window.showModalDialog 文法，showmodaldialogwindow.showModalDialog()方法用來建立一個顯示HTML內容的模態對話方塊。1、版本支援：showModalDialog() (IE 4+ 支援)2、文法：vReturnValue=window.showModalDialog(sURL[,vArguments][,sFeatures]);3、參數解釋：sURL 必選 字串
(十四)ng-if與ng-show、ng-hide指令的區別和注意事項，ng-ifng-hideangularJS中的ng-show、ng-hide、ng-if指令都可以用來控制dom元素的顯示或隱藏。ng-show和ng-hide根據所給運算式的值來顯示或隱藏HTML元素。當賦值給ng-show指令的值為false時元素會被隱藏，值為true時元素會顯示。ng-hide功能類似，使用方式相反。元素的顯示或隱藏是通過改變CSS
window.showModalDialog 應用，showmodaldialog1、參數傳遞 通過vArguments來傳遞參數，類型不限制，對於字串類型，最大為4096個字元，也可以傳遞對象。例： A.htm <script> window.showModalDialog("B.htm","這個是參數值","help:no;scroll:no");</script>B.htm<script>
angularjs學習筆記之一（ajax請求），angularjsajax實現ajax請求【HTML 程式碼】<!DOCTYPE html><html><head> <meta charset="utf-8"> <meta name="viewport" content="width=device-width,user-scalable=no,
百度面試題找到滿足條件的數組，百度面試題數組給定函數d(n)=n+n的各位之和，n為正整數，如d(78)=78+7+8=93。這樣這個函數可以看成一個產生器，如93可以看成由78產生。定義數A：數A找不到一個數B可以由d(B)=A，即A不能由其他數產生。現在要寫程式，找出1至10000裡的所有符合數A定義的數。回答：申請一個長度為10000的bool數組，每個元素代表對應的值是否可以有其它數產生。開始時將數組中的值都初始化為false。由於大於
【SICP練習】119 練習3.50，sicp練習1193.50 練習3-50原文Exercise 3.50. Complete the following definition, which generalizes stream-map to allow procedures that take multiple arguments, analogous to map in section 2.2.3, footnote 12. (define (stream-map
【SICP練習】120 練習3.51，sicp練習1203.51 練習3-51原文Exercise 3.51. In order to take a closer look at delayed evaluation, we will use the following procedure, which simply returns its argument after printing it:(define (show x) (display-line x)
NOJ 2033 一頁書的書 (組合數＋dp)，noj+dp一頁書的書時間限制(普通/Java) : 1000 MS/ 3000 MS 運行記憶體限制 : 65536 KByte總提交 : 55 測試通過 : 12 題目描述
JAVA學習筆記(四十)- 守護線程與中斷線程，java學習筆記 守護線程/* * Daemon線程，即守護線程 * 一般都在後台運行，為其他線程提供服務，不能單獨存在 */public class Test08 { public static void main(String[] args) { MyThread8 t1 = new MyThread8("守護線程");
Java IO兩大體系，javaio兩大區別一套是input/output stream 體系，一套是reader/writer體系，兩者的區別是stream負責的是位元組流的資料，reader/writer負責的是字元流。設計模式decorator模式該模式主要用於在不改變已有的component設計和代碼的情況下，不斷的增加附屬功能。打個比喻，就像你有一副畫，為它配一個或者多個不同的畫框。具體可以參考這裡：http://www.cnblogs.com/bastard/archive/2012
【SICP練習】127 練習3.58，sicp練習1273.58 練習3-58原文Exercise 3.58. Give an interpretation of the stream computed by the following procedure:(define (expand num den radix) (cons-stream (quotient (* num radix) den) (expand (remainder (*
git 合并兩個遠程庫，git合并環境：1. 本地是空的，什麼都沒有2. 有2個不同的項目（工程），在遠端git上3. 需要在本地合并這2個項目開工：1. 從遠端clone第一個工程的代碼下來，並切換到你要合并的分支上。git clone git@xxx.com/project_1.git //這個操作會在本地建一個project_1的檔案夾，裡面就是項目1的代碼。cd project_1git checkout project_1_v1
【SICP練習】131 練習3.62，sicp練習1313.62 練習3-62原文Exercise 3.62. Use the results of exercises 3.60 and 3.61 to define a procedure div-series that divides two power series. Div-series should work for any two series, provided that the denominator
【SICP練習】126 練習3.57，sicp練習1263.57 練習3-57原文Exercise 3.57. How many additions are performed when we compute the nth Fibonacci number using the definition of fibs based on the add-streams procedure? Show that the number of additions would be
【SICP練習】125 練習3.56，sicp練習1253.56 練習3-56原文Exercise 3.56. A famous problem, first raised by R. Hamming, is to enumerate, in ascending order with no repetitions, all positive integers with no prime factors other than 2, 3, or 5. One obvious way
【SICP練習】132 練習3.63，sicp練習1323.63 練習3-63原文Exercise 3.63. Louis Reasoner asks why the sqrt-stream procedure was not written in the following more straightforward way, without the local variable guesses:(define (sqrt-stream x) (cons-stream 1.
search，searchfor #include "stdio.h"#include "string.h"#include "malloc.h" #define MAX_LIST 50typedef struct _SqList { int data[MAX_LIST]; int length;}SqList;//The key difference between Fibonacci search and binary search //is how the
【SICP練習】123 練習3.54，sicp練習1233.54 練習3-54原文Exercise 3.54. Define a procedure mul-streams, analogous to add-streams, that produces the elementwise product of its two input streams. Use this together with the stream of integers to complete the
C語言結構體初始化的三種方法，初始化三種方法直接上樣本了#include <stdio.h>struct student_st{char c;int score;const char *name;};static void show_student(struct student_st *stu){printf("c = %c, score = %d, name = %s\n", stu->c, stu->score,
新手學JAVA（二）----String類與StringBuffer類的區別，javastringbuffer 在Java中有兩種字串的操作：String類和StringBuffer類（緩衝字串處理類）。 下面先簡單的說一下兩者的區別。 String類和StringBuffer類都提供了相應的方法實現字串的操作，但二者略有不同。（1） String類
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