jquery - ThinkPHP ajaxReturn multiple 2D arrays - Stack Overflow

<code>public function autorun(){
        $uid = session('uid');
        $map['pid'] = $uid;
        $User = M('land');
        $data = $User->field('land')->where($map)->select();

        print_r($data);
        //$this -> ajaxReturn($data);
    }
 </code>

<code> print_r($data);

Array ( [0] => Array ( [land] => 1 ) [1] => Array ( [land] => 2 ) ) //可能会有更多数据</code>

Please tell me how to write ajaxReturn in the brackets. In the past, it was a one-dimensional array and it was one. This won't happen. What the frontend receives is the value in [land]. If multiple values ​​are received, how does jquery separate these values.

Reply content:

<code>public function autorun(){
        $uid = session('uid');
        $map['pid'] = $uid;
        $User = M('land');
        $data = $User->field('land')->where($map)->select();

        print_r($data);
        //$this -> ajaxReturn($data);
    }
 </code>

<code> print_r($data);

Array ( [0] => Array ( [land] => 1 ) [1] => Array ( [land] => 2 ) ) //可能会有更多数据</code>

Please tell me how to write ajaxReturn in the brackets. In the past, it was always a one-dimensional array and it was one. This won't happen. What the frontend receives is the value in [land]. If multiple values ​​are received, how can jquery separate these values.

jquery is as follows

<code class="javascript">$.get(url, {}, function(d){
    if ($.isArray(d)) {
        $.each(d, function(i, v){
            $('.demo').append(v);
        })
    }
}, 'json');</code>

<code> $.ajax(  
    url: url,
    type: "get",
    dataType: "json",
    success:function(data) {
         for(var i in data){
             var obj = data[i];
             console.log(obj.land);
         }
    }  
 );</code>