Question

我正在使用PIC24F微控制器中的HTTP协议与M95 quectel调制解调器连接。

我正在尝试＆＃34; POST＆＃34;方法。我已经阅读了quectel的HTTP协议数据表。

以下是发送HTTP post Request AT命令：

AT+QIFGCNT
AT+QHTTPURL
AT+QHTTPPOST
AT+QHTTPREAD
AT+QIDEACT

我根据此流程序列编写代码。我把这些命令放在无限循环中。好的，我第一次成功地从服务器读取响应，但后来我收到类似＆＃34; http响应失败的错误＆＃34;要么＆＃34; http没有请求＆＃34;在AT + QHTTPREAD。

我是否必须重复整个HTTP post AT命令，或者我只是重复AT + QHTTPREAD？

CODE：

bool SendAtCommand_Response(char *command, char *response,int Delay);

    while(1)
{

                SendAtCommand_Response("AT\r","OK",500);
                uart1str("Modem is ready\r\n");
                Delayms(2000);
                uart1str("\r\n");
                clear();                        //clear response buffer
                SendAtCommand_Response("AT+CPIN?\r","OK",2000);
                Delayms(300);
                uart1str("\r\n");
                clear();
                SendAtCommand_Response("AT+QIFGCNT=0\r","OK",500);
                Delayms(300);
                uart1str("\r\n");
                clear();
                SendAtCommand_Response("AT+QICSGP=1,\"airtelGPRS.com\"\r","OK",500);
                Delayms(500);
                uart1str("APN is set\r\n");
                uart1str("\r\n");
                clear();
                SendAtCommand_Response("AT+QHTTPURL=37,30\r","CONNECT",500);
                uart1str("URL is set\r\n");
                clear();
                SendAtCommand_Response(HTTP_POST_URL_HS ,"OK",500);
                Delayms(500);
                uart1str("URL Sent\r\n");
                clear();
                SendAtCommand_Response("AT+QHTTPPOST=22,25,10\r","CONNECT",1000);
                Delayms(1000);
                uart1str("\r\n");
                clear();
                SendAtCommand_Response("imeino=865733024598757","OK",100);
                Delayms(1000);
                uart1str("HTTP data sent\r\n");
                uart1str("\r\n");
                clear();
                SendAtCommand_Response("AT+QHTTPREAD=30\r","CONNECT",3000);
                Delayms(200);
                clear();
                SendAtCommand_Response("AT+QIDEACT\r","DEACT OK",500);
                Delayms(200);
                clear();

}

Answer 1

如果您使用相同的URL，则无需使用，只需从AT + QHTTPPOST开始。

如果将另一个URL用于其他内容，则从AT + GHTTPURL开始

尝试在HTTP帖子中读取响应后获取错误

1 个答案: