each

$().each vs $.each vs for loop in jQuery?

由 坚强是说给别人听的谎言 提交于 2019-11-26 22:21:05
I Can't understand why it is happening. I read here that : The first $.each constitutes a single function call to start the iterator. The second $(foo.vals).each makes three function calls to start the iterator. The first is to the $() which produces a new jQuery wrapper set (Not sure how many other function calls are made during this process). Then the call to $().each. And finally it makes the internal call to jQuery.each to start the iterator. In your example, the difference would be negligible to say the least. However, in a nested use scenario, you might find performance becoming an issue

each方法

由 自古美人都是妖i 提交于 2019-11-26 21:06:10
<!DOCTYPE html><html lang="en"><head> <meta charset="UTF-8"> <title>Title</title></head><body> <div> <ul> <li>1</li> <li>1</li> <li>1</li> <li>1</li> <li>1</li> <li>1</li> <li>1</li> <li>1</li> </ul> </div> <script src="js/jquery-1.12.4.js"></script> <script> $(function () { $("ul>li").each(function (index,ele) { console.log(index); console.log(ele); /*$(ele).css("color","red");可以这样使用对象*/ console.dir(arguments[0]+"====="+arguments[1]); }) /*index 序号 ele 该对象<li>1</li>*/ }) </script></body></html> 来源： https://www.cnblogs.com/cycczh/p/11333353.html

cycling through a list of colors with sass

由 亡梦爱人 提交于 2019-11-26 20:46:55
It is possible to have list of three colors: $color-list: x y z; And then apply these three colors by cycling through them and adding them to on an unordered list item. I want: <li>row 1</li> (gets color x) <li>row 2</li> (gets color y) <li>row 3</li> (gets color z) <li>row 4</li> (gets color x) and so on and so forth. I had tried to use the @each ( http://sass-lang.com/docs/yardoc/file.SASS_REFERENCE.html#each-directive ) function but then it just stops applying color after the first time through the list. I want the colors to keep cycling until it runs out of list items to apply them to. is

Excel VBA For Each Worksheet Loop

由 别来无恙 提交于 2019-11-26 20:36:13
问题 I am working on code to basically go through each sheet in my Workbook, and then update column widths. Below is the code I wrote; I don't receive any errors, but it also doesn't actually do anything. Any help is greatly appreciated! Option Explicit Dim ws As Worksheet, a As Range Sub forEachWs() For Each ws In ActiveWorkbook.Worksheets Call resizingColumns Next End Sub Sub resizingColumns() Range("A:A").ColumnWidth = 20.14 Range("B:B").ColumnWidth = 9.71 Range("C:C").ColumnWidth = 35.86 Range

Return a value when using jQuery.each()?

由 坚强是说给别人听的谎言 提交于 2019-11-26 20:34:40
I want to return false and return from function if I find first blank textbox function validate(){ $('input[type=text]').each(function(){ if($(this).val() == "") return false; }); } and above code is not working for me :( can anybody help? You are jumping out, but from the inner loop, I would instead use a selector for your specific "no value" check, like this: function validate(){ if($('input[type=text][value=""]').length) return false; } Or, set the result as you go inside the loop, and return that result from the outer loop: function validate() { var valid = true; $('input[type=text]').each

Clone an undirected graph. Each node in the graph contains alabeland a list of itsneighbors. OJ's undirected graph serialization: Nodes are labeled uniquely. We use#as a separator for each node, an

由 生来就可爱ヽ(ⅴ＜●) 提交于 2019-11-26 20:21:45
/** * Definition for undirected graph. * class UndirectedGraphNode { * int label; * ArrayList<UndirectedGraphNode> neighbors; * UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); } * }; */ import java.util.*; public class Solution { public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) { if(node == null ){ return null; } Map<UndirectedGraphNode,UndirectedGraphNode> map = new HashMap<UndirectedGraphNode,UndirectedGraphNode>(); return clone(node,map); } public UndirectedGraphNode clone(UndirectedGraphNode node, Map<UndirectedGraphNode

