Question

我的程序的目标是自动将文本文件的内容发送到电子邮件地址。虽然发送工作完全正常，但我遇到的问题是我似乎无法找到一种方法来获取字符串“line”，其中我已将文本文件的内容存储到payload_text常量char中。我是业余程序员，对libcurl很新，但由于这是学校项目的一部分，任何帮助都将不胜感激。

我将不提及我试图解决这个问题的方法，因为回想起来，我所有的潜在解决方案都是愚蠢的。您将在下面找到相关代码。

#include <stdio.h>
#include <string.h>
#include <curl.h>
#include <iostream>
#include <fstream>
using namespace std;


#define FROM    "<example@gmail.com>"
#define TO      "example@gmail.com>"
#define CC      "<example@gmail.com>"





static const char *payload_text[] = {
  "Date: Mon, 29 Nov 2010 21:54:29 +1100\r\n",
  "To: " TO "\r\n",
  "From: " FROM "(Example User)\r\n",
  "Cc: " CC "(Another example User)\r\n",
  "Message-ID: <dcd7cb36-11db-487a-9f3a-e652a9458efd@"
  "rfcpedant.example.org>\r\n",
  "Subject: SMTP TLS example message\r\n",
  "\r\n", /* empty line to divide headers from body, see RFC5322 */
  "The body of the message starts here.\r\n",
  "\r\n",
  "It could be a lot of lines, could be MIME encoded, whatever.\r\n",
  "Check RFC5322.\r\n",
  NULL
};

struct upload_status {
  int lines_read;
};

static size_t payload_source(void *ptr, size_t size, size_t nmemb, void *userp)
{
  struct upload_status *upload_ctx = (struct upload_status *)userp;
  const char *data;

  if((size == 0) || (nmemb == 0) || ((size*nmemb) < 1)) {
    return 0;
  }

  data = payload_text[upload_ctx->lines_read];

  if(data) {
    size_t len = strlen(data);
    memcpy(ptr, data, len);
    upload_ctx->lines_read++;

    return len;
  }

  return 0;
}

int main(void)
{

  string line;
  ifstream myfile ("log.txt");
  if (myfile.is_open())
  {
    while ( getline (myfile,line) )
    {
      cout << line << '\n';
    }
    myfile.close();
  }
const char * lchr = line.c_str();


  CURL *curl;
  CURLcode res = CURLE_OK;
  struct curl_slist *recipients = NULL;
  struct upload_status upload_ctx;

  upload_ctx.lines_read = 0;

  curl = curl_easy_init();
  if(curl) {
    /* Set username and password */
    curl_easy_setopt(curl, CURLOPT_USERNAME, "example@gmail.com");
    curl_easy_setopt(curl, CURLOPT_PASSWORD, "********");
    curl_easy_setopt(curl, CURLOPT_URL, "smtp://smtp.gmail.com:587");
    curl_easy_setopt(curl, CURLOPT_USE_SSL, (long)CURLUSESSL_ALL);
    curl_easy_setopt(curl, CURLOPT_SSL_VERIFYPEER, 0L);
   curl_easy_setopt(curl, CURLOPT_SSL_VERIFYHOST, 0L);
    curl_easy_setopt(curl, CURLOPT_MAIL_FROM, FROM);
    recipients = curl_slist_append(recipients, TO);
    recipients = curl_slist_append(recipients, CC);
    curl_easy_setopt(curl, CURLOPT_MAIL_RCPT, recipients);
    curl_easy_setopt(curl, CURLOPT_READFUNCTION, payload_source);
    curl_easy_setopt(curl, CURLOPT_READDATA, &upload_ctx);
    curl_easy_setopt(curl, CURLOPT_UPLOAD, 1L);      
    curl_easy_setopt(curl, CURLOPT_VERBOSE, 1L);    
    res = curl_easy_perform(curl);

    if(res != CURLE_OK)
      fprintf(stderr, "curl_easy_perform() failed: %s\n",
              curl_easy_strerror(res));


    curl_slist_free_all(recipients);

    curl_easy_cleanup(curl);
  }

  return (int)res;
}

Answer 1

好吧，伙计，因为没有人回应我自己想出来了。如果有人遇到类似的问题，这里是解决方案。为了访问payload_text const char数组中的行，我直接访问了数组：

payload_text[9] = line;

但是，由于line是一个字符串，因此无法将其存储为payload_text数组的成员。为了解决这个问题，我只是简单地将其转换。

payload_text[9] = line.c_str();

对于草率的解释感到抱歉，正如我所说，我是一个业余爱好者，仍然在学校，希望这对某人有所帮助。

编辑 - 请记住，数组中的最后一行必须始终为NULL，否则程序将崩溃。

遇到libcurl中有效负载文本的问题

1 个答案: