Sendmail命令发送文件作为电子邮件正文以及附件

Question

我想在bash中使用sendmail命令发送电子邮件。电子邮件应该通过读取Input_file_HTML来获得它的身体，它也应该发送与附件相同的输入文件。要做到这一点我已经试过如下：

sendmail_touser() { 
cat - ${Input_file_HTML} << EOF | /usr/sbin/sendmail -oi -t 
From: ${MAILFROM} 
To: ${MAILTO} 
Subject: $1 
Content-Type: text/html; charset=us-ascii 
cat ${Input_file_HTML} 
Content-Transfer-Encoding: 7bit 
MIME-Version: 1.0 
Content-Disposition: attachment; filename: ${Input_file_HTML} 
EOF 
} 

上面的命令是给一封电子邮件，只有Input_file_HTML的附件，它不是在电子邮件正文中写它。你能帮助/指导我吗？我使用Outlook作为电子邮件客户端。我甚至在上面的命令中删除了cat命令，但它也不起作用。

Answer 1

改为使用mutt？

echo "This is the message body" | mutt -a "/path/to/file.to.attach" -s "subject of message" -- recipient@domain.com 

要在Debian系统上安装mutt：

sudo apt-get install -y mutt 

编辑试试这个，如果你只能使用sendmail：

sendmail_attachment() { 
    FROM="$1" 
    TO="$2" 
    SUBJECT="$3" 
    FILEPATH="$4" 
    CONTENTTYPE="$5" 

    (
    echo "From: $FROM" 
    echo "To: $TO" 
    echo "MIME-Version: 1.0" 
    echo "Subject: $SUBJECT" 
    echo 'Content-Type: multipart/mixed; boundary="GvXjxJ+pjyke8COw"' 
    echo "" 
    echo "--GvXjxJ+pjyke8COw" 
    echo "Content-Type: text/html" 
    echo "Content-Disposition: inline" 
    echo "<p>Message contents</p>" 
    echo "" 
    echo "--GvXjxJ+pjyke8COw" 
    echo "Content-Type: $CONTENTTYPE" 
    echo "Content-Disposition: attachment; filename=$(basename $FILEPATH)" 
    echo "" 
    cat $FILEPATH 
    echo "" 
    ) | /usr/sbin/sendmail -t 
} 

使用这样的：

sendmail_attachment "to@example.com" "from@example.com" "Email subject" "/home/user/file.txt" "text/plain"