var fileRef:FileReference =new FileReference()
var request:URLRequest = new URLRequest(" http://www.[yourdomain].com/fileUploadScript.php")
var variables:URLVariables = new URLVariables();

variables.参数名= "参数值" ;
request.data=variables;
request.method=URLRequestMethod.POST;
fileRef.fileList[i].upload(request);

相关文章:

  • 2021-06-17
  • 2022-12-23
  • 2022-12-23
  • 2022-12-23
  • 2021-06-30
  • 2022-12-23
  • 2021-11-04
猜你喜欢
  • 2022-12-23
  • 2022-12-23
  • 2022-12-23
  • 2022-12-23
  • 2021-08-07
  • 2022-12-23
  • 2021-12-21
相关资源
相似解决方案