Javascript: Is there a way to use a variable in a RegEx expression for what is being searched for?

Question

I used the slice() method to test to see if the contents of a variable was in a main string.

I was trying to use a regex. But, I could not figure out how to put the variable inside the regex.

Here is what I did:

    function verifySubstrs(mainStr, head, body, tail) {

    // Verify if the 1st string starts with the 2nd string, 
    var headLength = head.length;

    if(mainStr.slice(0, headLength) !== head){
        return "Incomplete.";
    }

Here is what I wanted to do instead. I wanted to use the variable "head" where I have the text "head".

function verifySubstrs(mainStr, head, body, tail) {

// Verify if the 1st string starts with the 2nd string, 
var TF = /^head/.test(mainStr);

if(TF === false){
    return "Incomplete.";
}

If I could use a variable in what is being searched for, I could use the same technique for finding the body inside the mainStr and the tail at the end. I am looking for the technique and not just the solution to this particular project.

**Note:** You could just do: `if(mainStr.indexOf(head) === 0)`. — ibrahim mahrir, Dec 20 '17 at 00:14
Why wouldn’t you want to use your first approach? It’s much cleaner than creating a dynamic regex. (`mainStr.startsWith(head)` also works in modern JavaScript.) — Ry-, Dec 20 '17 at 00:16
Ryan: I am interested in knowing if it is possible for future coding. — Planet.Megan, Dec 20 '17 at 00:18
ibrahim mahrir: Can you please provide a link to the previous question that I duplicated? Nothing relevant came up when I typed in the question. (I realize that the search on that page isn't comprehensive.) — Planet.Megan, Dec 20 '17 at 00:20

score 2 · Answer 1 · answered Dec 20 '17 at 00:16

2

You can compile a regexp from a string. Try this:

var regexp = new RegExp("^" + head);
var TF = regexp.test(mainStr);

answered Dec 20 '17 at 00:16

Derek Hopper

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Javascript: Is there a way to use a variable in a RegEx expression for what is being searched for?

1 Answers1