c - 使用 libcurl/smtp 和 gmail : login denied 发送电子邮件

Question

我知道那里有数百个帖子，但不知何故这对我不起作用。我正在尝试使用 libcurl 发送电子邮件。这是我的代码:

#include <stdio.h>
#include <string.h>
#include <curl/curl.h>

#define FROM    "<myemail@gmail.com>"
#define TO      "<someother@gmail.com>"
#define CC      "<someother2@hotmail.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 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)
{
  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) {
    curl_easy_setopt(curl, CURLOPT_URL, "smtp://smtp.gmail.com:587");
    curl_easy_setopt(curl, CURLOPT_USE_SSL, CURLUSESSL_ALL);
    curl_easy_setopt(curl, CURLOPT_USERNAME, "myemail@gmail.com");
    curl_easy_setopt(curl, CURLOPT_PASSWORD, "mypassword");
    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);

    res = curl_easy_perform(curl);

    /* Check for errors */
    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;
}

我也试过这个example除了文本通过不同的文件通过管道传输外，这非常相似。

但我总是得到

curl_easy_perform() failed: Login denied

然后 gmail 给我发了一条消息说有人试图侵入我的帐户。 (登录凭据正确)

这个curl-config --feature

给我

SSL IPv6 UnixSockets libz NTLM NTLM_WB

阅读几篇博文后，我还尝试使用 smtp://smtp.live.com:465

使用 hotmail

但是我明白了

curl_easy_perform() failed: Couldn't connect to server

我在这里做错了什么？

Answer 1

我不得不花费大量时间来让它工作，但我终于让它工作了。这是我发现的:

首先:

#define FROM    "<myemail@gmail.com>"
#define TO      "<someother@gmail.com>"
#define CC      "<someother2@hotmail.com>"

电子邮件地址中不应有任何“<”或“>”。我犯了重用定义的错误，我终生无法理解为什么登录失败。

第二:

您很可能需要转到此处并允许不安全的应用程序:https://www.google.com/settings/security/lesssecureapps

第三:

如果您使用端口 587，则需要以下行:

curl_easy_setopt(m_curl, CURLOPT_USE_SSL, CURLUSESSL_ALL);

465好像不需要这个

注意:

可以使用端口 465 或 587。465 需要 SSL，因此您需要使用 smtps 而不是 smtp 来指定它 - smtps://smtp.gmail.com:465。我在这里找到了一些相关信息:https://support.google.com/a/answer/176600?hl=en