由于某种原因,我转换时间格式,如:03:30 to seconds 3*3600 + 30*60, now.我想把它转换回它的第一个(相同)格式.怎么会这样？
我的尝试:
3*3600 + 30*60 = 12600
12600 / 60 = 210 / 60 = 3.5, floor(3.5) = 3 = hour
现在,分钟呢？
考虑到价值可能就像19:00 or 02:51.
我认为你得到的图片.
顺便说一句,如何转换 2:0 for example to 02:00 using RegEx?
我正在使用Mac OS x 10.10.3 Yosemite和Python 2.7.9 | Anaconda 2.2.0(x86_64)来处理很多python的东西.我正在使用eclipse和谷歌应用引擎.
我已经没有堆栈溢出帖子来读取很多人都有的这个错误,然后通过某种方式解决这个问题对我来说不起作用.
我收到这个错误:
import Image ImportError:没有名为Image的模块
从这段代码:
try:
from PIL import Image
except:
import Image
在我尝试了以下之后:
conda install pillow
sudo pip install pillow
pip install pillow
pip install pil
一些堆栈溢出帖子建议查看该路径是否适用于其他用户,所以我在终端中尝试了这个:
python
Python 2.7.9 |Anaconda 2.2.0 (x86_64)| (default, Dec 15 2014, 10:37:34)
[GCC 4.2.1 (Apple Inc. build 5577)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
Anaconda is brought to you by Continuum Analytics.
Please check out: http://continuum.io/thanks and https://binstar.org …从URL查询字符串获取参数的"需要更少代码"的方法是什么,格式如下？
我目前的网址
www.mysite.com/category/subcategory/#myqueryhash
我把这个代码
$url="http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];
它只返回 www.mysite.com/category/subcategory/
输出应该是:
www.mysite.com/category/subcategory/#myqueryhash
我在wordpress中开发了一个网站。但是现在我面临一个错误：
require(public_html/mywebsite/wp-includes/post.php):failed to open stream: permission denied in public_html/mywebsite/wp-settings.php on line 142
================================================== ==================
Fatal error: require(): Failed opening required '/public_html/mywebsite/wp-includes/post.php'(include_path='.:/opt/apt/php55/usr/share/pear:/opt/alt/php55/usr/share/php') in /public_html/indus/wp-settings.php on line 142
为了解决这个问题，我更改了post.php文件的权限。而且有效。但是第二天，当我试图打开它时，我又遇到了同样的问题。
如何解决呢？
我正在按照这里的说明操作:https://scotch.io/tutorials/easy-node-authentication-setup-and-local,当我在尝试运行我的代码时遇到此错误时,
Error: Failed to lookup view "index.ejs" in views directory "Directory/views"
at EventEmitter.render (Directory/node_modules/express/lib/application.js:579:17)
at ServerResponse.render (Directory/node_modules/express/lib/response.js:961:7)
at Directory/ourMeanJS/config/routes.js:10:13
at Layer.handle [as handle_request] (Directory/node_modules/express/lib/router/layer.js:95:5)
at next (Directory/node_modules/express/lib/router/route.js:131:13)
at Route.dispatch (Directory/node_modules/express/lib/router/route.js:112:3)
at Layer.handle [as handle_request] (Directory/node_modules/express/lib/router/layer.js:95:5)
at Directory/node_modules/express/lib/router/index.js:277:22
at Function.process_params (Directory/node_modules/express/lib/router/index.js:330:12)
at next (Directory/node_modules/express/lib/router/index.js:271:10)
at Directory/node_modules/connect-flash/lib/flash.js:21:5
at Layer.handle [as handle_request] (Directory/node_modules/express/lib/router/layer.js:95:5)
at trim_prefix (Directory/node_modules/express/lib/router/index.js:312:13)
at Directory/node_modules/express/lib/router/index.js:280:7
at Function.process_params (Directory/node_modules/express/lib/router/index.js:330:12)
at next (Directory/node_modules/express/lib/router/index.js:271:10)
这是我的路线设置:
module.exports = function(app, passport) {
// =====================================
// HOME PAGE (with login links) ======== …我想将价格向上(向上或向下)舍入到最接近的$ xx.99
例如:
$17.99 -> Stay as is
$22.03 -> $21.99
$33.85 -> $33.99
$45 -> $44.99
Jmeter：如何使用 BeanShell 预处理器从响应中提取和存储 accountID 变量？我们可以为此使用 JSON 或正则表达式。
我的回复看起来像：
{
"address":{
"street1":"6550 Vallejo Street",
"street2":"",
"city":"Emeryville",
"state":"CA",
"zipcode":"98456",
"country":"Uk"
},
"timezone":"US/Pacific-New",
"reference":"fe_qa001",
"token":"ns7h4rqVZegSZG6yZls6",
"login":"fe_qa001@qa.com",
"accountId":-9223372036762408565,
"firstName":"qa029",
"lastName":"svcqa",
"email":"fe_qa001@qa.com",
"component":[
{
"type":1,
"reference":"12-1",
"version":"1",
"base":"Plastic",
"zipcode":"94556",
"bedsize":"high",
"bed":false,
"componentId":0,
"code":"P5",
"sku":"abcd",
"serial":"1234",
"purchaseDate":1372723200000,
"chamber":2
}
]
}
我是 Jmeter 的新手。所以任何人都可以在这里帮助我。该要求就像从下面给出的响应中提取帐户 ID 并使用 BeanShell 预处理器将其存储在变量中。
我已经下载了Quickblox IOS SDK 2.6.2版.
SDK中有一些演示项目.喜欢"样本聊天"
我已经替换了"kApplicationID""kAuthKey""kAuthSecret""kAccountKey"
但是,当我运行该项目时,它给出了以下错误: -
[QBCore] Response error: Error Domain=com.alamofire.error.serialization.response Code=-1011 "Request failed: not found (404)" UserInfo={com.alamofire.serialization.response.error.response=<NSHTTPURLResponse: 0x7f95b2df65f0> { URL: https://api.quickblox.com/users/by_tags.json?page=1&per_page=100&tags=dev } { status code: 404, headers {
"Access-Control-Allow-Origin" = "*";
"Cache-Control" = "no-cache";
Connection = "keep-alive";
"Content-Length" = 1;
"Content-Type" = "application/json; charset=utf-8";
Date = "Wed, 20 Jan 2016 11:00:56 GMT";
"QB-Token-ExpirationDate" = "2016-01-20 12:59:20 UTC";
"QuickBlox-REST-API-Version" = "0.1.1";
Server = "nginx/1.6.2";
Status = "404 Not Found";
"X-Rack-Cache" = miss;
"X-Request-Id" = 8341005e68907408343438e49020e8dc;
"X-Runtime" = …我是 wordpress 新手，我使用的是 wordpress 4.5 版，我喜欢按子类别显示帖子，请任何人帮助我如何做到这一点。这就是我要的
家长类别名称
子类别 1
Post 1
Post 2
子类别 2
Post 3
Post 4
子类别 3
Post 5
Post 6
...
提前致谢
我正在尝试按时间值组合多个数组。我有一个这样的数组。
[
{
x:"Kasa",
y:" 8",
id:6,
a:0.019444444444444
},
{
x:"Kasa",
y:" 9",
id:6,
a:0.023611111111111
},
{
x:"Kasa",
y:"10",
id:6,
a:0.018055555555556
},
{
x:"Kasa",
y:"11",
id:6,
a:0.025
},
{
x:"Kasa",
y:"12",
id:6,
a:0.0097222222222222
}
],
[
{
x:"Kasap",
y:" 8",
id:7,
a:0.0013888888888889
},
{
x:"Kasap",
y:" 9",
id:7,
a:0.015277777777778
},
{
x:"Kasap",
y:"10",
id:7,
a:0.0027777777777778
},
{
x:"Kasap",
y:"11",
id:7,
a:0.0041666666666667
},
{
x:"Kasap",
y:"12",
id:7,
a:0.019444444444444
}
]
我需要将这 2 个数组与 y 值结合起来。例如组合数组后应该是这样的。
[
{
x:"Kasa",
y:" 8",
id:6, …