Nested jQuery.each() - continue/break

由 微笑、不失礼 提交于 2019-11-26 19:56:38
问题 Consider the following code: var sentences = [ 'Lorem ipsum dolor sit amet, consectetur adipiscing elit.', 'Vivamus aliquet nisl quis velit ornare tempor.', 'Cras sit amet neque ante, eu ultrices est.', 'Integer id lectus id nunc venenatis gravida nec eget dolor.', 'Suspendisse imperdiet turpis ut justo ultricies a aliquet tortor ultrices.' ]; var words = ['ipsum', 'amet', 'elit']; $(sentences).each(function() { var s = this; alert(s); $(words).each(function(i) { if (s.indexOf(this) > -1) {

LR(1)语法分析器生成器(生成Action表和Goto表)java实现(二)

由 倖福魔咒の 提交于 2019-11-26 19:41:40
　　本来这次想好好写一下博客的...结果耐心有限,又想着烂尾总比断更好些。于是还是把后续代码贴上。不过后续代码是继续贴在BNF容器里面的...可能会显得有些臃肿。但目前管不了那么多了。先贴上来吧hhh。说不定哪天觉得羞耻又改了呢。参考资料建议参考《编译器设计》一书。 　　目前完成进度 : 目前已经完成了表驱动,通过函数输出这个Action 和 Goto表。然后使用者就可以根据两个表来进行LR(1)语法分析。且经过比对,发现和书上的例子(括号语法)是完全吻合的。 　　 1 package cn.vizdl.LR1.version3; 2 3 import java.util.ArrayList; 4 import java.util.HashMap; 5 import java.util.HashSet; 6 import java.util.List; 7 import java.util.Scanner; 8 import java.util.Set; 9 10 /* 11 项目名 : LR(1) parser generator (LR(1)语法分析器生成器) 12 项目分析 : 13 输入 : 输入某文件内存地址。且内部采用 <Expr> ::= <Term> + <Factor>; 的结构输入的LR(1)语法。 14 这里的仅支持 BNF范式内部的 终结符,非终结符,或运算

python学习笔记--打印嵌套list中每个数据（遍历列表）

由 你。 提交于 2019-11-26 19:02:27
遍历new_list列表中元素 new_list = ["H1","H2",1999] for each_list in new_list: print (each_list); 若列表中包含嵌套列表,怎样处理？ 笨方法：判断列表中元素是不是列表;并继续使用for来循环打印, 缺点：多个嵌套列表时会使代码过长过重复 难读 new_list = ["H1","H2",1999,["hello","day"]] for each_list in new_list: if isinstance(each_list,list): for new_each in each_list: print (new_each) else: print (each_list); 简单方法：编写递归函数 def each_list(list_name): for yuansu in list_name: if isinstance(yuansu,list): each_list(yuansu) else: print (yuansu) new_list = ["H1","H2",1999,["hello","day"],["one","two"]] each_list(new_list) 如果想遇到列表就缩进一次怎么办？ 增加一个形参呗； def each_list(list_name,level=0):

what's different between each and collect method in Ruby [duplicate]

由 送分小仙女□ 提交于 2019-11-26 18:35:29
This question already has an answer here: Array#each vs. Array#map 6 answers From this code I don't know the difference between the two methods, collect and each . a = ["L","Z","J"].collect{|x| puts x.succ} #=> M AA K print a.class #=> Array b = ["L","Z","J"].each{|x| puts x.succ} #=> M AA K print b.class #=> Array Array#each takes an array and applies the given block over all items. It doesn't affect the array or creates a new object. It is just a way of looping over items. Also it returns self. arr=[1,2,3,4] arr.each {|x| puts x*2} Prints 2,4,6,8 and returns [1,2,3,4] no matter what Array
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