读谷歌回应JSON文件
首先,我是新来这个很抱歉,如果这是“常识”的一些你。
因此,使用该请求包的Node.js,我想送一个HTTP请求到谷歌的地方API和API与我有一个很难参与工作的JSON文件做出响应。
我的请求代码:
request(url_nearby, {json: true}, function(err, resp, body){
if(err){
console.log(err);
}else{
console.log(body);
}
});
例如响应JSON文件(从文档):
{
"html_attributions" : [],
"results" : [
{
"geometry" : {
"location" : {
"lat" : -33.870775,
"lng" : 151.199025
}
},
"icon" : ".png",
"id" : "21a0b251c9b8392186142c798263e289fe45b4aa",
"name" : "Rhythmboat Cruises",
"opening_hours" : {
"open_now" : true
},
"photos" : [
{
"height" : 270,
"html_attributions" : [],
"photo_reference" : "CnRnAAAAF-LjFR1ZV93eawe1cU_3QNMCNmaGkowY7CnOf-kcNmPhNnPEG9W979jOuJJ1sGr75rhD5hqKzjD8vbMbSsRnq_Ni3ZIGfY6hKWmsOf3qHKJInkm4h55lzvLAXJVc-Rr4kI9O1tmIblblUpg2oqoq8RIQRMQJhFsTr5s9haxQ07EQHxoUO0ICubVFGYfJiMUPor1GnIWb5i8",
"width" : 519
}
],
"place_id" : "ChIJyWEHuEmuEmsRm9hTkapTCrk",
"scope" : "GOOGLE",
"alt_ids" : [
{
"place_id" : "D9iJyWEHuEmuEmsRm9hTkapTCrk",
"scope" : "APP"
}
],
"reference" : "CoQBdQAAAFSiijw5-cAV68xdf2O18pKIZ0seJh03u9h9wk_lEdG-cP1dWvp_QGS4SNCBMk_fB06YRsfMrNkINtPez22p5lRIlj5ty_HmcNwcl6GZXbD2RdXsVfLYlQwnZQcnu7ihkjZp_2gk1-fWXql3GQ8-1BEGwgCxG-eaSnIJIBPuIpihEhAY1WYdxPvOWsPnb2-nGb6QGhTipN0lgaLpQTnkcMeAIEvCsSa0Ww",
"types" : [ "travel_agency", "restaurant", "food", "establishment" ],
"vicinity" : "Pyrmont Bay Wharf Darling Dr, Sydney"
}
],
"status" : "OK"
}
所以我的问题是我怎么能打印出JSON请求的只有“名”元素?
-谢谢
回答如下:假设results
阵列只包含2个元素,则可以使用results[0]
得到的第一个元素,然后name
为属性。
总之,这将是:results[0].name
为了得到results
,你很多需要从响应体抢吧:
request(url_nearby, {json: true}, function(err, response){
if(err){
console.log(err);
}else{
console.log(response.body.results[0].name);
}
});
读谷歌回应JSON文件
首先,我是新来这个很抱歉,如果这是“常识”的一些你。
因此,使用该请求包的Node.js,我想送一个HTTP请求到谷歌的地方API和API与我有一个很难参与工作的JSON文件做出响应。
我的请求代码:
request(url_nearby, {json: true}, function(err, resp, body){
if(err){
console.log(err);
}else{
console.log(body);
}
});
例如响应JSON文件(从文档):
{
"html_attributions" : [],
"results" : [
{
"geometry" : {
"location" : {
"lat" : -33.870775,
"lng" : 151.199025
}
},
"icon" : ".png",
"id" : "21a0b251c9b8392186142c798263e289fe45b4aa",
"name" : "Rhythmboat Cruises",
"opening_hours" : {
"open_now" : true
},
"photos" : [
{
"height" : 270,
"html_attributions" : [],
"photo_reference" : "CnRnAAAAF-LjFR1ZV93eawe1cU_3QNMCNmaGkowY7CnOf-kcNmPhNnPEG9W979jOuJJ1sGr75rhD5hqKzjD8vbMbSsRnq_Ni3ZIGfY6hKWmsOf3qHKJInkm4h55lzvLAXJVc-Rr4kI9O1tmIblblUpg2oqoq8RIQRMQJhFsTr5s9haxQ07EQHxoUO0ICubVFGYfJiMUPor1GnIWb5i8",
"width" : 519
}
],
"place_id" : "ChIJyWEHuEmuEmsRm9hTkapTCrk",
"scope" : "GOOGLE",
"alt_ids" : [
{
"place_id" : "D9iJyWEHuEmuEmsRm9hTkapTCrk",
"scope" : "APP"
}
],
"reference" : "CoQBdQAAAFSiijw5-cAV68xdf2O18pKIZ0seJh03u9h9wk_lEdG-cP1dWvp_QGS4SNCBMk_fB06YRsfMrNkINtPez22p5lRIlj5ty_HmcNwcl6GZXbD2RdXsVfLYlQwnZQcnu7ihkjZp_2gk1-fWXql3GQ8-1BEGwgCxG-eaSnIJIBPuIpihEhAY1WYdxPvOWsPnb2-nGb6QGhTipN0lgaLpQTnkcMeAIEvCsSa0Ww",
"types" : [ "travel_agency", "restaurant", "food", "establishment" ],
"vicinity" : "Pyrmont Bay Wharf Darling Dr, Sydney"
}
],
"status" : "OK"
}
所以我的问题是我怎么能打印出JSON请求的只有“名”元素?
-谢谢
回答如下:假设results
阵列只包含2个元素,则可以使用results[0]
得到的第一个元素,然后name
为属性。
总之,这将是:results[0].name
为了得到results
,你很多需要从响应体抢吧:
request(url_nearby, {json: true}, function(err, response){
if(err){
console.log(err);
}else{
console.log(response.body.results[0].name);
}
});