Question

我的项目需要帮助。这是我的代码：

void ConexaoServer(){
    uint8_t buffer[1024] = {0};

    if (wifi.createTCP(HOST_NAME, HOST_PORT)) {
        Serial.print("Conexao com o Host OK!\r\n");
    } else {
        Serial.print("Conexao com o Host com ERRO!\r\n");
    }

    char *ComandoGET = "GET /teste HTTP/1.1\r\nHost: SmartHome\r\nConnection: close\r\n\r\n";
    wifi.send((const uint8_t*)ComandoGET, strlen(ComandoGET));

    uint32_t len = wifi.recv(buffer, sizeof(buffer), 10000);
    if (len > 0) {
        Serial.print("Received:[");
        for(uint32_t i = 0; i < len; i++) {
            Serial.print((char)buffer[i]);
        }
        Serial.print("]\r\n\r\n");
    }
}

我从Serial

中的Node.js收到

Received:[HTTP/1.1 200 OK
X-Powered-By: Express
Content-Type: text/html; charset=utf-8
Content-Length: 4
ETag: W/"4-wArb1VtkpsN6In8g50pGNw"
Date: Wed, 31 Aug 2016 01:26:41 GMT
Connection: close

Luiz]

但我想收到Received:[Luiz]。我怎么能这样做？

Answer 1

您需要找到双换行符（\r\n\r\n）并在此之后使用字符。

char payload[1024] = "";
char *pl = strstr((char*)buffer, "\r\n\r\n");
if (pl != NULL)
{
    strcpy(payload, pl+4);
    Serial.println(payload);
}

从GET请求中跳过HTTP标头

1 个答案: