如何使用循环使用javascript检查输入中的值？

Question

如何使用循环使用javascript检查输入中的值？

首先，当用户填写char时。它会显示Your Price must be a number.

如果用户填写的数字小于1.5，则会显示Your Price must be at least $1.50 USD.

并点击Add more链接添加输入。

我尝试了我的代码，但没有工作，我该怎么做？

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>

<form onsubmit="return checkform(this);">
<a href="#" id="addScnt_price" style=" cursor: pointer; font-weight: bold; color: #0192B5; text-decoration: none; ">Add more</a>
<div id="p_scents_price">
    <p>
        <label>
            <input type="text" class="price" id="price0" size="20" name="price[]" onchange="myFunction0()"/><p id="demo0"></p>
        </label>
    </p>
</div>
<input type="submit" name="submit" value="OK">
</form>


<script>
var list = document.querySelectorAll(".price");
for (z = 0; z < list.length; ++z) {
    function myFunction'+z+'() {
            var x = document.getElementById("price'+z+'").value;
            var y = isNaN(x);
            if(y === true)
            {
                document.getElementById("demo'+z+'").innerHTML = "Your Price must be a number.";
            }
            else
            {
               if(x < 1.5)
                {
                    document.getElementById("demo'+z+'").innerHTML = "Your Price must be at least $1.50 USD.";
                }
               else
                {
                    document.getElementById("demo'+z+'").innerHTML = "";
                }
            }
        }
    }    
}
</script>


<script>
$(function() {
        var scntDiv = $('#p_scents_price');
        var i = 1;

        $('#addScnt_price').live('click', function() {
            $('<p><label><input type="text" class="price" id="price'+i+'" size="20" name="price[]" onchange="myFunction'+i+'()"/><a href="#" id="remScnt_price">Remove</a><p id="demo'+i+'"></p></label></p>').appendTo(scntDiv);
                i++;
                return false;
        });

        $('#remScnt_price').live('click', function() { 
                if( i > 2 ) {
                        $(this).parents('p').remove();
                }
                return false;
        });
});
</script>