1

数据库查询返回几行，我循环如下：

    foreach ($query->result() as $row) {

        $data[$row->post_id]['post_id']         = $row->post_id;
        $data[$row->post_id]['post_type']       = $row->post_type;
        $data[$row->post_id]['post_text']       = $row->post_text;
    }

如果我json_encode得到的数组 ( $a['stream']) 我得到

{
    "stream": {
        "1029": {
            "post_id": "1029",
            "post_type": "1",
            "post_text": "bla1",
        },
        "1029": {
            "post_id": "1030",
            "post_type": "3",
            "post_text": "bla2",
        },
        "1029": {
            "post_id": "1031",
            "post_type": "2",
            "post_text": "bla3",            
        }
    }
}

但json实际上应该是这样的：

{
    "stream": {
        "posts": [{
            "post_id": "1029",
            "post_type": "1",
            "post_text": "bla1",
        },
        {
            "post_id": "1030",
            "post_type": "3",
            "post_text": "bla2",
        },
        {
            "post_id": "1031",
            "post_type": "2",
            "post_text": "bla3",            
        }]
    }
}

我应该如何构建我的阵列才能做到这一点json？

4

3 回答 3

1

无论如何，这是你应该做的：

...
$posts = array();
foreach ($query->result() as $row) {
    $post = array();
    $post['post_id']         = $row->post_id;
    $post['post_type']       = $row->post_type;
    $post['post_text']       = $row->post_text;
    $posts[] = $post;
}
$data['posts'] = $posts;
...

一点解释：您必须根据从数据库中获取的信息构建一个对象，即$post. 这些对象中的每一个都需要一起添加到一个数组中，即$posts. 这个来自数据库的帖子数组被设置为 的键posts，$data即$data['posts'].

于 2012-09-15T21:28:12.037 回答
1

这个怎么样？

http://codepad.viper-7.com/zPHCm0

<?php
$myData = array();
$myData['posts'][] = array('post_id' => 3, 'post_type' => 343, 'post_text' => 'sky muffin pie');
$myData['posts'][] = array('post_id' => 4, 'post_type' => 111, 'post_text' => 'Mushroom chocolate banana');
$myData['posts'][] = array('post_id' => 231, 'post_type' => 888, 'post_text' => 'Cucumber strawberry in the sky');

$theStream['stream'] = $myData;

$json = json_encode($theStream);

echo 'JSON:<Br/>';
echo $json;

以上给了我：

{
    "stream":
    {   "posts":[
            {"post_id":3,"post_type":343,"post_text":"sky muffin pie"},
            {"post_id":4,"post_type":111,"post_text":"Mushroom chocolate banana"},
            {"post_id":231,"post_type":888,"post_text":"Cucumber strawberry in the sky"}]
    }
}
于 2012-09-15T21:38:33.657 回答
0 
foreach ($query->result() as $row) {
    $data['posts']['post_id']         = $row->post_id;
    $data['posts']['post_type']       = $row->post_type;
    $data['posts']['post_text']       = $row->post_text;
}
于 2012-09-15T21:27:41.073 回答