Php codeigniter中的json_编码格式
使用我现在的代码,我可以从数据库中获得正确的细节,并使用json_encodePhp codeigniter中的json_编码格式,php,json,codeigniter,model-view-controller,fullcalendar,Php,Json,Codeigniter,Model View Controller,Fullcalendar,使用我现在的代码,我可以从数据库中获得正确的细节,并使用json_encode [{ "tour_id": "1", "tour_name": "Test this", "start": "2016-01-12 13:21:32", "end": "2016-01-15 13:21:35", "itinerary_photo": "Desert.jpg" }][{ "tour_id": "3", "tour_name": "Everes
[{
"tour_id": "1",
"tour_name": "Test this",
"start": "2016-01-12 13:21:32",
"end": "2016-01-15 13:21:35",
"itinerary_photo": "Desert.jpg"
}][{
"tour_id": "3",
"tour_name": "Everest",
"start": "2016-02-10 13:23:00",
"end": "2016-02-18 13:23:07",
"itinerary_photo": "Chrysanthemum.jpg"
}]
预期的JSON提要我使用json提要向fullcalendar添加事件。我的json_编码结果的格式如下:
[{
"tour_id": "1",
"tour_name": "Test this",
"start": "2016-01-12 13:21:32",
"end": "2016-01-15 13:21:35",
"itinerary_photo": "Desert.jpg"
}, {
"tour_id": "3",
"tour_name": "Everest",
"start": "2016-02-10 13:23:00",
"end": "2016-02-18 13:23:07",
"itinerary_photo": "Chrysanthemum.jpg"
}]
不同之处在于,对象仅存储在一个数组中。随着im获得的输出,它显示多个数组
以下是我的控制器代码:
public function getBookedCal(){
if ($this->session->userdata('login')==true){
$data['twimembers'] = $this->session->userdata('twimembers');
$data['tour'] = $this->Model_MyBookedTours->getPaidTours($data['twimembers']['user_id']);
foreach ($data['tour'] as $tourConfirm){
$tourConfirm = $this->Model_MyBookedTours->getTourInfo($tourConfirm['Tour_packages_tourpkg_id']);
echo json_encode($tourConfirm);
}
//echo json_encode($tourConfirm); //this outputs only a single array
}//end if
}//end function
型号:
function getPaidTours($user_id){
$this->db->where('User_accounts_user_id',$user_id);
$this->db->where('payment_status','1'); //confirmed
$query = $this->db->get('user_tourpkg');
return $query->result_array();
} //calendar
function getTourInfo($tour_id){
$this->db->select('tour_id,tour_name, start, end, itinerary_photo');
$this->db->where('tour_id',$tour_id);
$query = $this->db->get('tour_packages');
return $query->result();
}
有没有办法通过对代码进行一些调整来获得预期的json提要?不要使用变量use array,而不要使用foreach值作为赋值
$tmptourConfirm = array();
foreach ($data['tour'] as $tourConfirm){
$tmptourConfirm[] = $this->Model_MyBookedTours->getTourInfo($tourConfirm['Tour_packages_tourpkg_id']);
}
echo json_encode($tmptourConfirm);
不要使用foreach值作为赋值,而要使用变量use array
$tmptourConfirm = array();
foreach ($data['tour'] as $tourConfirm){
$tmptourConfirm[] = $this->Model_MyBookedTours->getTourInfo($tourConfirm['Tour_packages_tourpkg_id']);
}
echo json_encode($tmptourConfirm);
您必须在每个循环之前将
$tourConfirm
声明为数组,并在循环之后将该数组编码为json。
例如
}/您必须在每个循环之前将
$tourConfirm
声明为数组,并在循环之后将该数组编码为json。
例如
}/更改
getBookedCal()
,如下所示:
public function getBookedCal(){
$tourConfirm = array();
if ($this->session->userdata('login')==true){
$data['twimembers'] = $this->session->userdata('twimembers');
$data['tour'] = $this->Model_MyBookedTours->getPaidTours($data['twimembers']['user_id']);
foreach ($data['tour'] as $tourConfirm){
$tourConfirm[] = $this->Model_MyBookedTours->getTourInfo($tourConfirm['Tour_packages_tourpkg_id']);
}
//echo json_encode($tourConfirm); //this outputs only a single array
}//end if
echo json_encode($tourConfirm);
}//end function
更改getBookedCal()
,如下所示:
public function getBookedCal(){
$tourConfirm = array();
if ($this->session->userdata('login')==true){
$data['twimembers'] = $this->session->userdata('twimembers');
$data['tour'] = $this->Model_MyBookedTours->getPaidTours($data['twimembers']['user_id']);
foreach ($data['tour'] as $tourConfirm){
$tourConfirm[] = $this->Model_MyBookedTours->getTourInfo($tourConfirm['Tour_packages_tourpkg_id']);
}
//echo json_encode($tourConfirm); //this outputs only a single array
}//end if
echo json_encode($tourConfirm);
}//end function
现在你的问题是什么?你试过
$array3=array\u merge($array1,$array2)
或$array1+$array2=$array3代码>?@devpro是否有任何方法可以通过对代码进行一些调整来获得预期的json提要?@Dray知道我从数据库获取json提要,我认为如果在数据库中添加额外的数据,合并单个数组将不起作用。但我明白你的意思@劳雷尔:为什么在向数据库中添加其他数据时它就不起作用了?现在你的问题是什么?你是否尝试过$array3=array\u merge($array1,$array2)
或$array1+$array2=$array3代码>?@devpro是否有任何方法可以通过对代码进行一些调整来获得预期的json提要?@Dray知道我从数据库获取json提要,我认为如果在数据库中添加额外的数据,合并单个数组将不起作用。但我明白你的意思@劳雷尔:为什么在向数据库添加额外数据时,它不起作用?