调度 python 脚本时出错：fcron[12036]: 无法执行 '/usr/bin/sendmail': 没有这样的文件或目录

Question

我有一个 python 脚本，我想用 fcrontab 安排它。此脚本根据其中一列中设置的状态字段发送电子邮件。当我调用以下内容时，脚本以预期的方式执行：

python2 mail.py

这会产生以下预期输出：

Successfully sent email... 
Update sent status on EdPost...  xxxxxx
Update sent status on EdPost...  xxxxxx
Update sent status on EdPost...  xxxxxx
Update sent status on EdPost...  xxxxxx
Update sent status on EdPost...  xxxxxx
Update sent status on EdPost...  xxxxxx
Update sent status on EdPost...  xxxxxx
Update sent status on EdPost...  xxxxxx

然而，当我尝试运行此脚本作为 cron 作业时，我最终遇到了上述错误。

计划任务：

0 15 * * * /home/x/JobParse/mail.py >> /home/x/JobParse/mailer.log

错误：

fcron[23191]: Couldn't exec '/usr/bin/sendmail': No such file or directory

我有另一个 python 脚本以类似的方式安排运行与 #!/usr/bin/env python2 shebang 一样。不过该脚本不使用 smtplib 库。

我要安排的脚本如下：

mail.py

#!/usr/bin/python2
import smtplib
from db import get_records,update_sent_status
import traceback
import datetime
configs = [line.rstrip('\n') for line in open('configs')]
usr = configs[0]
pwd = configs[1]
rcvr = configs[2]
receivers = [rcvr]
date = datetime.datetime.now().date()
posts = []
edpost_tbl_rows= ""
appli_tbl_rows= ""

ed_posts = get_records(0) # * non-appli and  unsent
ed_posts_ids = [row[0] for row in ed_posts]
ed_posts_sent = get_records(1) # post_id

uniq_appli = get_records(2) # unique unsent appli
uniq_appli_ids = [row[0] for row in uniq_appli]

all_appli_sent = [row[0] for row in get_records(4)] # all app_id sent
all_appli_unsent = [row[0] for row in get_records(5)] # all app_id unsent


# No jobs to send so exit
if len(all_appli_unsent) == 0 and len(ed_posts) == 0:
    print "Exit: No new unique posts.."
    exit()

# Transform db data into html table row format
for row in ed_posts:
    edpost_tbl_rows += \
            "<tr style=\"border:1px solid black\"><td style=\"border:1px solid black\">"\
            +str(row[0])\
            +"</td><td style=\"border:1px solid black\"><a href="\
            +'"'\
            +row[1]\
            +'"'\
            +">"\
            +row[4]\
            +"</a></td><td style=\"border:1px solid black\">"\
            +row[5]\
            +"</td><td style=\"border:1px solid black\">"\
            +row[6]\
            +"</td></tr>"

# Transform applitrack posts into html table
for row in uniq_appli:
    # check if sent before - app_id not unique on edpost
    if row[3] not in all_appli_sent:
        print "New unique & unsent applitrack post... ",row[3]
        appli_tbl_rows += \
            "<tr style=\"border:1px solid black\"><td style=\"border:1px solid black\">"\
            +str(row[0])\
            +"</td><td style=\"border:1px solid black\"><a href="\
            +'"'\
            +row[2]\
            +'"'\
            +">"\
            +str(row[3])\
            +"</a></td><td style=\"border:1px solid black\"><a href="\
            +'"'\
            +row[1]\
            +'"'\
            +">"\
            +row[4]\
            +"</a></td><td style=\"border:1px solid black\">"\
            +row[5]\
            +"</td><td style=\"border:1px solid black\">"\
            +row[6]\
            +"</td></tr>"
    else:
        print "Skipping applitrack post... ",row[3]


message = """From: JobUpdates <{usr}>
To: xxxx <{rcvr}>
MIME-Version: 1.0
Content-type: text/html; charset=utf-8
Subject: New Job Postings: {date}

<p>Check out these postings you may have not have seen yet :)</p>
<p>Searched: words

<h2>Jobs With Applitrack</h2>
<table style="width:100%;border:1px solid black">
    <tr style="border:1px solid black">
        <th style="border:1px solid black">EdPost ID</th>
        <th style="border:1px solid black">AppliTrack ID</th>
        <th style="border:1px solid black">Description</th>
        <th style="border:1px solid black">Post DT</th>
        <th style="border:1px solid black">Exp DT</th>
    </tr>
    {appli_tbl_rows}
</table>
</br>
</br>
<h2>Jobs Without Applitrack</h2>
<table style="width:100%;border:1px solid black">
    <tr style="border:1px solid black">
        <th style="border:1px solid black">EdPost ID</th>
        <th style="border:1px solid black">Description</th>
        <th style="border:1px solid black">Post DT</th>
        <th style="border:1px solid black">Exp DT</th>
    </tr>
    {edpost_tbl_rows}
</table>
""".format(usr=usr,rcvr=rcvr,date=date,edpost_tbl_rows=edpost_tbl_rows,appli_tbl_rows=appli_tbl_rows)

try:
    smtpObj = smtplib.SMTP('localhost',1025)
    smtpObj.login(usr,pwd)
    smtpObj.sendmail(usr, rcvr, message)
    print "Successfully sent email... "
    for post_id in ed_posts_ids:
        print "Update sent status on EdPost... ",post_id
        update_sent_status(post_id)

    for app_id in all_appli_unsent:
        print "Update sent status on Appli... ",app_id
        update_sent_status(app_id,1)

except smtplib.SMTPException:
    print "ERROR: unable to send email... "
    print traceback.format_exc()

我已经用 #!/usr/bin/env python2 shebang 试过了，但无济于事。

Answer 1

我能够在不安装 sendmail 的情况下解决问题。这是通过将 Python 路径移动到脚本开头的脚本来完成的，如下所示：

import os
abspath = os.path.abspath(__file__)
dname = os.path.dirname(abspath)
os.chdir(dname)

查看 Pitto 的回答。

我猜测从 crontab 本身更新 cron 环境的路径变量也是一个可行的解决方案。

调度 python 脚本时出错：fcron[12036]: 无法执行 '/usr/bin/sendmail': 没有这样的文件或目录

Error when scheduling python script: fcron[12036]: Couldn't exec '/usr/bin/sendmail': No such file or directory

python

cron

smtplib

python-2.7