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票数的问题是: 在控制器笨我有下一个代码:在一个foreach中调用第二个foreach!笨
$query_not_voted = "SELECT * FROM photos WHERE p_id NOT IN (SELECT distinct p_id FROM p_votes where u_id = ".$this->session->userdata('u_id').") LIMIT ".$this->db->escape_str($segment_url).", ".$config['per_page'];
$q = $this->db->query($query_not_voted);
$data['myphotos'] = $q->result_array();
foreach($data['myphotos'] as $key=>$val)
{
$query_g = "SELECT * FROM users WHERE u_id = (SELECT u_id FROM p_votes WHERE p_id = ".$val['p_id'].")";
$q_2 = $this->db->query($query_g);
$data['allvotess'] = $q_2->result_array();
$query_u = "SELECT * FROM users WHERE u_id = ".$val['u_id']." LIMIT 0, 5";
$q_1 = $this->db->query($query_u);
$data['author'] = $q_1->result_array();
}
所以现在我有$data['myphotos']
,这是可以考虑到与下一个代码被输出:
<?php foreach ($myphotos as $keys => $myphoto){ ?>
<div id="voteblock">
<div id="voteleft">
<a href="<?php echo base_url().$myphoto['p_small']; ?>" class="preview" title="<?php echo base_url().$myphoto['p_title']; ?>"><img src="<?php echo base_url().$myphoto['p_thumb']; ?>" /></a>
</div>
<?php } ?>
但如何在最后的foreach循环中输出$data['allvotess']
? 我试图做同样的事,但没有解决。
然后我试着把结果推到$ data ['myphotos']里面,没有什么好结果!
所以我在做什么?
只需要有人来帮助我foreachloop ins foreachloop! 一个星期我不能做,我甚至可以支付解决方案现在!
我也贴在codeingiter论坛,但他们说这很容易票数为link