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ok, so I've got this code: $(this).find('article.loading').each( function(i) { var el = this; setTimeout(function () { $(el).replaceWith($('#dumpster article:first')); }, speed); }); I want to replace each element with another but I want a delay between each replace. I can't figure out why this isn't working, it just replaces all of them after one timeout. Any ideas? Thanks. You are looping through the elements and adding a timer to each with the same configuration. Essentially a new timer is instantly set up for each element. On the first tick of all the timers the elements are updated. The

版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。 本文链接：https://blog.csdn.net/Innost/article/details/48228651 深入理解Android之Gradle 格式更加精美的PDF版请到：https://pan.baidu.com/s/1GfN6F8sOaKFAdz5y1bn3VQ下载 weibo分享失效，请各位到百度云盘下载 Gradle是当前非常“劲爆”得构建工具。本篇文章就是专为讲解Gradle而来。介绍Gradle之前，先说点题外话。 一、题外话 说实话，我在大法工作的时候，就见过Gradle。但是当时我一直不知道这是什么东西。而且大法工具组的工程师还将其和Android Studio大法版一起推送，偶一看就更没兴趣了。为什么那个时候如此不待见Gradle呢？因为我此前一直是做ROM开发。在这个层面上，我们用make，mm或者mmm就可以了。而且，编译耗时对我们来说也不是啥痛点，因为用组内吊炸天的神机服务器完整编译大法的image也要耗费1个小时左右。所以，那个时侯Gradle完全不是我们的菜。 现在，搞APP开发居多，编译/打包等问题立即就成痛点了。比如： 一个APP有多个版本，Release版、Debug版、Test版。甚至针对不同APP Store都有不同的版本

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I'd like to clean up my local repository, which has a ton of old branches: for example 3.2 , 3.2.1 , 3.2.2 , etc. I was hoping for a sneaky way to remove a lot of them at once. Since they mostly follow a dot release convention, I thought maybe there was a shortcut to say: git branch -D 3.2.* and kill all 3.2.x branches. I tried that command and it, of course, didn't work. 回答1: Not with that syntax. But you can do it like this: git branch -D 3.2 3.2.1 3.2.2 Basically, git branch will delete multiple branch for you with a single invocation.

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: This is my first attempting at creating my own EF model, and I'm finding myself stuck attempting to create a lookup table association using Code First so I can access: myProduct.Category.AltCategoryID I have setup models and mappings as I understand to be correct, but continue to get error 0019: Each property name in a type must be unique. Property name 'CategoryID' was already defined The following models are represented in my code: [Table("Product", Schema="mySchema")] public class Product { [Key, DatabaseGenerated(System.ComponentModel

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I'm getting that error, but I'm defining a key. Here's my App.js that it's complaining about. import React from 'react'; import Relay from 'react-relay'; import AccountTable from './AccountTable'; class App extends React.Component { render() { return ( <div> <h1>Account list</h1> {this.props.viewer.accounts.edges.map(edge => <AccountTable key={edge.node.id} account={edge.node} /> )} </div> ); } } export default Relay.createContainer(App, { fragments: { viewer: () => Relay.QL` fragment on User { accounts(first: 10) { edges { node { $

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I want to find the indices all the rows of a matrix which have duplicates. For example A = [1 2 3 4 1 2 3 4 2 3 4 5 1 2 3 4 6 5 4 3] The vector to be returned would be [1,2,4] A lot of similar questions suggest using the unique function, which I've tried but the closest I can get to what I want is: [C, ia, ic] = unique(A, 'rows') ia = [1 3 5] m = 5; setdiff(1:m,ia) = [2,4] But using unique I can only extract the 2nd,3rd,4th...etc instance of a row, and I need to also obtain the first. Is there any way I can do this? NB: It must be a method

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: So I thought the following code would run fine in TestNG, although it doesn't: public class Tests { int i = 0; @Test public void testA() { Assert.assertEquals(0, i); ++i; } @Test public void testB() { Assert.assertEquals(0, i); ++i; } } Is there a way to make TestNG fire up a new Tests class for each test method? 回答1: The common solution is to use an @BeforeMethod method to setup test state, @BeforeMethod public void setup() { i = 0; } 回答2: By far the most common solution to this issue I have found is to use ThreadLocal’s and just deal with

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I need a very large list, and am trying to figure out how big I can make it so that it still fits in 1-2GB of RAM. I am using the CPython implementation, on 64 bit (x86_64). Edit: thanks to bua's answer, I have filled in some of the more concrete answers. What is the space (memory) usage of (in bytes): the list itself sys.getsizeof([]) == 72 each list entry (not including the data) sys.getsizeof([0, 1, 2, 3]) == 104 , so 8 bytes overhead per entry. the data if it is an integer sys.getsizeof(2**62) == 24 (but varies according to integer size)

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Suppose I have a list of suits of cards as follows: suits = ["h","c", "d", "s"] and I want to add a type of card to each suit, so that my result is something like aces = ["ah","ac", "ad", "as"] is there an easy way to do this without recreating an entirely new list and using a for loop? 回答1: This would have to be the 'easiest' way >>> suits = ["h","c", "d", "s"] >>> aces = ["a" + suit for suit in suits] >>> aces ['ah', 'ac', 'ad', 'as'] 回答2: Another alternative, the map function: aces = map(( lambda x: 'a' + x), suits) 回答3: If you want to