我在linux上使用v4l2库,拍一张照片并想通过c程序将其发送到php服务器。 我想使用套接字来做到这一点。但我不知道如何将图像传递给请求。 这是我的示例代码:
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));
我可以使用这种方式将图像发布到服务器(包括其名称)吗? 对我有什么建议吗? 谢谢 PS:抱歉我的英语水平。
最佳答案
您仅包含文件名。您必须将整个图像文件内容包含到后期数据流中。使用 POST
请求提交二进制数据的表单应使用 multipart/form-data
内容类型。您不能使用 application/x-www-form-urlencoded
类型。
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.
您可以这样调整代码:
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));
发送数据后,您应该读取服务器响应,例如:
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);
如果您只是关闭套接字而不等待响应,服务器可能会提示并返回错误。您还应该检查响应是否确认有效请求。
关于php - 将图像从linux(使用不带curl的c)发布到php服务器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29386558/