0

我正在使用 javascript 检查文件大小。如果它大于 1m，它会显示一个警报，然后它会重定向到索引页面。

我想知道如何让它保持在同一个页面而不重定向和刷新，并保持用户插入的所有页面信息不变。

这是代码：

if(fileInput.files[0].size > 1050000) {
    alert('File size is bigger than 1Mb!');
    return false;
}

孔代码：

var handleUpload = function(event){
    event.preventDefault();
    event.stopPropagation();

    var fileInput = document.getElementById('file');

    var data = new FormData();

    data.append('javascript', true);

    if(fileInput.files[0].size > 1050000) {
        //document.getElementById("image_id").innerHTML = "Image too big (max 1Mb)";
        alert('File bigger than 1Mb!');
        //window.location="upload.php";
        return false;
        }

    for (var i = 0; i < fileInput.files.length; ++i){

        data.append('file[]', fileInput.files[i]);

    }   

    var request = new XMLHttpRequest();

    request.upload.addEventListener('progress', function(event){
        if(event.lengthComputable){
            var percent = event.loaded / event.total;
            var progress = document.getElementById('upload_progress');

            while (progress.hasChildNodes()){
                progress.removeChild(progress.firstChild);
            }
            progress.appendChild(document.createTextNode(Math.round(percent * 100) +' %'));
            document.getElementById("loading-progress-17").style.width= Math.round(percent * 100) +'%';
        }
    });
    request.upload.addEventListener('load', function(event){
        document.getElementById('upload_progress').style.display = 'none';
    });
    request.upload.addEventListener('error', function(event){
        alert('Upload failed');
    });
    request.addEventListener('readystatechange', function(event){
        if (this.readyState == 4){
            if(this.status == 200){
                var links = document.getElementById('uploaded');
                var uploaded = eval(this.response);
                var div, a;
                for (var i = 0; i < uploaded.length; ++i){
                    div = document.createElement('div');
                    a = document.createElement('a');
                    a.setAttribute('href', 'files/' + uploaded[i]);
                    a.appendChild(document.createTextNode(uploaded[i]));
                    div.appendChild(a);
                    links.appendChild(div);
                }
            }else{
                console.log('server replied with HTTP status ' + this.status);
            }
        }
    });
    request.open('POST', 'upload.php');
    request.setRequestHeader('Cache-Control', 'no-cache');
    document.getElementById('upload_progress').style.display = 'block';
    request.send(data);

}

window.addEventListener('load', function(event){

    var submit = document.getElementById('submit');
    submit.addEventListener('click', handleUpload);
});

带有 html 的 upload.php 代码

<?php
foreach($_FILES['file']['name'] as $key => $name){
        if ($_FILES['file']['error'][$key] == 0 && move_uploaded_file($_FILES['file']['tmp_name'][$key], "files/{$name}")){
            $uploaded[] = $name;
        }
    }
    if(!empty($_POST['javascript'])){
        die(json_encode($uploaded));
    }
?>

<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <script type="text/javascript" src="upload.js"></script>
</head>
<body>
        <div id="uploaded">
            <?php

            if (!empty($uploaded)){
                foreach ($uploaded as $name){
                    echo '<div><a href="files/',$name,'">',$name,'</a></div>';
                }
            }

            ?>
             </div>
<div id="upload_progress"></div>
        <div>
            <form action="" method="post" enctype="multipart/form-data">
<input type="file" id="file" name="file[]" />
<input type="submit" id="submit" value="upload" />
</form>

提前致谢

4

3 回答 3

0

你可以摆脱return它应该工作。否则，也许您应该尝试模式而不是警报。它们更整洁漂亮

于 2013-09-14T22:29:23.713 回答
0

Return false 阻止重定向。

       var val = $("#inputId").val();      

        if (val >= 0 || val <=9)
        {           
            return true;
        }
        else
        {
            alert("Enter Digits Only");
            return false;
        }
     ```
于 2019-11-04T13:00:11.080 回答
0

event.preventDefault();在下方添加警报功能。
这可以帮助您保持在同一页面上。

于 2020-06-24T06:04:26.733 回答