希望有人能阐明下面的代码,它总是提醒下面的错误。该代码旨在验证一个单选按钮是否被选中,否则返回 false 到当前页面,否则继续执行表单操作。提前谢谢你!!
function onDisplayItemsForm(){
var re = false; // used to determine when a button is checked
var radIdSelected = frmDisplayItems;
// traverse the radio buttons
// if one is checked sets re to true, and stops the iteration
for(var i=0; i<radIdSelected.length; i++)
{
if(radIdSelected[i].checked == true)
{
re = true;
break;
}
if (!radIdSelected[i].checked)
{
alert("Please select product");
return false;
}
return true;
}
};
表格如下:
<form name="frmDisplayItems" action="showItem.php" onsubmit="return onDisplayItemsForm();" >
<table width="50%" border="1">
<th>Country of Origin</th>
<th>Select</th>
</tr>
<td><input name=\"radId\" type=\"radio\" value=\"$id\" /></td>
</tbody>
</table>
<p><input name="btnSubmit" type="submit" value="Select"/> </p>
</form>
<