How to use String.match() in javascript

Question

I have this code here:

<!DOCTYPE html>
<html>
<body>

<p id="demo">Display the result here.</p> 
<input type = "text" value = "ood" id = "txt1"/>
<script>


var myString = "How much wood could a wood chuck chuck";
var myWord = document.getElementById("txt1").value;   // ood
var myPattern = /.myWord/;        // I want this to be /.ood/
var myResult = myString.match(myPattern);

document.getElementById("demo").innerHTML = myResult;

</script>

</body>
</html>

Now what I want to do here is that I want the value of the txt1 which is ood to be matched in myString.

But putting the variable myWord into the myPattern like this: /.myWord/ won't work. Need help please. Many thanks

UPDATE:

I did everything as what it is in the answers but it returns wood,wood instead of wood only, I just wanted to get only 1 match. Just like using for example /.ood/ - this only returns 1 match. Please help

And also, how can I get the word wood by having only od in my input text. I just wanted this for searching..

Use RegExp constructor syntax as var myPattern = new RegExp('\.' + myWord, 'gi'); — Tushar
– Tushar, Commented Sep 7, 2015 at 3:21

Spencer Wieczorek · Accepted Answer · 2015-09-07 05:31:52Z

3

You can use a string as a regular expression using the RegExp Object:

var myPattern = new RegExp('.'+myWord,'g');

Fiddle Example

Doing a single match in your case, is simply changed the second parameter in RegExp from g to m (which means to make one match per line for multi lines, but in this case a strings is simply all one line). For finding the word "wood" from "ood","od","d", or other cases. You can do this:

var myPattern = new RegExp("\\b\\w+"+myWord+"\\b",'m');

Note I have a solution in the comments below, but this one is better.

The items \\b ... \\b mean word boundaries. Basically ensuring that it matches a single word. Then the \\w means any valid "word character". So overall the regexp means (using myWord = "od"):

|word boundary| + (1 or more word characters) + "od" + |word boundary|

This will ensure that it matches any words in the string than end with the characters "od". Or in a more general case you can do:

var myPattern = new RegExp("\\b\\w*"+myWord+"\\w*\\b",'m');

edited Sep 7, 2015 at 5:31

answered Sep 7, 2015 at 3:22

Spencer Wieczorek

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10 Comments

Makudex Over a year ago

Thanks it worked and actually found the match but what i just wanted is to get only 1 match. Thanks.

Spencer Wieczorek Over a year ago

@wwwDELL In that case you can use RegExp('.'+myWord,'m'); to get a single match in this case. The m means to do a match for each line.

Makudex Over a year ago

Thanks! it really worked. I also updated my question. when the value of the input text is od it supposed to find wood but it could'nt match any word from that of myString :(

Spencer Wieczorek Over a year ago

@wwwDELL Edited my comment above with why that happens. So for example you can fix that by doing new RegExp('.{1,2}'+myWord,'g')., meaning you can have 1 to 2 characters before od.

Spencer Wieczorek Over a year ago

@wwwDELL Yes there is, but it has some overhead. You can split the string up by it's words. Then use new RegExp(".+"+myWord+"$",'g'); to mean: 1 or more characters + ends with "od" on each word. Here is an example. It's possible there is a better way to do that.

|

Carsten Massmann · Accepted Answer · 2015-09-07 03:25:53Z

2

Create a Regexp object like

new RegExp('.ood','g');

like in

var searchstring='ood' // this is the one you get in a variable ...

var myString = "How much wood could a wood chuck chuck";
var myPattern=new RegExp('.'+searchstring,'gi');
var myResult = myString.match(myPattern);

edited Sep 7, 2015 at 3:25

answered Sep 7, 2015 at 3:18

Carsten Massmann

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5 Comments

Makudex Over a year ago

would you site some example please?

Tushar Over a year ago

The string ood should be dynamic from the textbox value, see my comment above

Tushar Over a year ago

You also need to escape . by preceding \, in regex

Carsten Massmann Over a year ago

Well, yes, then do that too ;-) BuI I understood that OP actually wanted to look for an "arbitrary character" not the actual dot.

Makudex Over a year ago

Thanks it worked and actually found the match but what i just wanted is to get only 1 match. Thanks.

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