I have this array
(2) ['beginning=beginner', 'leaves=leave']
and this string
its beginner in the sounds leave
which i have converted to an array
var words = text.split(' ');
i want to replace beginner with beginning and leave with leaves it can be any dynamic words but for now it has only two elements i can replace it within for loop. Is it possible with map method.
this.words.map((words, i) => console.log(words));
Note: Only first instance should get replaced.
Any Solution Thanks
does this correct with your question ?
const arrString = ["beginning=beginner", "leaves=leave", "sound=sounds"];
let string = "its beginner in the sounds leave";
arrString.forEach((mapString) => {
const stringArr = mapString.split("=");
string = string.replace(stringArr[1], stringArr[0]);
});
console.log("string", string);
// "its beginning in the sound leaves"
You can do it in without a nested loop too if you compromise with space.
Here is a solution which creates a mapping object for replacement values and uses array methods like map(),forEach(),join() and string method like split()
const arrString = ["beginning=beginner", "leaves=leave", "sound=sounds"];
let string1 = "its beginner in the sounds leave beginner";
const replaceObj = {};
const arrBreaks = arrString.forEach(x => {
let els = x.split("=");
replaceObj[els[1]] = els[0]; });
const ans = string1.split(' ').map((x) => {
if(x in replaceObj) { let val = replaceObj[x]; delete val; return val; }
return x;
}).join(' ');
console.log(ans);
I have a string that comes to me like this: "[value1][value2]"
How can I get the values that are inside the square brackets?
NOTE: if the string is like this "[][value2]" the first bracket that has a space must return a "" to me...
I have been trying a lot of regex and split but none workd.
this is the last I tried:
var pattern = /[([^]]*)]/g;
var res = pattern.exec(datos[0].title);
Another one I tried is:
var res = datos[0].title.match(/^.*?[([^]]*)].*?[([^]]*)]/gm);
but none do what I need...
I'm trying to find a way "that does it all" a regex that gets anything inside the Square brackets (even white spaces)
As #HarryCutts stated, you don't need regex:
var x = "[value1][value2]";
console.log( x.slice(1,-1).split('][') );
You can try this regex and the brute force way to extract the contents.
var regex = /\[(.*?)\]/g;
var value = "[value1][value2][]";
var matches = value.match(regex);
var matchedValues = matches.map(match => {
return match.replace("[", "").replace("]", "");
}).join(" ");
console.log(matchedValues.toString())
You could just do this:
var str = "['value1']['value2']";
var value1 = str.split("]")[0].split("[")[1];
var value2 = str.split("]")[1].split("[")[1];
console.log(str);
console.log(value1);
console.log(value2);
You can easily expand it for more values.
const string = "[value1][value2]";
const removeBrackets = (stringWithBrackets) => {
return stringWithBrackets.split("][").map(s => s = s.replace(/\[*\]*/g, ""));
};
const [value1, value2] = removeBrackets(string);
console.log(value1, value2);
const getItems = (fullItemsString) => {
let items = fullItemsString.replace(/\[/g, "").split("]");
items.pop()
return items;
}
Using:
let items = getItems("[2][][34][1]");
result: [ '2', '', '34', '1' ]
I need to check each element of an array for a certain letter. if the element contains this letter it should stay within the array, if it does not it should be removed.
currently I'm able to remove elements if they are exact, but not sure how to go about checking each index of each element.
var exampleList = ['alpha', 'beta','dog']
exampleList.filter(function(word) {
return (word == 'dog');
})
my end goal would be something like this.
letterToInclude = 'a'
var exampleList = ['alpha', 'beta','dog', 'mouse']
exampleList.filter(function(word) {
return (word == letterToInclude);
})
// returned values = ['alpha', 'beta']
Instead of doing ==, you can use indexOf to see if letterToInclude occurs in word.
letterToInclude = 'a'
var exampleList = ['alpha', 'beta','dog', 'mouse']
exampleList.filter(function(word) {
return (word.indexOf(letterToInclude) > -1);
});
indexOf returns the position that letterToInclude occurs in word; in the event it is not found, indexOf will return -1.
