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Editorial Team
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Asked: 2026-06-15T05:21:11+00:00

I have the following javascript: <script type=text/javascript> function changeBet(parseFloat(bet)) { var moneyline = parseFloat(<?php

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I have the following javascript:

<script type="text/javascript">
function changeBet(parseFloat(bet)) {
var moneyline = parseFloat(<?php echo json_encode($win) ?>);
var gain = parseFloat(bet * moneyline);
document.getElementById("PotentialGain").value = gain;
}
</script>

The php variable $win is successfully var_dump’ed as a float. When the variable gain = bet, PotentialGain displays the user input from BetAmount as expected. Here is my echo’ed php code:

echo '<tr>';
echo '<td>type="text" name="BetAmount[]" id="BetAmount" onkeyup="changeBet(this.value);" ></td></tr><tr><td>Potential Gain:<input type="text" name="PotentialGain[]"  id="PotentialGain" ></td></tr><tr><td><input type="Submit" name="send" value="Submit"></td>';
echo '</tr>';

However, I want gain(which is inputted as the PotentialGain value) to be the user input bet * the var moneyline.

The result is NaN. Is there a var that I am not parsing correctly to display the correct numerical value of gain?

Thanks for any help.

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1 Answer

  1. Editorial Team

    Here is the final javascript that I used:

    <html>
<script type="text/javascript">
function changeBet(bet) {
var moneyline = <?php echo $win ?>;
var gain = (bet * moneyline).toFixed(2);
document.getElementById("PotentialGain").value = gain;
}
</script>
</html>

    My final notes are: 1. To remember to place the javascript function after the code with your php caclculations. 2. Thanks to everyone who clarified parseFloat is not necessary if you know the var is a int (or double in my case). 3. Thanks to everyone who clarified parseFloat did not belong in the function parameter.

    Hopefully this helps others out!

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