CakePHP中的图像链接(Image link in CakePHP)
我想创建一个图像链接,为此我编写了以下代码:
<?php echo $this->Html->link( $this->Html->image('logo.png', array('height' => '100', 'width' => '100','escape' => false)), array('controller'=>'officers','action'=>'home')); ?>
它不显示图像,而不是图像,它在链接中显示以下行:
<img src="/event_management/img/logo.png" height="100" width="100" alt="" />
链接正在运行。 我该如何显示图像?
I want to make an image link, for that I wrote the following code:
<?php echo $this->Html->link( $this->Html->image('logo.png', array('height' => '100', 'width' => '100','escape' => false)), array('controller'=>'officers','action'=>'home')); ?>
it doesn't show the image, instead of image it shows the following line in link:
<img src="/event_management/img/logo.png" height="100" width="100" alt="" />
the link is working. How can I show the image?
原文:https://stackoverflow.com/questions/26008403
更新时间:2023-03-09 18:03
最满意答案
对此进行修剪并最终找出结果
$start = new DateTime('2008-12-29 00:00'); $end = new DateTime('2009-01-7 00:00'); $point = $start; while($point <= $end){ echo $point->format('YW') . "\t"; echo $point->format('m-d-Y') . "\n"; $point = $point->modify('next week'); }
所以这里的第一次约会是
2008-12-29
。 因此Y
是正确的。 但2008-12-29
也是第1周 。 所以W
也是正确的Tinkered with this and finally figured it out
$start = new DateTime('2008-12-29 00:00'); $end = new DateTime('2009-01-7 00:00'); $point = $start; while($point <= $end){ echo $point->format('YW') . "\t"; echo $point->format('m-d-Y') . "\n"; $point = $point->modify('next week'); }
So the first date here is
2008-12-29
. ThusY
is correct. But2008-12-29
is also week 1. So theW
is also correct
相关问答
更多-
不,这就是它。 但是基于datetime.date.weekday()结果的列表理解应该很容易: today = datetime.date(2013, 06, 26) dates = [today + datetime.timedelta(days=i) for i in range(-7 - today.weekday(), 14 - today.weekday())] 请记住,范围不必从0开始。:-) 演示: >>> import datetime >>> from pprint import pp ...
-
对此进行修剪并最终找出结果 $start = new DateTime('2008-12-29 00:00'); $end = new DateTime('2009-01-7 00:00'); $point = $start; while($point <= $end){ echo $point->format('YW') . "\t"; echo $point->format('m-d-Y') . "\n"; $point = $point->modify('next week'); ...
-
应该像使用DateTime.AddDays一样简单添加7天。 var nextMonday = monday.AddDays(7); var nextFriday = friday.AddDays(7); Should be as simple as using DateTime.AddDays to add 7 days. var nextMonday = monday.AddDays(7); var nextFriday = friday.AddDays(7);
-
获得0周的年度范围(Get range of Week of Year from 0)[2023-08-27]
我可能会以下面的方式做到: SimpleDateFormat dateFormatForYear = new SimpleDateFormat("yyyy"); SimpleDateFormat yearWeekFormat = new SimpleDateFormat("w"); Integer weekFrom0 = Integer.valueOf(yearWeekFormat.format(date)) - 1; String s ... -
您对next_week语法需要调整: # Works <%= Date.current.next_week(:tuesday) %> 要理解为什么你的原始方法不起作用: monday和sunday恰好在DateAndTime :: Calculations上定义,但是一周中的其他日子不是。 Your syntax for next_week needs tweaking: # Works <%= Date.current.next_week(:tuesday) %> To understand why ...
-
本周和下一周的PHP(php for this week and next)[2023-01-12]
您可以使用更具可读性: echo date('F j, Y',strtotime('Monday this week')); echo date('F j, Y',strtotime('Monday next week')); You can use the far more readable: echo date('F j, Y',strtotime('Monday this week')); echo date('F j, Y',strtotime('Monday next week')); -
使用格式占位符'w'来获取当月第一天的星期几。 然后减去你想要获得的日期的相对数量(星期日= 7,星期六= 6,星期五= 5,依此类推)。 如果您想获得该月的第一个星期日: $paymentMonth = '2017-06-01'; $d = new DateTime($paymentMonth); $w = $d->modify('+1 month')->format('w'); // * Returns 6 as 1st July 2017 is Saturday $sun = (7-$w + 1)%7 ...
-
我面临着同样的问题。 我的解决方案 - 也许不是最好的解决方案,但它确实有效。 正如您已经提到的:大多数案例都是微不足道的,您只需知道年末/开始。 因此,我只需在下表中存储所需年数的相关周数: CREATE TABLE `NEXT_WEEK_TAB` ( `CUR_WEEK` varchar(8) NOT NULL, `NEXT_WEEK` varchar(8) NOT NULL, PRIMARY KEY (`CUR_WEEK`,`NEXT_WEEK`) ) 您现在使用当前算法获得以下一周: ...
-
Y是从该日期开始的年份 o是ISO-8601年号 W是ISO-8601周的年数 如果在周数中使用'W',则使用年份的'o'。 Y is year from the date o is ISO-8601 year number W is ISO-8601 week number of year if using 'W' for the week number use 'o' for the year.
-
这与#next_week和- 3.hours工作无关。 这是对#next_week如何运作的误解。 #next_week默认假设一周是星期一到星期日。 这意味着当只调用#next_week ,它将在00:00:00:00返回下一个星期一。 例: DateTime.now.next_week #=> Mon, 03 Aug 2015 00:00:00 -0400 好的,这样才有意义。 现在你正在3.hours再次成为星期天 DateTime.now.next_week - 3.hours #=> Sun, ...