我无法解析这个json:
var _json = [{"place_id":"18094048","licence":"Data \u00a9 OpenStreetMap contributors, ODbL 1.0. http:\/\/www.openstreetmap.org\/copyright","osm_type":"node","osm_id":"1695627257","boundingbox":[34.549406280518,34.569410095215,135.45611022949,135.47612548828],"lat":"34.5594098","lon":"135.4661246","display_name":"Singapore Embassy, \u583a\u72ed\u5c71\u7dda (Sakai-Sayama line), Sakai, Senboku District, Kinki Region, Giappone","class":"amenity","type":"embassy"},
"place_id":"17954461","licence":"Data \u00a9 OpenStreetMap contributors, ODbL 1.0. http:\/\/www.openstreetmap.org\/copyright","osm_type":"node","osm_id":"1695740584","boundingbox":
[35.647695770264,35.667699584961,139.72419189453,139.74420715332],"lat":"35.6576973","lon":"139.7341957","display_name":"Singapore Embassy, Gaien higashi dori, Roppongi, Minato, \u5317\u8db3\u7acb\u90e1, 1080074, Giappone","class":"amenity","type":"embassy"}]
我正在尝试JSON.parse(_json);
它在控制台中返回:
SyntaxError:JSON.parse:意外字符
console.log(JSON.parse(_json));
我需要lat&很长的价值观。
答案 0 :(得分:3)
这不是一个json字符串
它只是一个带有对象的JS数组
你可以这样简单地使用它:
for(var i = 0 ; i < _json.length ; i++ ){
var jsonObject = _json[i];
// then just use jsonObject['lat'] , jsonObject['license'] .....etc
}
答案 1 :(得分:1)
你var _json
不是JSON:它是一个标准的javascript数组,你可以在不解析的情况下使用。
如果你想要第一个纬度,只需做
var lat = _json[0].lat;
但我建议提供比_json
更好的名称,例如places
,因为此数组包含地点。
var firstPlace = _json[0]; // firstPlace has properties named lat and lon