我正在使用codeigniter。我从我的数据库中检索了一个数组。我需要将当前的图像编号(序列号)传递给视图,以便我可以在那里显示它。
说,在第一页中,我需要传递#1
,以便让用户知道它是第一张图片。并说如果他们正在访问第3张图片(array[2]
),则应显示#3
。
基本上我所拥有的是用户点击next #4
和#2 previous
来浏览图片的选项,我想在页面中显示当前图片的序列号。
array(5) {
[0]=>
object(stdClass)#21 (6) {
["id"]=>
string(2) "17"
["gallery_id"]=>
string(1) "5"
["title"]=>
string(6) "sample"
["filename"]=>
string(12) "n9a67681.jpg"
["description"]=>
string(10) "sdhdfhafdj"
["created"]=>
string(19) "2013-06-29 08:25:56"
}
[1]=>
object(stdClass)#22 (6) {
["id"]=>
string(2) "18"
["gallery_id"]=>
string(1) "5"
["title"]=>
string(6) "sample"
["filename"]=>
string(13) "n9a676811.jpg"
["description"]=>
string(10) "sdhdfhafdj"
["created"]=>
string(19) "2013-06-29 08:26:28"
}
[2]=>
object(stdClass)#23 (6) {
["id"]=>
string(2) "19"
["gallery_id"]=>
string(1) "5"
["title"]=>
string(6) "sample"
["filename"]=>
string(13) "n9a676812.jpg"
["description"]=>
string(10) "sdhdfhafdj"
["created"]=>
string(19) "2013-06-29 08:27:04"
}
[3]=>
object(stdClass)#24 (6) {
["id"]=>
string(2) "20"
["gallery_id"]=>
string(1) "5"
["title"]=>
string(7) "safdhfh"
["filename"]=>
string(35) "tumblr_mfyn3l81ft1qkfae2o1_1280.jpg"
["description"]=>
string(4) "dsgd"
["created"]=>
string(19) "2013-06-29 08:28:16"
}
[4]=>
object(stdClass)#25 (6) {
["id"]=>
string(2) "21"
["gallery_id"]=>
string(1) "5"
["title"]=>
string(8) "dshfdhsd"
["filename"]=>
string(36) "tumblr_mfyn3l81ft1qkfae2o1_12801.jpg"
["description"]=>
string(6) "dahadf"
["created"]=>
string(19) "2013-06-29 08:29:35"
}
}
答案 0 :(得分:1)
控制器:
$this->load->library('pagination');
$config['base_url'] = example.com/controller/action/uri_segment;
$config['total_rows'] = 5;
$config['per_page'] = 1;
$config['uri_segment'] = 3;
$this->pagination->initialize($config);
$position = ($this->uri->segment(3)) ? $this->uri->segment(3) : 0;
$data['links'] = $this->pagination->create_links();
$data['image'] = your_array[$position];
$this->load->view('your_view', $data);
放入您的视图文件“echo $ links”(创建下一个和上一个链接),并根据需要使用图像信息。