我有一个 html 表单，我首先希望使用 jQuery 验证库 jquery.validate.min.js 进行验证，如果表单有效，请将表单提交到某个位置。
我尝试了以下方法：
<html>
<head>
<link rel="stylesheet" href="../../common/view/css/bootstrap.min.css" type="text/css">
<script src="../../common/view/js/jquery.js"></script>
<script src="../../common/view/js/bootstrap.min.js"></script>
<script src="../../common/view/js/jquery.validate.min.js"></script>
</head>
<body>
<form method="post" action="stock-c.php" id="issueform" name="issueform">
<input type="text" name="pid"/>
<input type="button" name="button1" id="button1"/>
<script type="text/javascript" src="js/issueValidation.js"></script>
<!--Modal-->
<div id="myModal" class="modal fade" role="dialog">
<div class="modal-dialog">
<!--info to be displayed-->
<input type="submit" value="submit"/> <!--Explain1-->
</div>
</div>
</from>
</body>
</html>
问题验证.js
$( document ).ready( function () {
$validator= $( "#issueform" ).validate( {
rules: {
pid: "required",
},
messages: …我试图理解双链表的java实现.我有以下代码:
public class DLLNode{
//define variables
public int info;
public DLLNode next;
public DLLNode prev;
//Passing constructors
public DLLNode(int i){
info = i;
next = prev = null;
}
public DLLNode(int i, DLLNode n, DLLNode p){
info = i;
next = n;
prev = p;
}
}
以下内容:
public class DLL {
DLLNode head;
DLLNode tail;
public DLL(){
head = tail = null;
}
//Check whether list is empty or not
public boolean isEmpty(){
return head == null;
}
//Insert …我有一个特殊的问题.我正在开发一个需要Datatables使用它的php站点.我注意到当我在localhost上运行它时,Datatables样式不起作用,无论是Firefox还是Chrome.我的代码如下:
<!DOCTYPE html>
<!--
To change this license header, choose License Headers in Project Properties.
To change this template file, choose Tools | Templates
and open the template in the editor.
-->
<html>
<head>
<meta charset="UTF-8">
<title></title>
<link rel="text/css" href="//cdn.datatables.net/1.10.12/css/jquery.dataTables.min.css"/>
<script src="//code.jquery.com/jquery-2.2.3.js" type="text/javascript"></script>
<script src="//cdn.datatables.net/1.10.12/js/jquery.dataTables.min.js" type="text/javascript"></script>
<script>
$(document).ready(function() {
$('#example').DataTable();
} );
</script>
</head>
<body>
<table id="example" class="display" cellspacing="0" width="100%">
<thead>
<tr>
<th>Name</th>
<th>Position</th>
<th>Office</th>
<th>Age</th>
<th>Start date</th>
<th>Salary</th>
</tr>
</thead>
<tfoot>
<tr>
<th>Name</th>
<th>Position</th>
<th>Office</th>
<th>Age</th>
<th>Start date</th>
<th>Salary</th> …有人可以帮助纠正我面临的问题吗？我正在尝试在两个div之间设置一个空格，但是这样做会使两个div一起移动。我的代码如下：
@charset "utf-8";
/* CSS Document */
* {
box-sizing: border box;
}
body {
background-image: url(../images/nature_beach-1280x800.jpg);
background-repeat: no-repeat;
background-size: 100% 100%;
height: 100%;
margin: 0;
}
html {
height: 100%;
}
#container {
background-color: rgba(255, 255, 255, .50);
height: 65%;
width: 30%;
box-sizing: border-box;
transform: translate(200%, 20%);
font-family: Myriad Pro;
font-size: 20px;
}
#login {
text-align: center;
padding: 5%;
font-weight: bold;
}
#form {
margin-top: 5%;
margin-left: 10%;
}
.textfield {
height: 25px;
width: 250px;
background-color: rgba(109, 207, 246, …我正在为一个在线教程中找到的hailstone序列编写一个代码,但是这样做我遇到了一个无法访问的语句错误.我不知道我的代码是否正确,如果我错了,我不想要纠正它的建议(关于冰雹序列,我想自己做... :)).我只想帮助解决第19行的"无法访问的语句"错误.
class HailstoneSequence {
public static void main(String[] args) {
int[][] a = new int[10][];
a[0][0] = 125;
int number = 125;
for (int i = 0;; i++) {
for (int j = 1; j < 10; j++) {
if (number % 2 == 0) {
a[i][j] = number / 2;
number = number / 2;
} else {
a[i][j] = (number * 3) + 1;
number = (number * 3) + 1;
}
}
}
for (int i …我正在整理一个代码来输出以下模式:
000000000X
00000000XX
0000000XXX
000000XXXX
00000XXXXX
0000XXXXXX
000XXXXXXX
00XXXXXXXX
0XXXXXXXXX
(每行应该是一个接一个.我不太确定如何在论坛上显示模式...对不起)
我应该在代码中使用递归循环,但我最终在一个无限循环中,我真的不明白为什么..(可能可以确定我从来没有实际使用过递归循环.)是我的代码:
class Recursion {
//recursion should stop after 9 attempts
static int stopindex = 9;
public static void main(String[] args) {
//a=number of "O"s and b=number of "X"s
int a = 9;
int b = 1;
recursion(a, b);
}
public static void recursion(int a, int b) {
//start of recursion at index 1
int startindex = 1;
//stop condition of recursion
if (startindex == stopindex)
return;
//printing of pattern …