数据库查询返回几行,我按如下方式循环:
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'] = $row->post_text;
}
如果我json_encode
生成的数组($a['stream']
)我得到了
{
"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",
}
}
}
但是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数组以使json
正确?
答案 0 :(得分: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'] = $posts;
...
一点解释:您必须根据从数据库获取的信息构建一个对象,即$post
。需要将这些对象中的每一个添加到一个数组中,即$posts
。来自数据库的这一系列帖子设置为posts
的密钥$data
,即$data['posts']
。
答案 1 :(得分: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 $json;
上面给了我:
{
"stream":
{ "posts":[
{"post_id":3,"post_type":343,"post_text":"sky muffin pie"},
{"post_id":4,"post_type":111,"post_text":"Mushroom chocolate banana"},
{"post_id":231,"post_type":888,"post_text":"Cucumber strawberry in the sky"}]
}
}
答案 2 :(得分:0)
foreach ($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;
}