在这个代码表格提交,即使我点击否
document.querySelector('#from1').onsubmit = function(){
swal({
title: "Are you sure?",
text: "You will not be able to recover this imaginary file!",
type: "warning",
showCancelButton: true,
confirmButtonColor: '#DD6B55',
confirmButtonText: 'Yes, I am sure!',
cancelButtonText: "No, cancel it!",
closeOnConfirm: false,
closeOnCancel: false
},
function(isConfirm){
if (isConfirm){
swal("Shortlisted!", "Candidates are successfully shortlisted!", "success");
} else {
swal("Cancelled", "Your imaginary file is safe :)", "error");
}
});
};
Run Code Online (Sandbox Code Playgroud) 数据库:
`--> product table
id name cid uploadKey
1 Cemera 1 7365
2 Notebook 2 7222`
`--> category table
id catename
1 canon
2 toshiba`
`--> attactments table
id uploadKey filepath
1 7365 /img/jdf.png
2 7365 /img/sdsd.jpg`
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这段代码创建json文件:
$_GET['id']="1";
$json_response = array();
if(isset($_GET['id']))
{
$id=$_GET['id'];
$select = mysql_query("SELECT product.name,category.catename,attactments.filepath FROM product INNER JOIN category ON category.id = product.cid INNER JOIN attactments ON attactments.uploadKey = product.uploadKey where product.id='".$id."' ");
while ($row = mysql_fetch_array($select , MYSQL_ASSOC)) {
$json_response[] = $row;
}
} …Run Code Online (Sandbox Code Playgroud) 我想在来自URL参数的字符串上创建类的对象.
例如 :
http://localhost/CSWeb/api/search/Slideshare
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在上面的URL Slideshare是进入的参数apiController->indexAction.
slideshare.php class
<?php
namespace App\Http\API;
class slideshare
{
public function index()
{
return 'any data';
}
}
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apiController.php
namespace App\Http\Controllers\API;
use Illuminate\Http\Request;
use App\Http\Requests;
use App\Http\Controllers\Controller;
use Auth;
use App\Http\API\Slideshare;
class apiController extends Controller
{
public function index($source)
{
$controller= new $source;
return $controller->index();
// if i change code to $controller= new Slideshare; it works fine
}
}
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当我使用参数字符串创建类对象时laravel错误
apiController.php第17行中的FatalErrorException:找不到类'Slideshare'
如果我改变代码
$controller= new Slideshare; it works fine
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先感谢您
我是JS的新手jQuery,而且我正在努力制作一个使用它们的字幕播放器.不幸的是,我陷入了很早的阶段.
当我试图HTML通过.js文件选择一些元素时,它表现得无法找到我要求的元素,并且没有任何反应.如果我尝试提醒元素的值或HTML,它会发出警报undefined.
所以这是代码:
HTML
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
<script src="jquery-2.1.3.min.js"></script>
<script src="script.js"></script>
<style type="text/css">
body{
margin: 0px;
padding: 0px;
}
#wrapper{
width: 150px;
text-align: center;
background: -webkit-gradient(linear, left top, right bottom, color-stop(0%,#75bdd1), color-stop(14%,#75bdd1), color-stop(100%,#2294b3));
padding: 10px 2px;
}
h3{
font-family: Arial, Helvetica, sans-serif;
}
img{
width: 50px;
margin: 0 auto;
}
input{
display: none;
}
</style>
</head>
<body>
<div id="wrapper">
<h3>Select subtitles file</h3>
<img src="browse.png" alt="browse">
</div>
<input type="file" accept=".srt" id="file"> …Run Code Online (Sandbox Code Playgroud) $(function() {
myhtml = '<tr><td>1</td><td>Orange</td><td>1799.00 x <span>1</span></td><td>0.00</td><td>1799.00</td></tr><tr class="shippingRow"><td></td><td></td><td></td><td>Shipping</td><td>2.00</td></tr><tr class="totalRow"><td></td><td></td><td></td><td><strong>Total</strong></td><td><strong>1811.00</strong></td></tr>'
console.log($(myhtml).find('.shippingRow td').length);
});
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我得到零而不是一个,我错过了哪一部分?我想删除类名或为myhtml做一些样式.
如何制作<div A>和<div C>拥有与<div B>javascript/jquery 相同的高度?
<div id="wrapper">
<div id="divA"></div>
<div id="divB"></div>
<div id="divC"></div>
</div>
<script>
$('#divA').css('height', $('#divB').innerHeight());
$('#divC').css('height', $('#divB').innerHeight());
</script>
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我正在使用jQuery创建动态按钮,是否有一种在创建CSS类时向其中添加CSS类的方法,除非没有其他方法,否则我要避免编写另一行JavaScript以向其添加CSS类。在下面找到代码片段,
$.each(data, function (index, value) {
var button = $('<button/>', {
text: value,
id: 'btn_' + index,
click: function () { MNS.ButtonClick(this) }
});
$('#MyDiv').append(button);
});
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我们有这样的东西吗
$.each(data, function (index, value) {
var button = $('<button/>', {
text: value,
id: 'btn_' + index,
**css:'Myclass'**
click: function () { MNS.ButtonClick(this) }
});
$('#MyDiv').append(button);
});
Run Code Online (Sandbox Code Playgroud) 我一直试图让这个工作在过去的6个小时内工作,我只是无法做到这一点.我已经看过SO上的问题和解决方案,但也无济于事.我在做什么可怕的错误?!?
我只有一个没有背景的html5页面.在我的javascript代码中,我在文档准备就绪时调用此方法:
/*
* Initializes the landing page.
*/
function initializeLanding(){
$('body').css({
'backround':'url("http://40.media.tumblr.com/290e7a272b492735ddf8cd968e657d05/tumblr_nhmg6xYnv41u7ns0go1_1280.jpg")'
});
}
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它应该是一个非常简单的东西,但我在这里.我试过使用单引号,双引号,没有多个属性等等.我也宁愿不使用CSS或DOM属性来改变背景,因为我在页面上设置了方向点以在滚动过去时改变背景他们.
另外,如果我调用.css()方法并输入新属性,它会覆盖所有旧属性,还是只覆盖最新css()调用中声明的属性?
更新:这是我调用函数的地方:
var main = function(){
$(".owl-carousel").owlCarousel();
lorem();
initializeLanding();
}
$(document).ready(main);
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<!-- Included Script -->
<script src="https://unpkg.com/react@15/dist/react.min.js"></script>
<script src="https://unpkg.com/react-dom@15/dist/react-dom.min.js"></script>
<script src="https://unpkg.com/babel-standalone@6.15.0/babel.min.js"></script>
<script src="https://unpkg.com/react-router/umd/react-router.min.js"></script>
<div id="root"></div>
<script type="text/babel">
var Router = ReactRouter.Router;
var Route = ReactRouter.Route;
var IndexRoute = ReactRouter.IndexRoute;
var Link = ReactRouter.Link;
var browserHistory = ReactRouter.browserHistory;
class Page extends React.Component {
render() {
return (
<div>
<Link to='/about'>India </Link>
</div>
);
}
}
class About extends React.Component {
render() {
return (
<div>
to Router
</div>
);
}
}
class NoMatch extends React.Component {
render() {
return (
<div>
No …Run Code Online (Sandbox Code Playgroud)jquery ×7
javascript ×5
php ×2
arrays ×1
css ×1
css3 ×1
html ×1
html5 ×1
json ×1
laravel-5.1 ×1
mysql ×1
react-router ×1
reactjs ×1