收音机和复选框已检查的值不会在JavaScript中显示

Question

我尝试使用JavaScript获取单选按钮和多个复选框的值以显示在新窗口中。我似乎无法弄清楚为什么＆＃34;真的＆＃34;返回而不是检查的实际值。以下是结果的代码和截图。

＆＃13;

＆＃13;

function messagebox()
{
	var newWindow;
	var msg ="";
	var i = document.PizzaForm.state.options.selectedIndex;
	var text = document.PizzaForm.state.options[i].text;
	var value = document.PizzaForm.state.options[i].value;
	
	{ document.PizzaForm.customer.value = document.PizzaForm.customer.value.replace(/\w\S*/g, function(txt){return txt.charAt(0).toUpperCase() + txt.substr(1).toLowerCase();}); }
	
	{ document.PizzaForm.city.value = document.PizzaForm.city.value.replace(/\w\S*/g, function(txt){return txt.charAt(0).toUpperCase() + txt.substr(1).toLowerCase();}); }
	
	newWindow = window.open("","","status=no,height=500,width=500");
	message = "<ul><li>Name: " + document.PizzaForm.customer.value;
    message += "<li>Address: " + document.PizzaForm.address.value;
    message += "<li>City: " + document.PizzaForm.city.value;
    message += "<li>State: " + document.PizzaForm.state.value;
    message += "<li>Zip Code: " + document.PizzaForm.zip.value;
    message += "<li>Phone: " + document.PizzaForm.phone.value;
    message += "<li>Email: " + document.PizzaForm.email.value;
	message += "<li>Size: " + showSize();
	message += "<li>Toppings: " + showToppings();
	newWindow.document.write(message);

function showSize()
	{
	for(i=0;i<document.PizzaForm.sizes.length;i++)
        if(document.PizzaForm.sizes[i].checked)
		msg += document.PizzaForm.sizes[i].value;
	return true;
	}
	
function showToppings()
	{
	for(i=0;i<document.PizzaForm.toppings.length;i++)
        if(document.PizzaForm.toppings[i].checked)
            msg += (i==0?"":",")+document.PizzaForm.toppings[i].value;
	return true;
	}	
}

＆＃13;

＆＃13;

＆＃13;

Answer 1

showSize()和showToppings()方法最终都返回true，这就是你在字符串中获得真实的原因。你应该返回msg，你的问题就会得到解决。