我想通过capistrano从我的本地机器进入生产服务器上的rails控制台.我找到了一些要点,例如https://gist.github.com/813291,当我进入控制台时
cap production console
我得到以下结果
192-168-0-100:foldername username $ cap console RAILS_ENV=production
* executing `console'
* executing "cd /var/www/myapp/current && rails console production"
servers: ["www.example.de"]
[www.example.de] executing command
[www.example.de] rvm_path=$HOME/.rvm/ $HOME/.rvm/bin/rvm-shell '1.9.3' -c 'cd /var/www/myapp/current && rails console production'
/var/www/myapp/releases/20120305102218/app/controllers/users_controller.rb:3: warning: already initialized constant VERIFY_PEER
Loading production environment (Rails 3.2.1)
Switch to inspect mode.
这就是它...现在我可以输入一些文字,但没有任何反应......
有谁知道如何获得该工作或我的问题的另一种解决方案？
如果我通过curl向普通网站发出请求,我可以将输出定向到带有标题数据的文件,如下所示:
命令:
curl -is www.google.de > temp.txt
输出:
HTTP/1.1 200 OK
Date: Wed, 29 Feb 2012 16:49:09 GMT
Expires: -1
Cache-Control: private, max-age=0
Content-Type: text/html; charset=ISO-8859-1
Set-Cookie: PREF=ID=16ad7441d475580e:FF=0:TM=1330534149:LM=1330534149:S=2MpGe04BWQ8Kn3P7; expires=Fri, 28-Feb-2014 16:49:09 GMT; path=/; domain=.google.de
Set-Cookie: NID=57=BCgUcqNQU5dMWcBEMP85saDNUdXg2_LG4eZFmS-yr1W38UyTNuB8BU8iiOOsK2747WGJG99uJDW6t8N78D4QzBma7q2EmKQ69-gCjOysFNLGFggrvMxcqEOl7p1hemeQ; expires=Thu, 30-Aug-2012 16:49:09 GMT; path=/; domain=.google.de; HttpOnly
P3P: CP="This is not a P3P policy! See http://www.google.com/support/accounts/bin/answer.py?hl=en&answer=151657 for more info."
Server: gws
X-XSS-Protection: 1; mode=block
X-Frame-Options: SAMEORIGIN
Transfer-Encoding: chunked
<!doctype html><html itemscope itemtype="http://schema.org/WebPage"><head><meta http-equiv="content-type" content="text/html; charset=ISO-8859-1"><meta itemprop="image" content="/images/google_favicon_128.png"><title>Google</title><script>window.google={kEI:"BVdOT7OzE8_ktQaqk-WGDw",getEI:function(a){var d;while(a&&!(a.getAttribute&&(d=a.getAttribute("eid"))))a=a.parentNode;return d||google.kEI},https:function(){return window.location.protocol=="https:"},kEXPI:"31701,33551,34324,34904,35055,35091,36604,36683,36716,36946,37017,37055,37126",kCSI:{e:"31701,33551,34324,34904,35055,35091,36604,36683,36716,36946,37017,37055,37126",ei:"BVdOT7OzE8_ktQaqk-WGDw"},authuser:0,
ml:function(){},kHL:"de",time:function(){return(new Date).getTime()},log:function(a,d,f,h){var e=new Image,g=google,k=g.lc,i=g.li,m="";e.onerror=(e.onload=(e.onabort=function(){delete …添加OCMock框架后我得到了那个奇怪的错误...... :(
ld: file not found: -fobjc-arc clang: error: linker command failed with exit code 1 (use -v to see invocation)
看起来编译器标志是否试图由XCode加载？
任何的想法？
最好的问候,hijolan
这是我的一段代码:
NSUInteger indexArr[] = {1,2,3,4};
NSIndexSet *indexSet = [NSIndexPath indexPathWithIndexes:indexArr length:4];
有没有办法声明indexArr内联,比如
NSIndexSet *indexSet = [NSIndexPath indexPathWithIndexes:{1,2,3,4} length:4];
或者其他的东西？这是因为我正在编写带有一些参考路径的测试,我想保存对任何人都没用的代码行......
最好的问候,hijolen
在使用JS-Views时,在SAPUI5中为表或某事物分配模型或数据绑定非常容易.但是,在使用XML-Views时,我该怎么做呢？
<?xml version="1.0" encoding="UTF-8" ?>
<core:View
xmlns:core="sap.ui.core"
xmlns="sap.ui.commons"
xmlns:table="sap.ui.table"
xmlns:html="http://www.w3.org/1999/xhtml"
controllerName="view.Main">
<Panel text="Hello World from an XML view">
<Button text="Button" press="doSomething"></Button>
<table:Table width="100%" visibleRowCount="5" selectionMode="Single" editable="false">
<table:title><Label text="Wochentage"></Label></table:title>
<table:Column>
<Label text="ID" />
<table:template><TextField value="{id}"></TextField></table:template>
</table:Column>
</table:Table>
</Panel>
</core:View>
我不想给表一个修复id属性并通过调用实现它
sap.ui.getCore().getElementById("idProductsTable").setModel(
demoJSONModel);
在控制器...... :(
objective-c ×2
arrays ×1
bash ×1
capistrano ×1
console ×1
curl ×1
javascript ×1
linker ×1
sapui5 ×1
shell ×1
xcode ×1
xml ×1