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
但是我明白了
curl_easy_perform() failed: Couldn't connect to server
我在这里做错了什么?
最佳答案
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
关于c - 使用 libcurl/smtp 和 gmail : login denied 发送电子邮件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37092597/