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分享用SendStream发送图片,应如何接收?
客户端如下:
void __fastcall TForm1::SpeedButton7Click(TObject *Sender)
{
TMemoryStream * mem= new TMemoryStream;
Image1->Width=Screen->Width ;
Image1->Height= Screen->Height ;
BitBlt(screenimg->Canvas->Handle ,0,0, Screen->Width, Screen->Height ,GetDC(0),0,0,SRCCOPY);
Image1->Refresh();
Image1->Picture->Bitmap->SaveToStream(mem);
ClientSocket1->Socket->SendStream(mem);
}
服务器端应如何将收到的mem保存到Image2中?
void __fastcall TForm1::SpeedButton7Click(TObject *Sender)
{
TMemoryStream * mem= new TMemoryStream;
Image1->Width=Screen->Width ;
Image1->Height= Screen->Height ;
BitBlt(screenimg->Canvas->Handle ,0,0, Screen->Width, Screen->Height ,GetDC(0),0,0,SRCCOPY);
Image1->Refresh();
Image1->Picture->Bitmap->SaveToStream(mem);
ClientSocket1->Socket->SendStream(mem);
}
服务器端应如何将收到的mem保存到Image2中?
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sam200401 2003-02-19
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up
christopher_hp 2003-02-19
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先用SaveToStream()把图片转为FileStream
然后用SendStream发，
接受先接受长度ReceiveLength()
然后ReceiveBuf()收，收的时候先用TMemoryStream申明一块内存，然后接受完后SaveToFile()
然后用SendStream发，
接受先接受长度ReceiveLength()
然后ReceiveBuf()收，收的时候先用TMemoryStream申明一块内存，然后接受完后SaveToFile()
Jonix 2003-02-19
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用ReceiveBuf方法收到，收之前先用ReceiveLength取长度
pp616 2003-02-19
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慢满收啊。
