我试图在html5画布上绘制图像.问题是在画布中拉伸的图像
load('img_the_scream',
'http://www.w3schools.com/tags/img_the_scream.jpg',
function( img ){
console.log( img.width , img.height );
ctx.drawImage( img, 10, 10 , img.width , img.height );
}
);
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我在http://jsfiddle.net/75rAU/中添加了这个问题
该运算符如何在JavaScript中运行.来自https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/String#Non-native_methods
var s = 12;
s>>=1;
console.log( s );
s>>=1;
console.log( s );
s>>=1;
console.log( s );
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我得到结果"6,3,1".我很好奇它的工作.
我正在尝试创建,用于颜色输入字段的html集合..这将使用javascript动态添加
我的ColorFieldset
代码是
namespace Dashboard\Form;
use Zend\Form\Fieldset;
use Zend\InputFilter\InputFilterProviderInterface;
class ColorFieldset extends Fieldset implements InputFilterProviderInterface
{
public function __construct()
{
parent::__construct('color');
$this->add(array(
'name' => 'hash',
'options' => array(
'label' => 'Color'
),
'attributes' => array(
'required' => 'required',
'class' => 'input-mini'
)
));
}
/**
* @return array
\*/
public function getInputFilterSpecification()
{
return array(
'hash' => array(
'required' => true,
)
);
}
}
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并将其添加到表单中
$this->add(array(
'type' => 'Zend\Form\Element\Collection',
'name' => 'colors',
'options' => array(
'count' => 2 …
Run Code Online (Sandbox Code Playgroud) 我想知道,new
在JavaScript中使用运算符创建对象时的变量范围是什么?
function A(){
this.a = 1; // instance property
}
function B(){
this.a = 3; // instance property
}
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案例1:我理解这一点
// assign again prototype a property as 2
A.prototype.a = 2 ;// prototype property
var obj = new A();
console.log( obj instanceof A );
console.log( obj.a == 1 );
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情况2:将构造函数更改为B引用
A.prototype.constructor = B;
A.prototype.a = 2 ;// prototype property
var obj = new A();
console.log( obj instanceof B ); // false, as I expected
console.log( obj.a …
Run Code Online (Sandbox Code Playgroud) 我在zf2中有关于zend分页和路由的问题.我想在同一页面上显示反馈项的详细信息及其子项(操作)列表.我的路线代码如下
$routes['dashboard_inbox_actions'] = array(
'type' => 'segment',
'options' => array(
'route' => '/dashboard/inbox/detail[/:feedback[/actions/page/:page]]',
'constraints' => array(
'feedback' => '[0-9]+',
'page' => '[0-9]+',
),
'defaults' => array(
'__NAMESPACE__' => 'Dashboard\Controller',
'controller' => 'inbox' ,
'action' => 'detail',
'feedback' => 0 ,
'page' => 1
),
),
);
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我通过URL传递
/dashboard/inbox/detail/4
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在列表页面中,用于呈现子项目的提供页面.
<?php echo $this->paginationControl($this->paginator, 'Sliding' ); ?>
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创建分页网址,反馈ID为0(我的问题)
/dashboard/inbox/detail/0/actions/page/2
/dashboard/inbox/detail/0/actions/page/3
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我手动粘贴网址
/dashboard/inbox/detail/4/actions/page/2
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它将第2页显示为活动项.我的控制器代码工作正常并给我结果,但仍然paginationControl创建反馈ID为0的网址.
当我尝试安装时ZendSkeletonApplication
,它会抛出错误.我在windows命令行输入命令
>>composer create-project -sdev --repository-url="http://packages.zen
dframework.com" zendframework/skeleton-application E:\php_apps\mopinion-cp
我收到了错误
[InvalidArgumentException]
Could not find package zendframework/skeleton-application with stability st
able.
create-project [-s|--stability="..."] [--prefer-source] [--prefer-dist] [--repos
itory-url="..."] [--dev] [--no-custom-installers] [--no-scripts] [--no-progress]
[--keep-vcs] package [directory] [version]
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该应用程序位于https://github.com/zendframework/ZendSkeletonApplication ..并在那里使用相同的命令.
上传Zend\Db\Adapter\Exception\RuntimeException
文件
\library\Zend\Db\Adapter\Driver\Pdo\Result.php:159
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我的数据库设置代码是
array(
'db' => array(
'driver' => 'Pdo',
'pdodriver' => 'mysql',
'driver_options' => array(
PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES \'UTF8\'',
'buffer_results' => true
),
'username' => 'root',
'password' => '',
'host' => 'localhost',
'dbname' => 'db_name',
),
)
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结果提取代码是
$statement = $this->sql->prepareStatementForSqlObject($select);
$results = $statement->execute();
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我试过$results->buffer();
&& $results->rewind()
我的迭代代码就像
foreach ( $results as $result ){
// code goes here ..
}
....
foreach ( $results as $result ){
// code goes here ..
}
Run Code Online (Sandbox Code Playgroud) 我正在尝试在window.load上加载特定的twitter bootstrap选项卡
$(function () {
if (location.hash !== '') {
$('#myTab a[href="' + location.hash + '"]').tab('show');
} else {
$('#myTab a:first').tab('show');
}
$('#myTab a[data-toggle="tab"]').on('shown', function(e) {
location.hash = $(e.target).attr('href').substr(1);
return false ;
});
})
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这段代码工作正常.我的问题是跳转到特定的tab div内容,如何禁用浏览器的 twitter bootstrap选项卡的默认位置哈希更改事件.我应该对上面的代码做出哪些更改?