The reason I didn't use word.includes(letterToInclude) is just for compatibility purposes. includes is fairly new and isn't completely supported.
You can use indexOf() to check if each element contains a specific letter.
var letterToInclude = 'a'
var exampleList = ['alpha', 'beta', 'dog', 'mouse']
exampleList = exampleList.filter(function(word) {
return word.indexOf(letterToInclude) != -1
})
console.log(exampleList)
ES6 solution with String#includes(), also you can use match() or test() but those take regular expressions.
exampleList.filter(word => word.includes(letterToInclude))
Try using indexOf:
letterToInclude = 'a'
var exampleList = ['alpha', 'beta','dog', 'mouse']
console.log(exampleList.filter(function(word) {
return ~(word.indexOf(letterToInclude));
}));
It works well together with Unicode chars:
var lettersToInclude = 'aa'
var exampleList = ['a', 'aa','❤✓☀a★bAb', 'bbaa', 'aa☂']
var r = exampleList.filter(function(word) {
return (word.indexOf(lettersToInclude) > -1);
});
console.log(r);
I have a string that I want to split into an array. The string looks like this:
'O:BED,N:KET,OT,N:JAB,FA,O:RPT,'
The string can contain any number of objects eg
'O:BED,N:KET,OT,N:JAB,FA,O:RPT,X:BLA,GTO'
I want to split this string on the instance of \w: eg O:
So I'll end up with array like this:
['O:BED','N:KET, OT','N:JAB,FA','O:RPT']
I am using the following code:
var array = st.split(/^(\w:.+)(?=\w:)/g);
However I end up with array like this :
['','O:BED,N:KET,OT,N:JAB,FA,','O:RPT,']
It seems the regex is being greedy, what should I do to fix it?
Note I am using angularjs and eventually I want to end up with this :
var objs = [
{type:O,code: BED, suf: ''},
{type:N, code: KET, suf: OT},
{type:N, code: JAB, suf: FA},
{type:O, code: RPT, suf: ''}
]
It would be much easier if your string is formatted properly. But still we can achieve the task with extra effort. Hope the below code works for you.
var str = 'O:BED,N:KET,OT,N:JAB,FA,O:RPT,X:BLA,GTO';
var a = str.split(',');
var objs = [], obj, item, suf;
for(var i=0; i<a.length;){
item = a[i].split(':');
if(a[i+1] && a[i+1].indexOf(':') == -1){
suf = a[i+1];
i++;
}else{
suf = "";
}
obj = {
type: item[0],
code: item[1],
suf: suf
};
objs.push(obj);
i++;
}
console.log(objs);
You can use the RegExp.prototype.exec method to obtain successive matches instead of splitting the string with a delimiter:
var myStr = 'O:BED,N:KET,OT,N:JAB,FA,O:RPT,';
var myRe = /([^,:]+):([^,:]+)(?:,([^,:]+))??(?=,[^,:]+:|,?$)/g;
var m;
var result = [];
while ((m = myRe.exec(myStr)) !== null) {
result.push({type:m[1], code:m[2], suf:((m[3])?m[3]:'')});
}
console.log(result);
You want to do a string match and then iterate over that.
Full example inside AngularJS: http://jsfiddle.net/184cyspg/1/
var myString = 'O:BED,N:KET,OT,N:JAB,FA,O:RPT,';
$scope.myArray = [];
var objs = myString.match(/([A-Z])\:([A-Z]*)\,([A-Z]?)/g);
objs.forEach(function (entry) {
var obj = entry.replace(',', ':');
obj = obj.split(':');
$scope.myArray.push({type: obj[0], code: obj[1], suf: obj[2]});
});
I love regular expressions :)
This will match each object of your string, if you want to use the global flag and exec() through all the matches:
(\w):(\w+)(?:,((?!\w:)\w+))?
The only real trick is to only treat the next bit after the comma as the suffix to this one if it doesn't look like the type of the next.
