$(document).keypress(function(e)
{
alert(e.keyCode);
if(e.keyCode==27)
{
hide_menu();
}
});
我得到除了转义键之外的所有键的警报,并且if永远不会调用成功部分.为什么会这样？
我用lavarel和热情的包装.
当我想要更新一行时,我遇到了一些问题.
我有2个模型客户端和地址相关的morphone关系.
这种关系很好,当我想得到一个客户端这一行返回预期的结果:
Client::with('address')->find($id);
但我无法理解如何使用干净的解决方案更新客户端.有人可以回答这些问题:
当我在我的更新方法中记录Input :: all()时,我得到了:
[2014-05-31 15:52:56] production.INFO: {"id":983,"firstName":"Susanne","lastName":"Adam","birthDate":"18\/06\/1982","inscriptionDate":"08\/09\/2013","status":3,"created_at":"2014-05-31 14:26:25","updated_at":"2014-05-31 14:26:25","email":"bernard.alix@free.fr","address":{"id":983,"address":"avenue Etienne","address2":"","ville":"Cordierboeuf","cp":"25 10","phone":"0403983157","mobile":"+33 (0)3 0","addressable_id":983,"addressable_type":"Client","created_at":"2014-05-31 14:27:58","updated_at":"2014-05-31 14:27:58"}} [] []
如您所见,地址数据位于客户端数据中.
3.当我使用更新,保存或推送(eloquent的方法)时,雄辩不明白他应该更新地址模型然后更新相关的客户端模型.我的数据格式不是很好吗？
谢谢.
更新:
当我执行Log :: info(Input :: all())时,我在控制器中获得以下json数据:
[2014-06-01 18:10:46] production.INFO: {"id":284,"firstName":"Andr\u00e9e","lastName":"Adam","birthDate":"23\/07\/1944","inscriptionDate":"22\/11\/2013","status":2,"created_at":"2014-06-01 15:41:22","updated_at":"2014-06-01 18:06:44","email":"monique17@normand.com","address":{"id":284,"streetAddress":"93, avenue Lefort","streetAddress2":"","city":"Boulay-sur-Leger","zipCode":"14054","phone":"09 51 03 1","mobile":"+33 6 00 6","addressable_id":284,"addressable_type":"Client","created_at":"2014-06-01 15:42:50","updated_at":"2014-06-01 18:06:44"}} [] []
随着ardent的自动水化不起作用...客户端成功自动水合但地址不成功,可能是由于它们之间的多态关系(一对一).
我尝试用这种方式填充我的模型:
$client = Client::with('address')->find($id);
$client->update(Input::except('address'));
$client->address->update(Input::only('address'));
但是这不起作用,因为Input :: only('address')给出错误的形成数据,当我记录这个时我得到了:
Log::info(Input::except('address'));
Log::info(Input::only('address'));
//output
[2014-06-01 18:20:34] production.INFO: {"id":284,"firstName":"Andr\u00e9e","lastName":"Adam","birthDate":"23\/07\/1944","inscriptionDate":"22\/11\/2013","status":2,"created_at":"2014-06-01 15:41:22","updated_at":"2014-06-01 18:10:46","email":"monique17@normand.com"} [] []
[2014-06-01 18:20:34] production.INFO: {"address":{"id":284,"streetAddress":"93, avenue Lefort","streetAddress2":"","city":"Boulay-sur-Leger","zipCode":"14054","phone":"09 51 …我正在开发一个产品网站,其中有很多图像要放置.为图像提供服务,这是使用PHP的最佳方法imagejpeg()还是将图像放在直接路径上？
注意:存储中的文件是高分辨率,例如1000 x 1000像素,而显示器我可能需要使用CSS进行缩放.
在这个时刻imagecopyresampled,我认为采用将图像缩放到所需尺寸并在网站上显示的速度更快.事实是什么？
<?php
namespace security;
class Pbkdf2
{
const HASH_ITERATIONS = 6000;
const SALT_ITERATIONS = 10;
const POMPOUS_SECRET = <<<TOKEN
vT@sw6b7,GD#orY8iQG%CbHLyzeziWFNWGnew=X]QuFfUtc(vP
TOKEN;
public static function generateRandomSalt($iterationCount = Pbkdf2::SALT_ITERATIONS)
{
if ($iterationCount < 10)
{
$iterationCount = 10;
}
$rand = array();
for ($i = 0; $i < $iterationCount; ++$i) {
$rand[] = rand(0, 2147483647);
}
return strtolower(hash('sha256', implode('', $rand)));
}
public static function checklogin($password, $hash, $salt, $iterationCount = Pbkdf2::HASH_ITERATIONS)
{
$hashExpected = self::hash($password, $salt, $iterationCount);
return $hashExpected === $hash;
}
public static …<section class='parent' style='display: table-cell;'>
//<div class='enclose'>
<div class='child'></div>
<div class='child'></div>
<div class='child'></div>
<div class='child'></div>
//</div><!--enclose-->
</section><!--parent-->
JS:
<script>
$('.parent').owlCarousel(); // without children enclosure div
//$('.enclose').owlCarousel(); // If children are enclosed inside a block-div
</script>
现在<div class='parent'>扩展大约10000到20000像素的宽度取决于里面的孩子的数量.在检查时,我了解到owlCarousel <div>根据物品的数量设置宽度.由于父级设置为table-cell显示样式,因此也会扩展.所以我只是添加了另一个<div>来编码子节点并将其显示属性设置为block.但这也行不通.有任何想法吗？
考虑以下两个示例表
表1：“用户”
ID REGISTER_TIME FIRSTNAME LASTNAME OPERATION_ID
1 1401789877 John Doe 29
2 1401789879 Jack Doe 29
3 1401789878 Pete Doe 29
（请注意，register_time列存储为INT（11））
表2：“ calendar_days”
ID DAY
1 2011-01-01
... ....
n 2030-31-12
我有以下运行良好的MySQL查询：
SELECT d.day, COUNT(d.day)
FROM calendar_days AS d
LEFT OUTER JOIN users AS l ON DATE(FROM_UNIXTIME(l.`register_time`)) = d.day
WHERE l.`operation_id` = 29
GROUP BY d.day;
但这并不能保留count等于零的日子：
day;COUNT(d.day)
2014-05-07;1
2014-05-09;1
2014-05-12;11
2014-05-13;2713
2014-05-14;2631
我想要的是 ：
2014-05-07;1
**2014-05-08;0**
2014-05-09;1
2014-05-12;11
2014-05-13;2713
2014-05-14;2631
我觉得我离真相很近，我已经尝试过任何可能的左外连接，右外连接，但仍然没有空结果。