我刚安装sendmail在Linux操作系统上。
我似乎无法收到电子邮件,这是我在日志中看到的内容:
Aug 10 14:12:10 localhost sendmail[13316]: v7A9CAPC013316: from=root, size=238, class=0, nrcpts=1, msgid=<201708100912.v7A9CAPC013316@localhost.test>, relay=root@localhost
Aug 10 14:12:10 localhost sendmail[13317]: v7A9CAwD013317: from=<root@localhost.test>, size=484, class=0, nrcpts=1, msgid=<201708100912.v7A9CAPC013316@localhost.test>, proto=ESMTP, daemon=MTA, relay=localhost [127.0.0.1]
Aug 10 14:12:10 localhost sendmail[13316]: v7A9CAPC013316: to=sandeelo.danyal@gmail.com, ctladdr=root (0/0), delay=00:00:00, xdelay=00:00:00, mailer=relay, pri=30238, relay=[127.0.0.1] [127.0.0.1], dsn=2.0.0, stat=Sent (v7A9CAwD013317 Message accepted for delivery)
Aug 10 14:12:12 localhost sendmail[13319]: STARTTLS=client, relay=gmail-smtp-in.l.google.com., version=TLSv1/SSLv3, verify=FAIL, cipher=ECDHE-RSA-AES128-GCM-SHA256, bits=128/128
Aug 10 14:12:13 localhost sendmail[13319]: v7A9CAwD013317: to=<sandeelo.danyal@gmail.com>, ctladdr=<root@localhost.test> (0/0), delay=00:00:03, xdelay=00:00:03, mailer=esmtp, pri=120484, relay=gmail-smtp-in.l.google.com. [74.125.133.26], dsn=5.0.0, …Run Code Online (Sandbox Code Playgroud) 我想GridView::widget()在Yii2 中生成的表上应用类.
<?= GridView::widget([
'dataProvider' => $dataProvider,
'filterModel' => $searchModel,
'columns' => [
['class' => 'yii\grid\SerialColumn'],
'mobile_number',
'email:email',
'name',
'national_id',// and so....
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如何在桌面上应用CSS GridView::widget()?我想申请这个:
<table class="table tblSec">
Run Code Online (Sandbox Code Playgroud) Integer.valueOf接受String和int作为参数但是当我通过时Character,它不需要强制转换并且编译器不强制执行任何操作。
字符不扩展String类,它只是实现Serializable和Comparable
Character charc = '1';
System.out.println(Integer.valueOf(charc));
System.out.println(Integer.valueOf(charc.toString()));
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输出:
49
1
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这不是设计缺陷还是我在错误的方向思考?当您拒绝投票时,请在评论中写下原因。
我已经生成了一个sql来查询数据库表,其中保存了用户的帖子.但是我在执行它时遇到错误.我在这里测试了它,它还告诉我它不正确,问题出在users_user_id部分.我可以'弄清楚这里可能有什么问题.任何人都可以帮我解决这个错误.
我的userpost表结构如下:
POST_ID,岗位,职称,users_user_id
SQL:
SELECT title FROM userpost LIMIT 5 OFFSET 10 WHERE users_user_id=9
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