如何使用UnityWebRequest.Post()将多个文件上传到服务器；
如何使用UnityWebRequest.Post()将多个文件上传到服务器；
提问于 2019-05-16 02:28:23
我正在尝试使用UnityWebRequest.Post()上传多个文件，以下是我的代码。
public void UploadFiles()
{
string[] path = new string[3];
path[0] = "D:/File1.txt";
path[1] = "D:/File2.txt";
path[2] = "D:/File3.txt";
UnityWebRequest[] files = new UnityWebRequest[3];
WWWForm form = new WWWForm();
for (int i = 0; i < files.Length; i++)
{
files[i] = UnityWebRequest.Get(path[i]);
form.AddBinaryData("files[]", files[i].downloadHandler.data, Path.GetFileName(path[i]));
}
UnityWebRequest req = UnityWebRequest.Post("http://localhost/File%20Upload/Uploader.php", form);
yield return req.SendWebRequest();
if (req.isHttpError || req.isNetworkError)
Debug.Log(req.error);
else
Debug.Log("Uploaded " + files.Length + " files Successfully");
}但是，这些文件是在目标位置创建的，大小为0字节。
这是我的Uploader.php代码
<$php
$total = count($_FILES['files']['name']);
$uploadError = false;
for ( $i = 0; $i < $total; $i++)
{
$tmpFilePath = $_FILES['files']['tmp_name'][$i];
if ($tmpFilePath != "")
{
$newFilePath = "Uploads/".$_FILES['files']['name'][$i];
if (!move_uploaded_file($tmpFilePath, $newFilePath))
$uploadError = true;
}
}
if ($uploadError)
echo "Upload Error";
else
echo "Uploaded Successfully";
?>我使用了这个HTML示例作为参考。而在浏览器中，HTML代码可以完美地工作。Unity中存在问题。
<form enctype="multipart/form-data" action="Uploader.php" method="POST">
Choose a file to Upload:
<input type="file" name="files[]" multiple="multiple" /><br>
<input type="submit" value="Upload File" />
</form>回答 1
Stack Overflow用户
回答已采纳
发布于 2019-05-16 03:08:34
在for loop中，在C#代码中，在请求文件之后，我们必须在获取文件时让步。所以在请求文件后使用yield return files[i].SendWebRequest();就可以解决这个问题。修改后的代码如下：
IEnumerator UploadMultipleFiles()
{
string[] path = new string[3];
path[0] = "D:/File1.txt";
path[1] = "D:/File2.txt";
path[2] = "D:/File3.txt";
UnityWebRequest[] files = new UnityWebRequest[path.Length];
WWWForm form = new WWWForm();
for (int i = 0; i < files.Length; i++)
{
files[i] = UnityWebRequest.Get(path[i]);
yield return files[i].SendWebRequest();
form.AddBinaryData("files[]", files[i].downloadHandler.data, Path.GetFileName(path[i]));
}
UnityWebRequest req = UnityWebRequest.Post("http://localhost/File%20Upload/Uploader.php", form);
yield return req.SendWebRequest();
if (req.isHttpError || req.isNetworkError)
Debug.Log(req.error);
else
Debug.Log("Uploaded " + files.Length + " files Successfully");
}代码的其余部分就可以了。PHP代码没有变化。HTML代码仅供参考。
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原文链接：
https://stackoverflow.com/questions/56155603
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