使用 php 从 dart 将多个文件上传到服务器端答案

【问题标题】：Upload multiple files to server side with php from dart使用 php 从 dart 将多个文件上传到服务器端
【发布时间】：2021-08-01 14:09:33
【问题描述】：

.dart

  Future<http.Response> uploadFile(String fileName, List<int> fileBytes) async {
try {
  var request = new http.MultipartRequest("POST", Uri.parse("https://****/***/fileupload.php"));

  for (var i = 0; i < uploadedImage.length; i++) {
    selectedFilesBytes = List.from(uploadedImage[i]);
    request.files.add(http.MultipartFile.fromBytes('file', selectedFilesBytes, contentType: MediaType('application', 'octet-stream'), filename: files[i].name));
  }

  print("request.files.length");
  print(request.files.length);
  var streamedResponse = await request.send();

  return await http.Response.fromStream(streamedResponse);
} catch (e) {
  print(e);
}

文件上传.php

    <?php 

 // Count total files
 $countfiles = count($_FILES['file']);
 error_log($countfiles);
 // Looping all files
 for($i=0;$i<$countfiles;$i++){
   $filename = $_FILES['file']['name'][$i];
   
   // Upload file
   move_uploaded_file($_FILES['file']['tmp_name'][$i],'upload/'.$filename);
    
 }

?>

嗨，我对从 dart 到 php 的文件上传有问题。当我打印出print(request.files.length); 时，它会给出我选择上传的文件数。但在 php 端 $countfiles = count($_FILES['file']); 总是返回 1。为什么会这样？

【问题讨论】：

可能是因为 PHP 无法处理同名的多个参数，除非使用带方括号的“语法”。尝试使用file[] 作为参数名称。

标签： php dart flutter-web

【解决方案1】：

感谢@CBroe 的回答。有用。我添加了最近的代码。

.dart

  Future<http.Response> uploadFile(String fileName, List<int> fileBytes) async {
try {
  var request = new http.MultipartRequest("POST", Uri.parse("https://****/***/fileupload.php"));

  for (var i = 0; i < uploadedImage.length; i++) {
    selectedFilesBytes = List.from(uploadedImage[i]);
    request.files.add(http.MultipartFile.fromBytes('file[]', selectedFilesBytes, contentType: MediaType('application', 'octet-stream'), filename: files[i].name));
  }

  print("request.files.length");
  print(request.files.length);
  var streamedResponse = await request.send();

  return await http.Response.fromStream(streamedResponse);
} catch (e) {
  print(e);
}

.php

<?php 

 // Count total files
 $countfiles = count($_FILES['file']['name']);
 error_log($countfiles);
 // Looping all files
 for($i=0;$i<$countfiles;$i++){
   $filename = $_FILES['file']['name'][$i];
   
   // Upload file
   move_uploaded_file($_FILES['file']['tmp_name'][$i],'upload/'.$filename);
    
 }

?>

【讨论】：