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...";
var rex = /&path_of_code=([^&]+)/;
var match = rex.exec(theString);
if (match) {
var text = match[1];
console.log(text);
}
Use combination of String.indexOf() and String.substr()
var contentJS= "123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...";
var index = contentJS.indexOf("&path_of_code"),
substr = contentJS.substr(index+1),
res = substr.substr(0, substr.indexOf("&"));
console.log(res)
but the second task I didn't.
You can use filter() and startsWith()
splittedResult = splittedResult.filter(i => i.startsWith('path_of_code='));
I have a text which goes like this...
var string = '~a=123~b=234~c=345~b=456'
I need to extract the string such that it splits into
['~a=123~b=234~c=345','']
That is, I need to split the string with /b=.*/ pattern but it should match the last found pattern. How to achieve this using RegEx?
Note: The numbers present after the equal is randomly generated.
Edit:
The above one was just an example. I did not make the question clear I guess.
Generalized String being...
<word1>=<random_alphanumeric_word>~<word2>=<random_alphanumeric_word>..~..~..<word2>=<random_alphanumeric_word>
All have random length and all wordi are alphabets, the whole string length is not fixed. the only text known would be <word2>. Hence I needed RegEx for it and pattern being /<word2>=.*/
This doesn't sound like a job for regexen considering that you want to extract a specific piece. Instead, you can just use lastIndexOf to split the string in two:
var lio = str.lastIndexOf('b=');
var arr = [];
var arr[0] = str.substr(0, lio);
var arr[1] = str.substr(lio);
http://jsfiddle.net/NJn6j/
I don't think I'd personally use a regex for this type of problem, but you can extract the last option pair with a regex like this:
var str = '~a=123~b=234~c=345~b=456';
var matches = str.match(/^(.*)~([^=]+=[^=]+)$/);
// matches[1] = "~a=123~b=234~c=345"
// matches[2] = "b=456"
Demo: http://jsfiddle.net/jfriend00/SGMRC/
Assuming the format is (~, alphanumeric name, =, and numbers) repeated arbitrary number of times. The most important assumption here is that ~ appear once for each name-value pair, and it doesn't appear in the name.
You can remove the last token by a simple replacement:
str.replace(/(.*)~.*/, '$1')
This works by using the greedy property of * to force it to match the last ~ in the input.
This can also be achieved with lastIndexOf, since you only need to know the index of the last ~:
str.substring(0, (str.lastIndexOf('~') + 1 || str.length() + 1) - 1)
(Well, I don't know if the code above is good JS or not... I would rather write in a few lines. The above is just for showing one-liner solution).
A RegExp that will give a result that you may could use is:
string.match(/[a-z]*?=(.*?((?=~)|$))/gi);
// ["a=123", "b=234", "c=345", "b=456"]
But in your case the simplest solution is to split the string before extract the content:
var results = string.split('~'); // ["", "a=123", "b=234", "c=345", "b=456"]
Now will be easy to extract the key and result to add to an object:
var myObj = {};
results.forEach(function (item) {
if(item) {
var r = item.split('=');
if (!myObj[r[0]]) {
myObj[r[0]] = [r[1]];
} else {
myObj[r[0]].push(r[1]);
}
}
});
console.log(myObj);
Object:
a: ["123"]
b: ["234", "456"]
c: ["345"]
(?=.*(~b=[^~]*))\1
will get it done in one match, but if there are duplicate entries it will go to the first. Performance also isn't great and if you string.replace it will destroy all duplicates. It would pass your example, but against '~a=123~b=234~c=345~b=234' it would go to the first 'b=234'.
.*(~b=[^~]*)
will run a lot faster, but it requires another step because the match comes out in a group:
var re = /.*(~b=[^~]*)/.exec(string);
var result = re[1]; //~b=234
var array = string.split(re[1]);
This method will also have the with exact duplicates. Another option is:
var regex = /.*(~b=[^~]*)/g;
var re = regex.exec(string);
var result = re[1];
// if you want an array from either side of the string:
var array = [string.slice(0, regex.lastIndex - re[1].length - 1), string.slice(regex.lastIndex, string.length)];
This actually finds the exact location of the last match and removes it regex.lastIndex - re[1].length - 1 is my guess for the index to remove the ellipsis from the leading side, but I didn't test it so it might be off by 1.