我有这个代码的PHP脚本
$tmp_image = $_FILES["file"]["tmp_name"];
$data = mysql_real_escape_string(file_get_contents($tmp_image));
$sql = "update userinfo set Picture = '$data' where ID = '$ID'";
$returnArray['Data'] = $sql.$ID;
$result=mysql_query($sql) or die(mysql_error());
我有.Net代码
MultipartFormDataContent content = new MultipartFormDataContent();
HttpContent data = new ByteArrayContent(x.PixelBuffer.ToArray());//x=WriteableBitmap with image
data.Add(content, '"'+ "file"+'"', '"'+"userphoto.jpg"+'"');
HttpClient client = new HttpClient();
var post = await client.PostAsync(ServerPath, data);
var result = await post.Content.ReadAsStringAsync();
并且在代码运行完成后,我看到结果具有此值 {"数据":空} 并且表格没有更新 据我所知,sql查询没有运行,因为我发送了错误的图像字节数组。但无法理解我发送的错误。