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在 Windows 商店应用程序上上传文件时遇到一些问题。我得到了状态 200，但是当我检查服务器时没有上传文件..我使用的是 apache 服务器，我检查了 Microsoft后台传输示例的示例，但它适用于 ASP .Net 服务器。

任何解决方案？

4

2 回答 2

1

我需要一次上传一个文件，它对我有用，请告诉我：

上传.php：

<?php
    $target = "data/"; 
    $target = $target . basename( $_FILES['Filename']['name']) ; 
    if(move_uploaded_file($_FILES['Filename']['tmp_name'], $target)) 
    {
    echo "The file ". basename( $_FILES['Filename']['name']). " has been uploaded";
    } 
    else {
    echo "Sorry, there was a problem uploading your file.";
    }       
?>

C#代码：

    private async void StartMultipartUpload_Click(object sender, RoutedEventArgs e)
    {
        Uri uri;
        if (!Uri.TryCreate(serverAddressField.Text.Trim(), UriKind.Absolute, out uri))
        {
            rootPage.NotifyUser("Invalid URI.", NotifyType.ErrorMessage);
            return;
        }

        // Verify that we are currently not snapped, or that we can unsnap to open the picker.
        if (ApplicationView.Value == ApplicationViewState.Snapped && !ApplicationView.TryUnsnap())
        {
            rootPage.NotifyUser("File picker cannot be opened in snapped mode. Please unsnap first.", NotifyType.ErrorMessage);
            return;
        }

        FileOpenPicker picker = new FileOpenPicker();
        picker.FileTypeFilter.Add("*");
        IReadOnlyList<StorageFile> files = await picker.PickMultipleFilesAsync();

        if (files.Count == 0)
        {
            rootPage.NotifyUser("No file selected.", NotifyType.ErrorMessage);
            return;
        }

        List<BackgroundTransferContentPart> parts = new List<BackgroundTransferContentPart>();
        for (int i = 0; i < files.Count; i++)
        {
            BackgroundTransferContentPart part = new BackgroundTransferContentPart("Filename", files[i].Name);
            part.SetFile(files[i]);
            parts.Add(part);
        }

        BackgroundUploader uploader = new BackgroundUploader();
        UploadOperation upload = await uploader.CreateUploadAsync(uri, parts);

        String fileNames = files[0].Name;
        for (int i = 1; i < files.Count; i++)
        {
            fileNames += ", " + files[i].Name;
        }

        Log(String.Format("Uploading {0} to {1}, {2}", fileNames, uri.AbsoluteUri, upload.Guid));

        // Attach progress and completion handlers.
        await HandleUploadAsync(upload, true);
    }

我使用了 SDK 示例的相同代码。data/文件夹位于upload.php的同一文件夹中，我的 Uri 是http://mySite/myApp/upload.php

于 2013-02-13T11:12:35.820 回答
0

好的，所以我找到了一个完整的解决方案。基本上你需要在你的 SDK 示例上作为多部分上传（不管它是 C# 还是 C++），关键是 php 文件。这是我处理所有选定文件的 PHP 文件。作为魅力！此外，如果您收到错误 1，则意味着您的 php.ini 中的最大文件上传大小太小！所以也要注意这一点。希望它有所帮助，我确实花了一些精力来完成这项工作。

<?php

$myFile = "testFile.txt";
$fh = fopen($myFile, 'w') or die("can't open file");

foreach ($_FILES as $name) {
    if ($_FILES[$name]["error"] > 0){
        fwrite($fh, "Error: " . $_FILES[$name]['error'] . " on file: " . $_FILES[$name]['name']); // log the error
    }
    $target = "data/"; 
    $target = $target . basename( $_FILES[$name]['name']) ; 
    if(move_uploaded_file($_FILES[$name]['tmp_name'], $target)) 
    {
        echo "The file ". basename( $_FILES[$name]['name']). " has been uploaded"; // succes, do whatever you want
    } 
    else {
        fwrite($fh, "File movement error on file: " . $_FILES[$name]['name']); // log the error
    }       
}

fclose($fh);

?>

于 2013-03-06T15:17:14.487 回答