tryjsref_fileupload_name 在线代码实例

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运行代码

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<!DOCTYPE html>
<html>
<head>
<script>
function displayResult()
{
var x=document.getElementById("fname").name;
alert(x);
}
</script>
</head>
<body>
<form>
Select a file to upload:
<input type="file" name="filename" id="fname" size="50"/>
</form>
<button type="button" onclick="displayResult()">Display name</button>
</body>
</html>
 
 
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运行结果
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