Get indices of characters in string for interpolation

Question

I have create a method that given a string will find all interpolation points relative to a certain pattern some string {{point}} this is working however its not very elegant and I'm wondering if anyone knows of a cleaner more minified way of doing this?

Here is my method:

_interoplationPoints: function(string){
    var startReg = /{{/gi,
        endReg = /}}/gi,
        indices = {start: [],end: []},
        match;
    while (match = startReg.exec(string)) indices.start.push(match.index);
    while (match = endReg.exec(string)) indices.end.push(match.index);
    return indices;
},

Given a string it will check for all start and end points {{ & }} it will then return an object with the start and end points of each occurance of {{}}.

The reason for me doing this is I will later substring() these indexes with there relevant value.

Is there a reason why you're not simply String.replaceing the placeholders in one go? The offsets will change once you start to substr in new values, so this is all quite messy it'd seem. — deceze
– deceze ♦, Commented Feb 12, 2015 at 14:14
I can't replace as the string will keep changing, the original string some string {{value}} will be stored alongside the binding data which is referenced each time value is changed. The interpolation points are assigned on compile and reused with the original value to set textNode.nodeValue — Simon Staton
– Simon Staton, Commented Feb 12, 2015 at 14:38

radiaph · Accepted Answer · 2015-02-12 15:00:48Z

1

Not much simpler, but:

_interoplationPoints: function(string){
    var reg = /{{[^}]*}}/gi,
        indices = {start: [],end: []},
        match;
    while (match = reg.exec(string)) {
        indices.start.push(match.index);
        indices.end.push(match.index + match[0].length - 2);
    }
    return indices;
},

This regular expression matches {{ followed by an expression of any length that does not contain a closing brace [^}]* followed by }}. the end index is computed by adding the length of the match (which would put it just beyond the second closing brace) then subtracting 2 since there are two closing braces.

answered Feb 12, 2015 at 15:00

radiaph

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1 Comment

Simon Staton Over a year ago

Thanks this is what I was trying to achieve

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