我的数据库结构:
人
id | name |
1 | John |
2 | Doe |
3 | Marc |
任务
task_id | task_name| person_id
1 | Get milk | 1
2 | Play cs | 1
3 | Walk dog | 2
4 | Eat fruit | 3
评论
id | comment | task
1 | Wich one | 1
2 | When? | 2
我试过这个:
function get_shapes2() {
$this->db->select('person.name,person.id,')
->select('GROUP_CONCAT(DISTINCT comments.id separator " -/r/- ") as "commentid" ')
->select('GROUP_CONCAT(DISTINCT comments.comment separator " -/r/- ") as "comment" ')
->select('GROUP_CONCAT(DISTINCT tasks.task_name separator " -/r/- ") as "tname"')
->select('GROUP_CONCAT(DISTINCT tasks.task_id separator " -/r/- " ) as "id2"');
$this->db->from('person');
$this->db->join('tasks', 'tasks.person_id = person.id', 'left');
$this->db->join('comments', 'comments.task = tasks.task_id', 'left ');
$this->db->group_by('id');
$query = $this->db->get();
$res = array();
foreach ($query->result() as $row) {
$posts[] = $row->name;
$posts[] = (int) $row->id;
$posts[] = array_map(function($tname, $tid){
return array('tname'=>$name,'tid'=>$tid);
},
explode(" -/r/- ",$row->tname),
explode(" -/r/- ",$row->id2));
array_push($res, $posts);
unset($posts);
}
return $res;
}
我得到的是:
{name: John, id: 1, task =[ {tname: "Get milk", tid: "1"},{tname: "Play cs", tid: "2"}]}
我想要做的是为每个人获取所有任务并且只有 1 个最新的
comment, comment_id
如果 评论存在 并将其存储在与任务相同的数组中{name: John, id: 1, task =[ {tname: "Get milk", tid: "1" comment: " Wich one", commentid: 1 },{tname: "Play cs", tid: "2" comment: " When?", commentid: 2 }]}
我遇到的问题是我不知道如何做到这一点..
我尝试将它添加到
array_map
但它随机存储评论它没有接缝工作 最佳答案
请使用 jquery 来捕获 json 数据——这是一种简单的捕获方法
$( document ).ready(function() {
$.getJSON("YOUR_PHP_LINK",function(data)
{
var tb = $("#tab");
$.each(data,function(i,value)
{
tb.append("<tr><td>Name: " + value.name + "</td><td>ID: " + value.id+ " </td></tr>");
});
});
});
关于php - 如何从数据库中获取数据,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/48611776/