从Linux（使用不带curl的c）将图像发布到php服务器

Question

I am using v4l2 library on linux, take a picture and want to send it to a php server via c program. I want to using a socket to do it. But i don't know how to pass the image to request . This is my sample code:

int portno =        80;
struct sockaddr_in serv_addr;
int sockfd, bytes, sent, received, total;
sockfd = socket(AF_INET, SOCK_STREAM, 0);
char message[1024],response[4096];
if (sockfd < 0){
    printf("ERROR opening socket");
}
memset(&serv_addr,0,sizeof(serv_addr));
serv_addr.sin_family = AF_INET;
serv_addr.sin_port = htons(portno);
if(inet_pton(AF_INET, CONST_DOMAIN, &serv_addr.sin_addr)<=0){
    printf("\n inet_pton error occured\n");
    return 1;
}
if (connect(sockfd,(struct sockaddr *)&serv_addr,sizeof(serv_addr)) < 0) {
    printf("ERROR connecting");
}
char content[1024];
char *contentTemp="image_name=%s";
sprintf(content,contentTemp,imageName);
char *headerTemp="POST %supload.php HTTP/1.0\r\nHost: %s\r\nContent-Type: application/x-www-form-urlencoded\r\nContent-length: %d\r\n\r\n%s";
sprintf(message,headerTemp,SERVICE_PATH,SERVICE_HOST,strlen(content),content);
write(sockfd,message,strlen(message));

Can i using this way to post an image to server (include its name) ? Any suggest for me ? Thanks PS: sorry about my english skill.

Answer 1

You are including only file name. You have to include the whole image file contents into post data stream. Forms submitting binary data with POST request should use multipart/form-data content type. You can't use application/x-www-form-urlencoded type.

From HTML 4.01 specification :

The content type "application/x-www-form-urlencoded" is inefficient for sending large quantities of binary data or text containing non-ASCII characters. The content type "multipart/form-data" should be used for submitting forms that contain files, non-ASCII data, and binary data.

You could adjust your code like this:

char *filename="file.jpg"; // this example uses jpeg

// optionally load file from filesystem
// though I think you have it in a buffer, don't you?
FILE *file = fopen(filename, "rb");
char binary[1024]; // adjust buffer size to your needs
size_t filesize = fread(binary, 1, sizeof(binary), file);
// check for error here to make sure read succeeded
fclose(file);

// multipart/form-data POST header
const char *headerTemp = "POST %supload.php HTTP/1.0\r\n"
                 "Host: %s\r\n"
                 "Content-Type: multipart/form-data; boundary=BoUnDaRy\r\n"
                 "Content-Length: %lu\r\n"
                 "\r\n";
// first and only part beginning
const char *bodyTemp =
                 "--BoUnDaRy\r\n"
                 "Content-Disposition: form-data; name=\"file\"; filename=\"%s\"\r\n"
                 "Content-Type: image/jpeg\r\n"
                 "Content-Transfer-Encoding: binary\r\n"
                 "\r\n";
// and ending
const char body2[] = "\r\n"
                 "--BoUnDaRy--\r\n";
char body1[1024]; // adjust buffer size to your needs
// calculate body size, will be included in Content-Length header
size_t body_size = strlen(body1) + strlen(body2) + filesize;
snprintf(header, 1024, headerTemp, SERVICE_PATH, SERVICE_HOST, body_size);
snprintf(body1, 1024, bodyTemp, filename);
// you should add checking for each write return value
write(sockfd, header, strlen(header));
write(sockfd, body1, strlen(body1));
write(sockfd, binary, filesize);
write(sockfd, body2, strlen(body2));

After sending data you should read server response, for example:

while (1) {
    ssize_t result = recv(sockfd, response, sizeof(response), 0);
    if (result == 0) {
        break;
    } else if (result < 0) {
        perror("reading socket failed");
        break;
    }
    printf("%s\n", response);
}
close(sockfd);

If you just close socket without waiting for the response server may complain and return error. You should also check if the response confirms valid request.