Each match captures the groups:
type
code
suf
If you just want to split as you said, then the solution to your greedy problem is to tell it to split on commas which are followed by those matching objects, eg:
,(?=(\w):(\w+)(?:,((?!\w:)\w+))?)
The following does not solve your regex issue however is an alternative approach to introduce underscorejs to handle from simple to more complex operations. Although an overkill in this case;
// ie. input string = 'O:BED,N:KET,OT,N:JAB,FA,O:RPT,';
.controller('AppCtrl', [function() {
/**
* Split by comma then (chain) eval each (map)
* element that (if-else) contains '0:' is pushed
* into array as a new element, otherwise concat element
*
* :#replace hardcoded values with params
*
* #param String string - a string to split
* #param String prefix - prefix to determine start of new array element ie. '0:'
* #param String delimiter - delimiter to split string ie ','
* #return Array array of elements by prefix
*/
$scope.splitter = function(string) {
var a = [];
var tmp = "";
_.chain(string.split(','))
.map(function(element) {
if(element.indexOf('O:') >= 0) {
element += tmp;
a.push(element);
tmp = "";
} else {
tmp += element;
}
});
return a;
};
}]);
Output:
array: Array[2]
0: "O:BED"
1: "O:RPTN:KETOTN:JABFA"
length: 2
Updated: Just read your requirements on Objects. underscorejs allows chaining operations. For example, the code above could be tweaked to handle Objects, chained to .compact().object().value() to produce output as Object k:v pairs;
Hope this helps.
Pretty straight forward. In javascript, I need to check if a string contains any substrings held in an array.
There's nothing built-in that will do that for you, you'll have to write a function for it, although it can be just a callback to the some array method.
Two approaches for you:
Array some method
Regular expression
Array some
The array some method (added in ES5) makes this quite straightforward:
if (substrings.some(function(v) { return str.indexOf(v) >= 0; })) {
// There's at least one
}
Even better with an arrow function and the newish includes method (both ES2015+):
if (substrings.some(v => str.includes(v))) {
// There's at least one
}
Live Example:
const substrings = ["one", "two", "three"];
let str;
// Setup
console.log(`Substrings: ${substrings}`);
// Try it where we expect a match
str = "this has one";
if (substrings.some(v => str.includes(v))) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
// Try it where we DON'T expect a match
str = "this doesn't have any";
if (substrings.some(v => str.includes(v))) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
Regular expression
If you know the strings don't contain any of the characters that are special in regular expressions, then you can cheat a bit, like this:
if (new RegExp(substrings.join("|")).test(string)) {
// At least one match
}
...which creates a regular expression that's a series of alternations for the substrings you're looking for (e.g., one|two) and tests to see if there are matches for any of them, but if any of the substrings contains any characters that are special in regexes (*, [, etc.), you'd have to escape them first and you're better off just doing the boring loop instead. For info about escaping them, see this question's answers.
Live Example:
const substrings = ["one", "two", "three"];
let str;
// Setup
console.log(`Substrings: ${substrings}`);
// Try it where we expect a match
str = "this has one";
if (new RegExp(substrings.join("|")).test(str)) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
// Try it where we DON'T expect a match
str = "this doesn't have any";
if (new RegExp(substrings.join("|")).test(str)) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
One line solution
substringsArray.some(substring=>yourBigString.includes(substring))
Returns true\false if substring exists\does'nt exist
Needs ES6 support
var yourstring = 'tasty food'; // the string to check against
var substrings = ['foo','bar'],
length = substrings.length;
while(length--) {
if (yourstring.indexOf(substrings[length])!=-1) {
// one of the substrings is in yourstring
}
}
function containsAny(str, substrings) {
for (var i = 0; i != substrings.length; i++) {
var substring = substrings[i];
if (str.indexOf(substring) != - 1) {
return substring;
}
}
return null;
}
var result = containsAny("defg", ["ab", "cd", "ef"]);
console.log("String was found in substring " + result);
For people Googling,
The solid answer should be.
