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Javascript FileUpload:获取JS不工作的文件名_Javascript_C#_Asp.net_Webforms - Fatal编程技术网

Javascript FileUpload:获取JS不工作的文件名

Javascript FileUpload:获取JS不工作的文件名,javascript,c#,asp.net,webforms,Javascript,C#,Asp.net,Webforms,我想获取文件名并将其放在labellbName中,但此代码不起作用。我使用了asp:FileUpload。这是我的代码: 和JS代码: function upload() { var name = ""; var files = document.getElementById("<%= FileUpload1.ClientID %>"); for (var i = 0; i < files.length; i+

我想获取文件名并将其放在label
lbName
中,但此代码不起作用。我使用了
asp:FileUpload
。这是我的代码:

和JS代码:

function upload() {
    var name = "";

    var files = document.getElementById("<%= FileUpload1.ClientID %>");
    for (var i = 0; i < files.length; i++) {
        name = name + (files[i].name) + ";";
    }
    document.getElementById("lbName").value = "1: " + name;
}
函数上传(){
var name=“”;
var files=document.getElementById(“”);
对于(var i=0;i
您可以使用jQuery并将更改事件绑定到
输入类型=文件

<script type="text/javascript">

    $('input[type="file"]').change(function (e) {
        var files = [];
        for (var i = 0; i < $(this)[0].files.length; i++) {
            files.push($(this)[0].files[i].name);
        }
        $(this).next('span').html(files.join(', '));
    });

</script>

$('input[type=“file”]”)。更改(函数(e){
var文件=[];
对于(var i=0;i<$(this)[0].files.length;i++){
files.push($(this)[0].files[i].name);
}
$(this.next('span').html(files.join(','));
});
或者,如果您想让javascript正常工作:

<script type="text/javascript">

    function upload() {
        var name = "";
        var files = document.getElementById("<%= FileUpload1.ClientID %>");
        for (var i = 0; i < files.files.length; i++) {
            name = name + (files.files[i].name) + ";";
        }
        document.getElementById("<%= lbName.ClientID %>").innerHTML = "1: " + name;
    }

</script>

函数上传(){
var name=“”;
var files=document.getElementById(“”);
对于(var i=0;i
哪个特定位不起作用?是否有错误?文件是否已上载但被服务器忽略?更多详细信息betterI无法将所有文件名放入my
lbName