Random number between negative and positive value

Here's a generalized solution that will let you set the boundaries, and opt in/out of including the 0.

var pos = 99,
    neg = 99,
    includeZero = false,
    result;

do result = Math.ceil(Math.random() * (pos + neg)) - neg;
while (includeZero === false && result === 0);

The pos and neg values are inclusive.

This way there's no requirement that the positive and negative ranges be balanced.

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Or if you're worried about the rerun due to a single excluded value, you can just make the initial range less by one, and add 1 to any result greater than or equal to 0.

var pos = 5,
    neg = 5,
    result;

result = Math.floor(Math.random() * (pos + neg)) - neg;
result = result < 0 ? result : result + 1;

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That last line could be shorter if you prefer:

result += (result >= 0)

This returns what you want

function getNonZeroRandomNumber(){
    var random = Math.floor(Math.random()*199) - 99;
    if(random==0) return getNonZeroRandomNumber();
    return random;
}

Here's a functional fiddle

EDIT

To contribute for future readers with a little debate happened in the comments which the user @MarkDickinson made a indeed relevant contribution to my first code posted, I've decided to make another fiddle with a fast comparison between using Math.floor() and Math.round() functions to return the value the op wanted.

First Scenario: Using var random = Math.round(Math.random()*198) - 99; (My first suggestion)

function getNonZeroRandomNumberWithMathRound(){
    var random = Math.round(Math.random()*198) - 99;
    if(random==0) return getNonZeroRandomNumber();
    return random;
}

Second scenario: Using var random=Math.floor(Math.random()*199) - 99; (Mark suggestion)

function getNonZeroRandomNumberWithMathFloor(){
    var random = Math.floor(Math.random()*199) - 99;
    if(random==0) return getNonZeroRandomNumber();
    return random;
}

Methodology

Since it's a short debate I've chosen fiddle.net to do the comparison.

The test consists of running the above functions 100.000 times and then retrieving how much times the extreme numbers 99 and -99 would appear against a other number, let's say 33 and -33.

The test will then give a simple output consisting of the percentage of appearances from 99 and -99 and the percentage of appearances of 33 and -33.

It'll be used the Webkit implementation from Safari 6.0.2 to the give the output from this answer but anyone can test with your favourite browser late on fiddle.net

Result from first scenario:

• Percentage of normal ocurrences:0.97%
• Percentage of extreme ocurrences:0.52%
• Percentage of extreme ocurrences relative to normal ocurrences:53.4% // Half the chances indeed

Result from second scenario:

• Percentage of normal ocurrences:1.052%
• Percentage of extreme ocurrences:0.974%
• Percentage of extreme ocurrences relative to normal ocurrences:92% //Closer of a fair result with a minimal standard deviation

The result can be seen here: http://jsfiddle.net/brunovieira/LrXqh/

var num = Math.floor(Math.random()*99) + 1; // this will get a number between 1 and 99;
num *= Math.round(Math.random()) ? 1 : -1; // this will add minus sign in 50% of cases

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Altogether

var ranNum = Math.ceil(Math.random() * 99) * (Math.round(Math.random()) ? 1 : -1)