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I need to know how to insert an asterix before an open bracket.

I am doing multiplication using javascript's eval, for example, eval("2(3)"), but I am getting the error "unexpected input value".

To solve this I need the following: 

var str = "2(3)";

to be changed to this: 

output = "2*(3)";
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5 Answers 5

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3

You can use String.prototype.replace().

Example:

var str = '2(3)'
var output = str.replace(/\(/g, '*(')
console.log(output)

That will output: 2*(3)

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4 Comments

Congratulations on your first post. A good tip when answering javascript/HTML/CSS questions is to use a code snippet. Not only does it allow others to test the code, but it allows you to debug it before posting. (Unfortunately, editing snippets isn't supported well/at all for mobile devices.) Apart from my answer ;), yours is the next best one as it works for multiple values and negative numbers as well. It is also the simplest.
Thanks @robinCTS !
I appreciate the thanks. Try not to use "thanks" only comments too often, though. See this FAQ post for further info. The usual way to say thanks on StackOverflow/Stack Exchange sites is to upvote comments, accept answers, and when you have at least 15 rep, to upvote answers/questions. For exceptional answers you can also award a bounty (min 75 rep required).
Also note that whilst, technically, trailing semi-colons are not required in javascript, it is a very good standard practice to always include them.
2

If you want to perform more complex/nested multiplication you need to use this more complicated regex (\)|\d+(?!\d)\.?)(?=[(\d.+-]) with the replace() function. It will work for any combination of nested brackets, as well as plus signs, negative numbers and numbers starting or ending with a decimal point:

var str, output;
str = '2(3)';
output = str.replace(/(\)|\d+(?!\d)\.?)(?=[(\d.+-])/g, '$1*');
console.log(output);
str = '2(3)(-4)';
output = str.replace(/(\)|\d+(?!\d)\.?)(?=[(\d.+-])/g, '$1*');
console.log(output);
str = '((.22(33))((-44)-55.(66)+77))';
output = str.replace(/(\)|\d+(?!\d)\.?)(?=[(\d.+-])/g, '$1*');
console.log(output);

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3 Comments

Dear robinCTS, I like to learn regex but was confuse while learning regex. Do you have any nice tutorial to easily learn regex. thanks
@Vijaykarthik The site I learned about regexes, and still use if I need to check something, is regular-expressions.info/tutorial.html. Another useful site is regex101. There you can write, debug and learn regexes all at the same time.
Yes sure, useful link. Thanks @robinCTS
1

You can use the replace() function combined with a little regex ([0-9])\(([0-9]) :

var str = "2(3)";
var output = str.replace(/([0-9])\(([0-9])/g,'$1*($2');
console.log(output);

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2 Comments

Thanks for your reply, your answer will suit for constant value, but if the value getting dynamically how we can proceed For example : var str = "2(3)(3)"; or var str = "2(3)3";
Zenoo & @Vijaykarthik Plus this answer doesn't work for negative numbers.
0

You can slice the string and add the asterisk sign like this:

var str = "2(3)";
var new_str = str.slice(0, str.indexOf("(")) + "*" + str.slice(str.indexOf("("));
console.log(new_str)

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0

Try this:

var str = "2(3)";
var a = str.split('');
a.splice(1,0,"*");
a = a.join(''); 
console.log(a);

Works good for me :)

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