Question

问它为什么在一次点击中装载,而是在两个点上装载?

<input id="btnPaymentAdd" type="button" value="Add Payment Info" />
</p>
<div id="paymentSection"></div>
<br />    

....

$("#btnPaymentAdd").click(function () {
    $("#paymentSection").load( /Donation/AddPaymentInfo );
    $("#paymentSection").show( slow );
});

....

public ActionResult AddPaymentInfo()
{

    var vModel = new PaymentViewModel();
    vModel.Payment = new Payments();
    vModel.PaymentType = new WCCDentalApp.Models.PaymentType();

    ViewBag.PaymentTypes = new SelectList(dbEntities.PaymentTypes.OrderBy(pt => pt.PaymentTypeID),
    "PaymentTypeID", "PaymentType1", vModel.Payment.PaymentTypeID);

    return PartialView("PaymentPView", vModel);
}

Answer 1

You need to bind your .show() function after the page loads...

$("#btnPaymentAdd").click(function(){
    $("#paymentSection").load( /Donation/AddPaymentInfo , function(){
         $(this).show( slow );
    });
});

or you can use .get(), like this:

$("#btnPaymentAdd").click(function(){
    $.get( /Donation/AddPaymentInfo , function(data){
         $("#paymentSection").html(data).show( slow );
    });
});

Otherwise, the way you have it will just do the .show() without waiting for the page to load, will make you think like clicking it twice works.

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