const substrings = ['connect', 'ready'];
const str = 'disconnect';
if (substrings.some(v => str === v)) {
// Will only return when the `str` is included in the `substrings`
}
Here's what is (IMO) by far the best solution. It's a modern (ES6) solution that:
is efficient (one line!)
avoids for loops
unlike the some() function that's used in the other answers, this one doesn't just return a boolean (true/false)
instead, it either returns the substring (if it was found in the array), or returns undefined
goes a step further and allows you to choose whether or not you need partial substring matches (examples below)
Enjoy!
const arrayOfStrings = ['abc', 'def', 'xyz'];
const str = 'abc';
const found = arrayOfStrings.find(v => (str === v));
Here, found would be set to 'abc' in this case. This will work for exact string matches.
If instead you use:
const found = arrayOfStrings.find(v => str.includes(v));
Once again, found would be set to 'abc' in this case. This doesn't allow for partial matches, so if str was set to 'ab', found would be undefined.
And, if you want partial matches to work, simply flip it so you're doing:
const found = arrayOfStrings.find(v => v.includes(str));
instead. So if str was set to 'ab', found would be set to 'abc'.
Easy peasy!
var str = "texttexttext";
var arr = ["asd", "ghj", "xtte"];
for (var i = 0, len = arr.length; i < len; ++i) {
if (str.indexOf(arr[i]) != -1) {
// str contains arr[i]
}
}
edit:
If the order of the tests doesn't matter, you could use this (with only one loop variable):
var str = "texttexttext";
var arr = ["asd", "ghj", "xtte"];
for (var i = arr.length - 1; i >= 0; --i) {
if (str.indexOf(arr[i]) != -1) {
// str contains arr[i]
}
}
substringsArray.every(substring=>yourBigString.indexOf(substring) === -1)
For full support ;)
For full support (additionally to #ricca 's verions).
wordsArray = ['hello', 'to', 'nice', 'day']
yourString = 'Hello. Today is a nice day'.toLowerCase()
result = wordsArray.every(w => yourString.includes(w))
console.log('result:', result)
If the array is not large, you could just loop and check the string against each substring individually using indexOf(). Alternatively you could construct a regular expression with substrings as alternatives, which may or may not be more efficient.
Javascript function to search an array of tags or keywords using a search string or an array of search strings. (Uses ES5 some array method and ES6 arrow functions)
// returns true for 1 or more matches, where 'a' is an array and 'b' is a search string or an array of multiple search strings
function contains(a, b) {
// array matches
if (Array.isArray(b)) {
return b.some(x => a.indexOf(x) > -1);
}
// string match
return a.indexOf(b) > -1;
}
Example usage:
var a = ["a","b","c","d","e"];
var b = ["a","b"];
if ( contains(a, b) ) {
// 1 or more matches found
}
This is super late, but I just ran into this problem. In my own project I used the following to check if a string was in an array:
["a","b"].includes('a') // true
["a","b"].includes('b') // true
["a","b"].includes('c') // false
This way you can take a predefined array and check if it contains a string:
var parameters = ['a','b']
parameters.includes('a') // true
Best answer is here:
This is case insensitive as well
var specsFilter = [.....];
var yourString = "......";
//if found a match
if (specsFilter.some((element) => { return new RegExp(element, "ig").test(yourString) })) {
// do something
}
const str = 'Does this string have one or more strings from the array below?';
const arr = ['one', 'two', 'three'];
const contains = arr.some(element => {
if (str.includes(element)) {
return true;
}
return false;
});
console.log(contains); // true
Not that I'm suggesting that you go and extend/modify String's prototype, but this is what I've done:
String.prototype.includes()
String.prototype.includes = function (includes) {
console.warn("String.prototype.includes() has been modified.");
return function (searchString, position) {
if (searchString instanceof Array) {
for (var i = 0; i < searchString.length; i++) {
if (includes.call(this, searchString[i], position)) {
return true;
}
}
return false;
} else {
return includes.call(this, searchString, position);
}
}
}(String.prototype.includes);
console.log('"Hello, World!".includes("foo");', "Hello, World!".includes("foo") ); // false
console.log('"Hello, World!".includes(",");', "Hello, World!".includes(",") ); // true
console.log('"Hello, World!".includes(["foo", ","])', "Hello, World!".includes(["foo", ","]) ); // true
console.log('"Hello, World!".includes(["foo", ","], 6)', "Hello, World!".includes(["foo", ","], 6) ); // false
building on T.J Crowder's answer
using escaped RegExp to test for "at least once" occurrence, of at least one of the substrings.
