Php 如何在codeigniter中从数据库中提取JSON数组
我在数据库中插入了json数组,没有任何键值,现在我想将json数组提取为单个值 像这样存储的数据库表Php 如何在codeigniter中从数据库中提取JSON数组,php,mysql,codeigniter,Php,Mysql,Codeigniter,我在数据库中插入了json数组,没有任何键值,现在我想将json数组提取为单个值 像这样存储的数据库表 13 53 76 3 71 17 9 <?php $members_id=$this->db->get_where('ag_matched_members', array('staff_id' =>$msgdata->staff_id))->row()->members_id; $members
13
53
76
3
71
17
9
<?php
$members_id=$this->db->get_where('ag_matched_members', array('staff_id' =>$msgdata->staff_id))->row()->members_id;
$members_ids = json_decode($members_id, true);
?>
<li>
<a href="#">
<i class="fa fa-envelope text-aqua"></i> <?php echo $members_ids[0]; ?>
</a>
</li>
我想这样展示
13
53
76
3
71
17
9
<?php
$members_id=$this->db->get_where('ag_matched_members', array('staff_id' =>$msgdata->staff_id))->row()->members_id;
$members_ids = json_decode($members_id, true);
?>
<li>
<a href="#">
<i class="fa fa-envelope text-aqua"></i> <?php echo $members_ids[0]; ?>
</a>
</li>
我试过这样做
13
53
76
3
71
17
9
<?php
$members_id=$this->db->get_where('ag_matched_members', array('staff_id' =>$msgdata->staff_id))->row()->members_id;
$members_ids = json_decode($members_id, true);
?>
<li>
<a href="#">
<i class="fa fa-envelope text-aqua"></i> <?php echo $members_ids[0]; ?>
</a>
</li>
使用此代码:<?php
$res = $this->db->query("SELECT members_id FROM `ag_matched_members` WHERE staff_id = ? " , $msgdata->staff_id )->result_array();
$members_ids = explode(',', json_decode($res[0]['members_id'])[0]);
foreach($members_ids as $id){
echo('<li><a href="#"><i class="fa fa-envelope text-aqua"></i> '.$id.'</a></li>');
}
?>
使用json\u decode()
并打印阵列您所说的显示是什么意思?你是说打印到控制台?HTML?你到底有什么问题?仅将格式设置为JSON?检查上面的linkAdd循环以打印$members\u ids数组的itemshow以逐个显示?您可以使用foreach($members\u ids as$id){echo($id);}我将更新答案,检查它!