Question

当我使用以下方法将一些图像发布到我的服务器时：

ASIFormDataRequest *request = [[ASIFormDataRequest alloc] initWithURL:url];
[request setShouldStreamPostDataFromDisk:YES];
[request setData:image withFileName:@"file(0)" andContentType:@"image/jpg" forKey:fileString];
[request setData:image withFileName:@"file(1)" andContentType:@"image/jpg" forKey:fileString];
[request setData:image withFileName:@"file(2)" andContentType:@"image/jpg" forKey:fileString];
[request startAsynchronous];

如何在PHP脚本中捕获这个？

我在做

for($i = 0; $i<$imageCount; $i++){

$target_path = "files/";

$target_path = $target_path . basename( $_FILES['file($i)']['name']); 

if(move_uploaded_file($_FILES['file($i)']['tmp_name'], $target_path)) {
  echo "Image: ".  basename( $_FILES['uploadedfile']['name']). 
" has been uploaded";
 } else{
       echo "error uploading";
  }

 }

使用imageCount将发送的图像总数以不同的变量发送到服务器，但每次都会引发上传错误。

Answer 1

echo $target_path;
$fileVar = "file($i)";
$target_path = $target_path.basename($_FILES[$fileVar]['name']);

我认为该错误与该文件有关。

Answer 2

解决方案：

for ($i = 0; $i<$imageCount; $i++){
$target_path = "files/".basename($_FILES['file('.$i.')']['name']); 

if(copy($_FILES['file('.$i.')']['tmp_name'], $target_path)) {
   echo "Image has been uploaded";
}else{
    echo "error uploading";
    print_r($_FILES) ;
 }
}

ASIHTTPRequest和PHP;如何将多个图像流式传输到服务器

2 个答案: