英文:
make duplicate elements if their number is less than 4 and they are not empty
问题
我有一个用于浏览图片的滑块，并通过PHP将所有图片添加到滑块中。
情况是，滑块至少需要4张图片才能正常工作，如果我们只添加一两张图片，我编写了相应的代码来复制和粘贴图片。
这是可工作的代码：
<?php if (count($images) < 4) { ?>
<?php
$images = array_merge($images, $images);
$images = array_merge($images, $images);
$images = array_slice($images, 0, 4);
foreach ($images as $image) {
if ($image->image_id) { ?>
<div class="slider-image">
<img src="<?= $image->url ?>" alt="<?= $image->name ?>">
</div>
<?php } ?>
<?php } ?>
<?php } ?>
还有这一行：
if ($image->image_id)
也就是说，我可能会遇到这样一种情况，即数据库中没有这样的图片，为了避免错误，我添加了适当的检查。
因此，如果我有一张不在数据库中的图片，那么这段复制代码不起作用，因为它会复制一个空白的图片，然后不会显示它。
我该如何考虑图片不在数据库中的因素并修正复制的代码？或者我需要在这里使用另一种实现方法吗？
英文:
I have a slider for scrolling through images, and I add all the images to the slider through php
The situation is that the slider needs at least 4 images to work, and I wrote the corresponding code that copies and pastes the images, if we add only one or two
Here is the working code
<?php if (count($images) < 4) { ?>
<?php
$images = array_merge($images, $images);
$images = array_merge($images, $images);
$images = array_slice($images, 0, 4);
foreach ($images as $image) {
if ($image->image_id) { ?>
<div class="slider-image">
<img src="<?= $image->url ?>" alt="<?= $image->name ?>">
</div>
<?php } ?>
<?php } ?>
<?php } ?>
And there is this line
if ($image->image_id)
That is, I may have a situation that such an image is not in the database, and so that there are no errors, I added the appropriate check
So, if I have an image that is not in the database, then this copy code does not work, since it copies an empty image, which is then not displayed
$images = array_merge($images, $images);
$images = array_merge($images, $images);
$images = array_slice($images, 0, 4);
How can I take into account the factor that the picture is not in the database and correct the code for copying? Or do I need to use another implementation here?
答案1
得分: 3
我假设你的`$images`不为空且是一个常规列表（具有顺序的从零开始的索引）。在循环中将图像数组填充如下：
```php
$images = [0,1];
$i = 0;
while( count( $images ) < 4 )
$images[] = $images[ $i++ ];
print_r($images); # [0,1,0,1]
/*
对于给定的初始数组，我们将得到以下结果：
[0] -> [0,0,0,0]
[0,1] -> [0,1,0,1]
[0,1,2] -> [0,1,2,0]
[0,1,2,3] -> [0,1,2,3]
*/
如果你有一些虚拟的图像对象 - 先将它们过滤掉：
$images = array_filter($images, fn($image) => !empty($image->image_id) );
英文:
I presume your $images is not empty and is a regular list (has sequental zero-based indices). Just populate the images array in the loop as:
$images = [0,1];
$i = 0;
while( count( $images ) < 4 )
$images[] = $images[ $i++ ];
print_r($images); # [0,1,0,1]
/*
For given initial arrays we will get following results:
[0] -> [0,0,0,0]
[0,1] -> [0,1,0,1]
[0,1,2] -> [0,1,2,0]
[0,1,2,3] -> [0,1,2,3]
*/
If you have some dummy images objects - filter them out first:
$images = array_filter($images, fn($image) => !empty($image->image_id) );
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