Question

我有数组链接。每个链接都呈现一个XML文件。如何通过一次调用迭代每个XML并将其保存到文件夹。

GetMesssageAttachments(userId) 返回数组6链接，但是当前代码仅保存第一个文件。怎么了谢谢

 public async void SaveXMLMessages(string userId)
    {
                try
                {
                    if (_responseMessage.IsSuccessStatusCode)
                    {
                        string messagesFolder = @"C:\XMLMessages";
                        Directory.CreateDirectory(messagesFolder);

                        string messageFileName = Path.GetRandomFileName();
                        string messagesPath = Path.Combine(messagesFolder, messageFileName);

                        foreach (string xmlMessage in await GetMesssageAttachments(userId))
                        {
                            var xmlMessageResponse = await _client.GetAsync(xmlMessage);
                            using (FileStream fileStream = new FileStream(messagesPath, FileMode.Create))
                            {
                                await xmlMessageResponse.Content.CopyToAsync(fileStream);
                            }
                        }
                    }
                }
                catch (Exception e)
                {
                    throw e.InnerException;
                }
            }

已更新

这是工作。

    public async void SaveXMLMessages(string userId)
    {
        try
        {
            if (_responseMessage.IsSuccessStatusCode)
            {
                string messagesFolder = @"C:\XMLMessages";
                Directory.CreateDirectory(messagesFolder);

                foreach (string xmlMessage in await GetMesssageAttachments(userId))
                {
                    string messageFileName = Path.GetRandomFileName();
                    string messagesPath = Path.Combine(messagesFolder, messageFileName);

                    var xmlMessageResponse = await _client.GetAsync(xmlMessage);
                    using (FileStream fileStream = new FileStream(messagesPath, FileMode.Create))
                    {
                        await xmlMessageResponse.Content.CopyToAsync(fileStream);
                    }
                }
            }
        }
        catch (Exception e)
        {
            throw e.InnerException;
        }
    }

Answer 1

与messagesPath中使用的foreach相同。这意味着循环中仅创建一个文件

您必须像这样在循环中重新初始化它：

foreach (string xmlMessage in await GetMesssageAttachments(userId))
{
    string messageFileName = Path.GetRandomFileName();
    string messagesPath = Path.Combine(messagesFolder, messageFileName);
    var xmlMessageResponse = await _client.GetAsync(xmlMessage);
    using (FileStream fileStream = new FileStream(messagesPath, FileMode.Create))
    {
        await xmlMessageResponse.Content.CopyToAsync(fileStream);
    }
}

Answer 2

也许您在每次迭代中都覆盖文件，请尝试将此块移到您的foreach内部：

string messageFileName = Path.GetRandomFileName();
string messagesPath = Path.Combine(messagesFolder, messageFileName);

Answer 3

'thierry v'代码和rsb55所说的只有很小的变化是正确的。您的代码应如下所示

foreach (string xmlMessage in await GetMesssageAttachments(userId))
{
string messageFileName = Path.GetRandomFileName();
string messagesPath = Path.Combine(messagesFolder, messageFileName);

    var xmlMessageResponse = await _client.GetAsync(xmlMessage);
    using (FileStream fileStream = new FileStream(messagesPath, FileMode.Create))
    {
        await xmlMessageResponse.Content.CopyToAsync(fileStream);
    }
}

FileStream不保存数组文件

3 个答案: