本文介绍了将json转换为变量的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我不会从数据库为jvectormap创建MapData,像这样
i wont create MapData for jvectormap from database,like this
{"AP":3,"DE":1,"GB":2,"ID":24,"US":10}
但是我的代码总是这样返回
but my code always return like this
["{\"AP\":3}","{\"DE\":1}","{\"GB\":2}","{\"ID\":24}","{\"US\":10}"]
这是我的javascript
this my javascript
$(document).ready(function () {
var mapData ={}
$.ajax({ url: '/orders/map', type: 'GET', dataType: 'json' }).success(function (data) {
mapData = (data);
});
var dd = JSON.stringify( mapData, null, '\t');
$('#world-map').vectorMap({
map: 'world_mill_en',
regionStyle: {
initial: {
fill: '#e4e4e4',
"fill-opacity": 1,
stroke: 'none',
"stroke-width": 0,
"stroke-opacity": 0
}
},
series: {
regions: [{
values: dd,
scale: ["#1ab394", "#22d6b1"],
normalizeFunction: 'polynomial'
}]
},
onRegionTipShow: function (e, el, code) {
console.log(dd[code]);
var mapcode = ( dd[code] ) ? ' ( ' + dd[code] + ' Visitor/Visitors )' : ' ( No visitor yet! )';
el.html(el.html() + mapcode );
}
});
});
this my codeigniter controllers
function map(){
$rows = array();
$query = $this->db->query('SELECT country, COUNT(1) AS rpt_count FROM vistors GROUP BY country');
foreach($query->result() as $row)
{
$rows[] = '{"'.$row->country.'":'.$row->rpt_count.'}';
}
$this->output
->set_content_type('application/json')
->set_output(json_encode($rows));
}
同志可以帮助我
推荐答案
您需要解析数据以获取可以从中获取数据的对象.
You need to parse the data to get an object you can get data from.
var mapData = JSON.parse(data);
这篇关于将json转换为变量的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!