我有一个HTML格式,我想将其提交给控制器
@using (Html.BeginForm("RegisterApartmentOwner", "Home", FormMethod.Post,
new { enctype = "multipart/form-data" }))
{
<p>
<label>First Name</label>
<input type="text" placeholder="Enter your first Name" name="firstName" />
<span class="errorMessage"></span>
</p>
<p>
<label>Last Name</label>
<input type="text" placeholder="Enter your last Name" />
<span class="errorMessage"></span>
</p>
<p>
<label>Password</label>
<input type="text" placeholder="Enter your password" name="Password"/>
<span class="errorMessage"></span>
</p>
<p>
<label>Password Again</label>
<input type="text" placeholder="Enter your password again" name="Password2"/>
<span class="errorMessage"></span>
</p>
<p>
<label>Mobile Number</label>
<input type="text" placeholder="Enter your mobile number" />
<span class="errorMessage"></span>
</p>
<p>
<input type="submit" …我有这个json数据
{
"default": [
[
1325876000000,
0
],
[
1325876000000,
0
],
[
1325876000000,
0
],
[
1325876000000,
0
]
],
"direct": [
[
1328196800000,
0
],
[
1328196800000,
100
],
[
1328196800000,
0
],
[
1328196800000,
0
]
],
"Sales": [
[
1330517600000,
0
],
[
1330517600000,
0
],
[
1330517600000,
91
],
[
1330517600000,
0
]
],
"Support": [
[
1332838400000,
0
],
[
1332838400000,
0
],
[
1332838400000,
0
],
[
1332838400000,
0
]
]
}
我想将其更改为以下格式: …
我想每 120 秒运行一些 python 代码。
我试过这个：
class AppServerSvc :
def f(self):
# call f() again in 120 seconds
spider = FantasySerieaSpider()
settings = get_project_settings()
crawler = Crawler(settings)
crawler.signals.connect(reactor.stop, signal=signals.spider_closed)
crawler.configure()
crawler.crawl(spider)
crawler.start()
log.start()
reactor.run() # the script will block here until the spider_closed signal was sent
threading.Timer(120, f).start()
if __name__ == '__main__':
AppServerSvc().f();
我得到了threading is not defined错误
这是我的进口：
import pythoncom
import win32serviceutil
import win32service
import win32event
import servicemanager
import socket
from twisted.internet import reactor
from scrapy.crawler import Crawler …这是路线
Route::get('contact', 'Pages@contact');
是控制器:
<?php
class PagesController extends BaseController {
public function contact()
{
return View::make('hello');
}
}
这是错误消息
500 Internet error
ReflectionException
Class Pages does not exist
我正在使用laravel 4.2
我像这样接受明天的约会
tomorrow = datetime.date.today() + datetime.timedelta(days=1)
self.FirstDateString = str(tomorrow.strftime("%d %b %Y"))
结果是 11 Jun 2014
我像这样解析它:
datetime.strptime('11 Jun 2014', "%d %B %Y").date()
我收到了这个错误:
ValueError: time data '11 Jun 2014' does not match format '%d %B %Y'
但是当我改变Jun时June,它会起作用.
那么,我怎么能告诉tomorrow = datetime.date.today() + datetime.timedelta(days=1)我给我June而不是Jun
这是我的代码
spTAC.InnerText = ((Tabn / TinBCalls) * 100) + "%";
我发现了那个代码而且Tabn是204
TinBCalls是1162年
spTAC保持值为0%
告诉我为什么？我要放松对这件事的看法.
(该行之后没有执行代码).
<td class="informationTableValue" runat="server" id="spTAC">
</td>
要获取CustsExport其名称中包含的所有文件,请执行以下操作:
foreach (string file in Directory.EnumerateFiles(xmlFolder, "*CustsExport*.xml"))
但有一种情况我不想提取其名称Empty以此结尾的文件:
Erb3PCustsExport-303_20080319_063109_000_Empty
请问你能帮帮我吗？
非常感谢
这是我的代码
$conn = mysql_connect($serverName, $username, $password)
or die("Unable to connect to MySQL");
$selected = mysql_select_db($databaseName,$conn)
or die("Could not find the database");
$result = mysql_query("INSERT INTO `user` () VALUES ()");
return mysqli_insert_id($result)
这是个例外
Warning: mysqli_insert_id() expects parameter 1 to be mysqli, boolean given in C:\xampp2\htdocs\XXX\controllers\User.php on line 24
而第24行是 return mysqli_insert_id($result)
我想使用 R studio 在 R 中构建一个包
\n\n当按下时我收到此错误build source package
==> R CMD build CLASS_Projectcompany\n\n* checking for file \'CLASS_Projectcompany/DESCRIPTION\' ... OK\nDuring startup - Warning messages:\n1: Setting LC_CTYPE failed, using "C" \n2: Setting LC_TIME failed, using "C" \n3: Setting LC_MESSAGES failed, using "C" \n4: Setting LC_MONETARY failed, using "C" \n* preparing \'CLASS_Projectcompany\':\n* checking DESCRIPTION meta-information ... ERROR\nMalformed package name\n\nSee section \'The DESCRIPTION file\' in the \'Writing R Extensions\'\nmanual.\n该文件的内容DESCRIPTION是：
Package: CLASS_Projectcompany\nType: Package\nTitle: Class Projectcompany\nVersion: 1.0\nDate: …这是我的弦
2011-01-01T00：00：00.000Z
我将其更改为日期时间对象 org.joda.time.DateTime
这是我的代码
DateTimeFormatter formatter = DateTimeFormat
.forPattern("yyyy-MM-ddTHH:mm:ss.000Z");
DateTime dt = formatter.parseDateTime("2011-01-01T00:00:00.000Z");
我有例外
非法图案成分：T
因此，Tand .000Z始终存在，该如何格式化？
我能够像这样获得电影的演员阵容：
#!/usr/bin/env python
import imdb
ia = imdb.IMDb()
s_result = ia.search_movie('The Untouchables')
the_unt = s_result[0]
print the_unt['cast']
然而，这给了所有演员，我只是在寻找电影中的明星。例如，阿尔帕西诺是教父中的明星
这是我的代码：
Stream.of("d2", "a2", "b1", "b3", "c")
.map(s -> {
System.out.println("map: " + s);
return s.toUpperCase();
})
.anyMatch(s -> {
System.out.println("anyMatch: " + s);
return ((String) s).startsWith("A");
});
输出是：
map: d2
anyMatch: D2
map: a2
anyMatch: A2
我的问题是为什么第一个元素进入流而第三个和第四个没有？
任何详细的解释表示赞赏。