我是MySQL的新手.我正在尝试编写一个在表中插入行的查询,但前提是在5分钟前插入了上一行.另外,我希望我的PHP代码能够知道是否插入了数据.这是我的尝试:
IF (SELECT max(EntryDate) from MyTable) < DATE_SUB(CURRENT_DATE() INTERVAL 5 MINUTE) THEN
INSERT INTO MyTable (...) values (...)
ELSE
Select false
END IF
不幸的是,但并不奇怪,这会产生语法错误(错误对于问题的位置非常模糊).
我做了些蠢事吗？
在Java中是generic,cast和threadsafe保留关键字？我知道java中的52个关键字,但正在寻找新的保留关键字列表.
下面是一些CSS代码.
.form-field {min-height: 20px; margin-bottom: 10px; padding-top: 4px; width: 80px;}
.form-field TEXTAREA, INPUT[type='text'] {position: absolute; left: 100px; top: 0px; height: 15px;}
.form-field TEXTAREA {height: 80px;}
所以每次有一个input或者textarea在div.form-field该CSS应适用.
虽然只有一个INPUT[type='text'](甚至在外面.form-field)css得到应用.我如何阻止它这样做？
我正在尝试根据特定字符拆分字符串,然后计算每个部分中的字符数.有没有办法做到这一点？
所以我有:
<a href="#" class="splitMe" title="Testing | this out">blah</a>
$(document).ready(function() {
$('.splitMe').each(function() {
var item = $(this).attr('title');
var characters = item.split("|");
// Here's where I get stuck...
// Tried various methods of length, but haven't been able to get it to work
// Most recent version that failed miserably...
var first = characters[0].text().length;
var second = characters[1].text().length;
alert(first+" "+second); //Yields characters[0] is not a function
});
});
我想在我的urls.py中添加一些网址.
urls.py
urlpatterns = patterns('',
# Examples:
# url(r'^$', 'Permissions.views.home', name='home'),
# url(r'^Permissions/', include('Permissions.foo.urls')),
#url(r'^Permissions$', include('Permissions.Authenication.urls')),
url(r'^admin/doc/', include('django.contrib.admindocs.urls')),
url(r'^admin/', include(admin.site.urls)),
url(r'^$',RedirectView.as_view(url='/welcome/'),
)
urlpatterns += patterns('',
url(r'^register/$', 'drinker.views.DrinkerRegistration',name='DrinkerRegistration'),
url(r'^welcome/$','drinker.views.DrinkerWelcome',name='DrinkerWelcome'),
url(r'^home/(?P<userpk>[^/]+)/$','drinker.views.DrinkerHome', name='DrinkerHome'),
)
错误:
Traceback:
File "/Users/cnnlakshmen_2000/Projects/env/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
101. request.path_info)
File "/Users/cnnlakshmen_2000/Projects/env/lib/python2.7/site-packages/django/core/urlresolvers.py" in resolve
298. for pattern in self.url_patterns:
File "/Users/cnnlakshmen_2000/Projects/env/lib/python2.7/site-packages/django/core/urlresolvers.py" in url_patterns
328. patterns = getattr(self.urlconf_module, "urlpatterns", self.urlconf_module)
File "/Users/cnnlakshmen_2000/Projects/env/lib/python2.7/site-packages/django/core/urlresolvers.py" in urlconf_module
323. self._urlconf_module = import_module(self.urlconf_name)
File "/Users/cnnlakshmen_2000/Projects/env/lib/python2.7/site-packages/django/utils/importlib.py" in import_module
35. __import__(name)
Exception Type: SyntaxError at /
Exception Value: invalid …我正在尝试基于属性列表构建一个sparql查询,但是我得到一个错误,说明查询结果不好.问题是我不知道该怎么做才能修复它.
这是功能:
def create_query(dbpedia_uri, props):
#props are something like this ('dbpedia-owl', 'birthdate')
filters = ''
for prop in QUERIES_DICT[ename]:
filters += ' OPTIONAL { ?x %s:%s ?%s. } \n' % (corresp_dict[prop[0]], prop[1], prop[1])
query = u"""
SELECT * WHERE {
. <%s>.
?x dbpedia-owl:abstract ?abstract.
%s
FILTER (LANG(?abstract) = 'en')
}
""" % (dbpedia_uri, filters)
return query
这是我得到的查询:
u"\n SELECT * WHERE { <http://dbpedia.org/resource/Tim_Cook>\n ?x dbpedia-owl:abstract ?abstract.\n OPTIONAL { ?x dbpedia-owl:birthDate ?birthDate. }\n OPTIONAL { ?x dbpedia-owl:birthPlace ?birthPlace. }\n OPTIONAL …编写一个函数compute_bill,它将参数food作为输入,并通过循环遍历食物列表并汇总列表中每个项目的成本来计算您的账单.
现在,请继续并忽略您要结算的商品是否有货.
请注意,您的功能应适用于任何食物清单.
给定的代码是
groceries = ["banana", "orange", "apple"]
stock = { "banana": 6,
"apple": 0,
"orange": 32,
"pear": 15
}
prices = { "banana": 4,
"apple": 2,
"orange": 1.5,
"pear": 3
}
我写过:
def compute_bill(food):
total = 0;
for f in food:
if stock[f] > 0:
total+=prices[f]
stock[f] -=1
return total
compute_bill(groceries)
错误信息是
哎呀,再试一次!当['apple']用作输入时,您的代码似乎不起作用 - 它返回0而不是2.
我知道python中有许多魔术方法会影响对象在某些情况下的行为方式(比如定义__cmp__(self, other)改变它与其他实例相比时的工作方式),但我想知道,有没有办法让改变在'in'运算符中如何调用对象的行为？
if thing in custom_object:
call_the_object_in_a_customized_way()
有没有办法做到这一点？
如果我在列表中有m个项目,那么检查列表中的n个项目是否满足特定条件的最快方法是什么？例如:
l = [1,2,3,4,5]
如何检查列表中的任何两项是否符合条件x%2 == 0？
天真的方法是使用嵌套for循环:
for i in l:
for j in l:
if not i%2 and not j%2:
return True
但这是一种非常低效的检查方式,如果我想在200万到1000万件物品的清单中检查任何50,000件商品,那将变得特别难看.
为什么这不合法？
class Base {
public:
virtual void bar() = 0;
virtual void barc() const = 0;
};
class Derived: public Base {
public:
// deliberately omitted
// virtual void bar()
virtual void barc() const override { };
};
int main() {
const Derived b;
b.barc();
// b.bar();
// fails as expected with "passing 'const Bar' as 'this' argument discards qualifiers"
}
这失败并出现以下错误,因为我没有定义virtual void bar():
prog.cpp: In function 'int main()':
prog.cpp:16:12: error: cannot declare variable 'd' to be of …