1
我有一個簡單的表單提交所選位置的地址的詳細信息。在提交之前，我希望允許用戶預覽將要發送的地址。地址存儲在JSON對象中。我有以下代碼：訪問JSON對象的詳細信息
HTML：
<html>
<form action="something.asp" >
<select id="selectAddress">
<option value="Cheonan2">Cheonan 2</option>
<option value="Cheonan3">Cheonan 3</option>
</select>
<input type="submit"/>
</form>
<input type="button" value="Display addresses" onClick="showAddress()"/><br>
<span id="addressField1"></span>
<span id="addressField2"></span>
<span id="addressField3"></span>
</html>
JS：
<script>
function showAddress()
{
var JSONAddress =
[
{ "id":"Cheonan2",
"Field1": "96, 3Gongdan1-ro",
"Field2": "Seobuk-gu, Cheonan-si",
"Field3": "Chungcheongnam-do, 31093, Korea"
},
{
"id":"Cheonan3",
"Field1": "80, 3Gongdan6-ro,",
"Field2": "Seobuk-gu, Cheonan-si,",
"Field3": "Chungcheongnam-do, 31085, Korea"
}
];
var e=document.getElementById("selectAddress");
var selectedAddress = e.options[e.selectedIndex].value;
for (var i;i<JSONAddress.length;i++){
if (JSONAddress[i].id===selectedAddress){
document.getElementById('addressField1').innerHTML=JSONAddress[i].Field1;
document.getElementById('addressField2').innerHTML=JSONAddress[i].Field2;
document.getElementById('addressField3').innerHTML=JSONAddress[i].Field3;
}
}
}
</script>
我可能訪問的對象JSONAddress錯誤的，因爲該功能不顯示任何東西......可以你幫忙？
你能在小提琴上提供代碼嗎？ – amoeba