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Best practices to handle routes for STI subclasses in rails
...and controllers are littered with redirect_to , link_to , and form_for method calls. Sometimes link_to and redirect_to are explicit in the paths they're linking (e.g. link_to 'New Person', new_person_path ), but many times the paths are implicit (e.g. link_to 'Show', person ).
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How to change the button text of ?
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Doesn't seem to be working for me. It still shows the brorser's default buttons. Also, the demo at github.com/markusslima/bootstrap-filestyle.git doesn't work either - all examples show the default buttons. Am I missing something?
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IntelliJ and Tomcat.. Howto..?
... Netbeans it was "Install, write hit Run and it works"
How do I pull the same thing off in IntelliJ?
7 Answers
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Similar to jQuery .closest() but traversing descendants?
... Unlike the jQuery .closest() function, this matches the element it was called on as well. See my jsfiddle. By changing it to $currentSet = this.children(); // Current place it will start with it's children instead jsfiddle
– allicarn
Nov 5 '1...
Reset Entity-Framework Migrations
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@Tood. +1 to that. Saved me a lot of time. Now if only the EF team could incorporate all that into a Reset-Migrations command. Maybe EF 6...
– Gerald Davis
Jan 8 '14 at 16:28
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jQuery on window resize
...e events.
(css if your best bet if you're just stylizing things on resize (media queries))
http://jsfiddle.net/CoryDanielson/LAF4G/
css
.footer
{
/* default styles applied first */
}
@media screen and (min-height: 820px) /* height >= 820 px */
{
.footer {
position: absolute;
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AngularJS - Any way for $http.post to send request parameters instead of JSON?
I have some old code that is making an AJAX POST request through jQuery's post method and looks something like this:
13 A...
How can I extend typed Arrays in Swift?
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For extending typed arrays with classes, the below works for me (Swift 2.2). For example, sorting a typed array:
class HighScoreEntry {
let score:Int
}
extension Array where Element == HighScoreEntry {
func sort() -> [HighScoreEntry] {
return sort { $0.score < $1....
Python mysqldb: Library not loaded: libmysqlclient.18.dylib
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How to stop flask application without using ctrl-c
I want to implement a command which can stop flask application by using flask-script.
I have searched the solution for a while. Because the framework doesn't provide "app.stop()" API, I am curious about how to code this. I am working on Ubuntu 12.10 and Python 2.7.3.
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