Question

我正在编写一个脚本，让用户上传文件并查看上传状态，然后将该文件保存在服务器上。

问题是我无法使用php api。所以我想用javascript上传文件，我可以很容易地获得上传状态，但后来我想将文件传递给php并保存。

这可能吗？

这是要上传的JavaScript代码。

<!DOCTYPE html>
<html>
<head >
<title>Upload Files using XMLHttpRequest</title>

<script type="text/javascript">
    function fileSelected() {
        var file = document.getElementById('fileToUpload').files[0];
        if (file) {
            var fileSize = 0;
            if (file.size > 1024 * 1024)
                fileSize = (Math.round(file.size * 100 / (1024 * 1024)) / 100).toString() + 'MB';
            else
                fileSize = (Math.round(file.size * 100 / 1024) / 100).toString() + 'KB';

            document.getElementById('fileName').innerHTML = 'Name: ' + file.name;
            document.getElementById('fileSize').innerHTML = 'Size: ' + fileSize;
            document.getElementById('fileType').innerHTML = 'Type: ' + file.type;
        }
    }

    function uploadFile() {
        var fd = new FormData();
        fd.append("fileToUpload", document.getElementById('fileToUpload').files[0]);
        var xhr = new XMLHttpRequest();
        xhr.upload.addEventListener("progress", uploadProgress, false);
        xhr.addEventListener("load", uploadComplete, false);
        xhr.addEventListener("error", uploadFailed, false);
        xhr.addEventListener("abort", uploadCanceled, false);
        xhr.open("POST", "UploadHandler.ashx");
        xhr.send(fd);
    }

    function uploadProgress(evt) {
        if (evt.lengthComputable) {
            var percentComplete = Math.round(evt.loaded * 100 / evt.total);
            document.getElementById('progressNumber').innerHTML = percentComplete.toString() + '%';
            document.getElementById('prog').value = percentComplete;
        }
        else {
            document.getElementById('progressNumber').innerHTML = 'unable to compute';
        }
    }

    function uploadComplete(evt) {
        /* This event is raised when the server send back a response */
        alert(evt.target.responseText);
    }

    function uploadFailed(evt) {
        alert("There was an error attempting to upload the file.");
    }

    function uploadCanceled(evt) {
        alert("The upload has been canceled by the user or the browser dropped the connection.");
    }
</script>

</head>
<body>
    <form id="form1">
    <div>
        <label for="fileToUpload">
            Select a File to Upload</label>
        <input type="file" name="fileToUpload[]" id="fileToUpload" onchange="fileSelected();" />
    </div>
    <div id="fileName">
    </div>
    <div id="fileSize">
    </div>
    <div id="fileType">
    </div>
    <div>
        <input type="button" onclick="uploadFile()" value="Upload" />
    </div>
    <div id="progressNumber">
    </div>
    <progress id="prog" value="0" max="100.0"></progress>
    </form>
</body>
</html>

Answer 1

我对此有一个Gist。它使用xhr.send(FormData)并显示最少的代码用PHP处理文件。

Answer 2

向PHP文件发出AJAX请求，发送上传的文件属性，如名称等，然后打开它，移动它，重命名或者需要的任何内容。它应该有效，呵呵？

我是否正确理解了您的问题？

示例代码：

在Javascript中：

// xhr
var http = new XMLHttpRequest();
var url = "file_handler.php";
var file_data = "name=mypic.jpg&size=123&othe=etc";
http.open("POST", url, true);

// headers
http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http.setRequestHeader("Content-length", file_data.length);
http.setRequestHeader("Connection", "close");

http.onreadystatechange = function() {
    if(http.readyState == 4 && http.status == 200) {
        alert(http.responseText);
    }
}

http.send(file_data);

在file_handler.php：

// file data
$file_data = $_POST['file_data'];

// working on the file
$temp_dir = 'usr/upload/';
$new_dir = 'usr/photos/';

// new unique name
$new_name = time() . '_' . $file_data['name'];

// copy?
if (copy($temp_dir . $file_data['name'], $new_dir . $new_name)) {
    unlink($temp_dir . $file_data['name']);
}

// rename?
rename($temp_dir . $file_data['name'], $temp_dir . $new_name);

// delete old file?
unlink($temp_dir . $file_data['name']);

// do whatever else needed here ...
// echo some JSON data to interact with your client-side JS code, maybe ...

将文件从Javascript上传到PHP

2 个答案: