jQuery, performing a function on page load

Question

I'm building a finance calculator with jQuery which is working well. The calculator uses jQuery sliders. I have the following function which is called when a slider is activated.

However I need to replicate this so the same actions are called on page load, could anyone suggest how I might achieve this?

jQuery(document).ready(function($) {
    $("#slider-deposit").slider({
        value: 100,
        min: <?php echo json_encode($minDeposit); ?>,
        max: <?php echo json_encode($maxDeposit); ?>,
        step: 1,
        slide: function(event, ui) {
            $("#amount-deposit").val(Number(ui.value).toFixed(2));
            totalPayableFunc();
            monthlyPaymentsFunc();
        }
    });
    $("#amount-deposit").val("" + $("#slider-deposit").slider("value"));
});

You are doing right, but what is the issue you are facing? Any error(s) in the console. — Parkash Kumar
– Parkash Kumar, Commented Jan 6, 2016 at 11:26
just call the function totalPayableFunc();monthlyPaymentsFunc(); right after $("#amount-deposit").val("" + $("#slider-deposit").slider("value")); i.e. last line of above code. — vijayP
– vijayP, Commented Jan 6, 2016 at 11:34
Essentially I need to call the two functions, totalPayableFunc(); and monthlyPaymentsFunc(); — Liam Fell
– Liam Fell, Commented Jan 6, 2016 at 11:34

Steve Harris · Accepted Answer · 2016-01-06 11:43:44Z

1

Is it this?

jQuery(document).ready(function($) {
    $("#slider-deposit").slider({
        value: 100,
        min: <?php echo json_encode($minDeposit); ?>,
        max: <?php echo json_encode($maxDeposit); ?>,
        step: 1,
        slide: function(event, ui) {
            $("#amount-deposit").val(Number(ui.value).toFixed(2));
            totalPayableFunc();
            monthlyPaymentsFunc();
        }
    });
    $("#amount-deposit").val("" + $("#slider-deposit").slider("value"));
    totalPayableFunc();
    monthlyPaymentsFunc();
});

answered Jan 6, 2016 at 11:43

Steve Harris

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