function buildSearch(substrings) {
return new RegExp(
substrings
.map(function (s) {return s.replace(/[.*+?^${}()|[\]\\]/g, '\\$&');})
.join('{1,}|') + '{1,}'
);
}
var pattern = buildSearch(['hello','world']);
console.log(pattern.test('hello there'));
console.log(pattern.test('what a wonderful world'));
console.log(pattern.test('my name is ...'));
Drawing from T.J. Crowder's solution, I created a prototype to deal with this problem:
Array.prototype.check = function (s) {
return this.some((v) => {
return s.indexOf(v) >= 0;
});
};
Using underscore.js or lodash.js, you can do the following on an array of strings:
var contacts = ['Billy Bob', 'John', 'Bill', 'Sarah'];
var filters = ['Bill', 'Sarah'];
contacts = _.filter(contacts, function(contact) {
return _.every(filters, function(filter) { return (contact.indexOf(filter) === -1); });
});
// ['John']
And on a single string:
var contact = 'Billy';
var filters = ['Bill', 'Sarah'];
_.every(filters, function(filter) { return (contact.indexOf(filter) >= 0); });
// true
If you're working with a long list of substrings consisting of full "words" separated by spaces or any other common character, you can be a little clever in your search.
First divide your string into groups of X, then X+1, then X+2, ..., up to Y. X and Y should be the number of words in your substring with the fewest and most words respectively. For example if X is 1 and Y is 4, "Alpha Beta Gamma Delta" becomes:
"Alpha" "Beta" "Gamma" "Delta"
"Alpha Beta" "Beta Gamma" "Gamma Delta"
"Alpha Beta Gamma" "Beta Gamma Delta"
"Alpha Beta Gamma Delta"
If X would be 2 and Y be 3, then you'd omit the first and last row.
Now you can search on this list quickly if you insert it into a Set (or a Map), much faster than by string comparison.
The downside is that you can't search for substrings like "ta Gamm". Of course you could allow for that by splitting by character instead of by word, but then you'd often need to build a massive Set and the time/memory spent doing so outweighs the benefits.
convert_to_array = function (sentence) {
return sentence.trim().split(" ");
};
let ages = convert_to_array ("I'm a programmer in javascript writing script");
function confirmEnding(string) {
let target = "ipt";
return (string.substr(-target.length) === target) ? true : false;
}
function mySearchResult() {
return ages.filter(confirmEnding);
}
mySearchResult();
you could check like this and return an array of the matched words using filter
I had a problem like this. I had a URL, I wanted to check if the link ends in an image format or other file format, having an array of images format. Here is what I did:
const imagesFormat = ['.jpg','.png','.svg']
const link = "https://res.cloudinary.com/***/content/file_padnar.pdf"
const isIncludes = imagesFormat.some(format => link.includes(format))
// false
You can check like this:
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
var list = ["bad", "words", "include"]
var sentence = $("#comments_text").val()
$.each(list, function( index, value ) {
if (sentence.indexOf(value) > -1) {
console.log(value)
}
});
});
</script>
</head>
<body>
<input id="comments_text" value="This is a bad, with include test">
</body>
</html>
let obj = [{name : 'amit'},{name : 'arti'},{name : 'sumit'}];
let input = 'it';
Use filter :
obj.filter((n)=> n.name.trim().toLowerCase().includes(input.trim().toLowerCase()))
var str = "A for apple"
var subString = ["apple"]
console.log(str.includes(subString))