HBitmap字节数组到php

我正试图以以下方式在我的控制台win32应用程序中制作屏幕截图：

char* getScreenshot(){
    HDC hScreenDC = CreateDC("DISPLAY", NULL, NULL, NULL);
    HDC hMemoryDC = CreateCompatibleDC(hScreenDC);

    int x = GetDeviceCaps(hScreenDC, HORZRES);
    int y = GetDeviceCaps(hScreenDC, VERTRES);

    HBITMAP hBitmap = CreateCompatibleBitmap(hScreenDC, x, y);
    size_t width = GetSystemMetrics(SM_CXVIRTUALSCREEN);
    size_t height = GetSystemMetrics(SM_CYVIRTUALSCREEN);
    HBITMAP hOldBitmap = (HBITMAP)SelectObject(hMemoryDC, hBitmap);

    BitBlt(hMemoryDC, 0, 0, width, height, hScreenDC, 0, 0, SRCCOPY);

    hBitmap = (HBITMAP)SelectObject(hMemoryDC, hOldBitmap);
    char* buffer = new char[x*y]();
    GetBitmapBits(hBitmap,x*y,buffer);
    DeleteDC(hMemoryDC);
    DeleteDC(hScreenDC);
    return buffer;
}

char *buff = getScreenshot();
string a(buff);
send(hClientSocket, buff, a.length(), 0);
if (hClientSocket!=INVALID_SOCKET)
    closesocket(hClientSocket);     

$sock = socket_create(AF_INET, SOCK_STREAM, 0);
set_time_limit(15);
$succ = socket_connect($sock, $ip, $port) or die("timeout");
$output = "";
while($resp = socket_read($sock, 1000)) {
   $output .= $resp;
}
socket_close($sock);
$base64 = base64_encode($output);
echo "<img src='data:image/bmp;base64,$base64' />";

缓冲区应该是包含hbitmap屏幕截图的字节数组

然后我通过一个插座以这种方式发送：

char* getScreenshot(){
    HDC hScreenDC = CreateDC("DISPLAY", NULL, NULL, NULL);
    HDC hMemoryDC = CreateCompatibleDC(hScreenDC);

    int x = GetDeviceCaps(hScreenDC, HORZRES);
    int y = GetDeviceCaps(hScreenDC, VERTRES);

    HBITMAP hBitmap = CreateCompatibleBitmap(hScreenDC, x, y);
    size_t width = GetSystemMetrics(SM_CXVIRTUALSCREEN);
    size_t height = GetSystemMetrics(SM_CYVIRTUALSCREEN);
    HBITMAP hOldBitmap = (HBITMAP)SelectObject(hMemoryDC, hBitmap);

    BitBlt(hMemoryDC, 0, 0, width, height, hScreenDC, 0, 0, SRCCOPY);

    hBitmap = (HBITMAP)SelectObject(hMemoryDC, hOldBitmap);
    char* buffer = new char[x*y]();
    GetBitmapBits(hBitmap,x*y,buffer);
    DeleteDC(hMemoryDC);
    DeleteDC(hScreenDC);
    return buffer;
}

char *buff = getScreenshot();
string a(buff);
send(hClientSocket, buff, a.length(), 0);
if (hClientSocket!=INVALID_SOCKET)
    closesocket(hClientSocket);     

$sock = socket_create(AF_INET, SOCK_STREAM, 0);
set_time_limit(15);
$succ = socket_connect($sock, $ip, $port) or die("timeout");
$output = "";
while($resp = socket_read($sock, 1000)) {
   $output .= $resp;
}
socket_close($sock);
$base64 = base64_encode($output);
echo "<img src='data:image/bmp;base64,$base64' />";

我通过以下方式从php捕捉if：

char* getScreenshot(){
    HDC hScreenDC = CreateDC("DISPLAY", NULL, NULL, NULL);
    HDC hMemoryDC = CreateCompatibleDC(hScreenDC);

    int x = GetDeviceCaps(hScreenDC, HORZRES);
    int y = GetDeviceCaps(hScreenDC, VERTRES);

    HBITMAP hBitmap = CreateCompatibleBitmap(hScreenDC, x, y);
    size_t width = GetSystemMetrics(SM_CXVIRTUALSCREEN);
    size_t height = GetSystemMetrics(SM_CYVIRTUALSCREEN);
    HBITMAP hOldBitmap = (HBITMAP)SelectObject(hMemoryDC, hBitmap);

    BitBlt(hMemoryDC, 0, 0, width, height, hScreenDC, 0, 0, SRCCOPY);

    hBitmap = (HBITMAP)SelectObject(hMemoryDC, hOldBitmap);
    char* buffer = new char[x*y]();
    GetBitmapBits(hBitmap,x*y,buffer);
    DeleteDC(hMemoryDC);
    DeleteDC(hScreenDC);
    return buffer;
}

char *buff = getScreenshot();
string a(buff);
send(hClientSocket, buff, a.length(), 0);
if (hClientSocket!=INVALID_SOCKET)
    closesocket(hClientSocket);     

$sock = socket_create(AF_INET, SOCK_STREAM, 0);
set_time_limit(15);
$succ = socket_connect($sock, $ip, $port) or die("timeout");
$output = "";
while($resp = socket_read($sock, 1000)) {
   $output .= $resp;
}
socket_close($sock);
$base64 = base64_encode($output);
echo "<img src='data:image/bmp;base64,$base64' />";

并发送到套接字：

int length;
char *buff = getScreenshot(&length);
send(hClientSocket, buff, length, 0);

---

但是它继续避免显示…

BMP图像不仅仅是图像中的“位”（这是您的

缓冲区中的位）。BMP图像有一个标题，上面有宽度、高度、每像素位数和其他一些类似的项目

最简单的方法是将HBITMAP保存为图像文件，然后将文件数据加载到套接字代码中。如果不想将其另存为文件，则必须构造一个兼容的头文件，并在实际图像数据之前发送该头文件

至少为了确认您的原则，在编写代码生成标题之前，请尝试“将位图保存到文件”方法。
这肯定会导致问题：

char *buff = getScreenshot();
string a(buff);

您正在获取一个任意二进制数据数组，并试图将其转换为
字符串
对象。
char*
构造函数将需要以null结尾的字符串。这意味着当图像中的一个像素在任何RGB组件中的值为0时，它将停止

为了传输原始像素数据，最好避免转换为
字符串
对象。只需将
char*
指针直接传递到
send（）
函数。不过，您显然需要告诉它正确的大小（任何类似于
strlen（）
的东西都会遇到与
字符串
构造函数相同的问题）。您可以通过引用参数将大小信息从
getScreenshot（）
函数中传递出去

我怀疑在以HTML格式输出图像时，也会出现标题信息缺失的问题。客户端浏览器目前无法确定图像的尺寸（即宽/高多少像素）或数据格式。
感谢您的回复。我试着照你说的做，但结果还是一样