我正在尝试使用带有php的ajax jquery从数据库中获取json值。 我的代码...... 目前我没有收到任何错误,我在此处发布之前已经尝试过此How do I return the response from an asynchronous call?此链接。当我在ajax中添加使用它时,我没有获得返回值的数据,
-------------------- code 1 -------------------
$.ajax({
type: "Post",
url: site_url + "admin/venue_booking_list/" + val,
//dataType: 'html',
data: {
van_id: val
},
async: true,
cache: false,
//dataType: 'json',
success: function(data) {
return [{
"title": "Booked",
"start": "19-01-2016 08:00:00",
"end": "19-01-2016 08:30:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}, {
"title": "Booked",
"start": "19-01-2016 08:00:00",
"end": "19-01-2016 08:30:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}, {
"title": "Booked",
"start": "04-01-2016 08:00:00",
"end": "04-01-2016 08:30:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}, {
"title": "Booked",
"start": "01-01-2016 07:30:00",
"end": "01-01-2016 08:00:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}, {
"title": "Booked",
"start": "13-01-2016 07:30:00",
"end": "13-01-2016 08:00:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}];
}
});
}
------------- code 2 ----------------------
$.ajax({
type: "Post",
url: site_url + "admin/venue_booking_list/" + val,
//dataType: 'html',
data: {
van_id: val
},
async: true,
cache: false,
//dataType: 'json',
success: function(data) {
// removed code
}
return [{
"title": "Booked",
"start": "19-01-2016 08:00:00",
"end": "19-01-2016 08:30:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}, {
"title": "Booked",
"start": "19-01-2016 08:00:00",
"end": "19-01-2016 08:30:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}, {
"title": "Booked",
"start": "04-01-2016 08:00:00",
"end": "04-01-2016 08:30:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}, {
"title": "Booked",
"start": "01-01-2016 07:30:00",
"end": "01-01-2016 08:00:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}, {
"title": "Booked",
"start": "13-01-2016 07:30:00",
"end": "13-01-2016 08:00:00",
"backgroundColor": "#12CA6B",
"textColor": "#FFF"
}];
在从数据库中检索数据的2个代码段中的第一个中,它不会在控制台栏中显示为Object json,如下图所示。
如何得到第二张图片的结果,js文件中有静态代码,如code2所示?
请帮助