如何从JavaScript对象数组中获取键值对？
如何从JavaScript对象数组中获取键值对？
提问于 2015-06-30 00:47:41
我有过
var results = {};我编写了使用以下格式的数据填充结果(来自HTTP GET请求)的代码：
({
"result":[
{
"Longitude" : "-097.722382",
"Zipcode" : "78751",
"ZipClass" : "STANDARD",
"County" : "TRAVIS",
"City" : "AUSTIN",
"State" : "TX",
"Latitude" : "+30.310606"
}
]}
)但是，我希望结果以TRAVIS作为关键字，然后添加另一个名为count的变量，该变量计算该县的总数。
我在访问键和值时遇到了问题；我似乎总是得到undefined。我该如何访问这些密钥？
这是我的代码。从本质上讲，我正在处理一堆邮政编码，只过滤掉德克萨斯州的邮政编码。
var i = 0;
var results = {};
/*
var results = {
'TRAVIS': 10,
'DALLAS': 15,
};
*/
function getValues(obj, key) {
var objects = [];
for (var i in obj) {
if (!obj.hasOwnProperty(i)) continue;
if (typeof obj[i] == 'object') {
objects = objects.concat(getValues(obj[i], key));
} else if (i == key) {
objects.push(obj[i]);
}
}
return objects;
}
callback = function(response) {
//console.log('callback('+i+')');
var str = '';
response.on('data', function (chunk) {
str += chunk;
});
response.on('end', function () {
console.log('Processing: ' + i);
// TODO: Parse JSON
str = str.replace(/['\(\)]/g, "");
if(str.substring(0,1) == "{"){
JSON.parse(str);
}
if(str.substring(0,1) == "{"){
if( (str.substring(str.search("\"State\"") + 10, str.search("\"State\"") + 14)) == "\"TX\"")
{ //console.log("THIS IS FROM TEXAS ");
results[i] = str;
}
}
setTimeout(function() {
i++;
if (i >= data.length) {
console.log(results);
} else {
fetch();
}
}, 1000)
});
}
function fetch() {
//console.log('fetch('+i+')');
var options = {
host: 'gomashup.com',
path: '/json.php?fds=geo/usa/zipcode/'+ JSON.parse(data[i].zip)
};
http.request(options, callback).end();
}
fetch();回答 5
Stack Overflow用户
发布于 2015-06-30 01:06:05
var response = ({
"result":[
{
"Longitude" : "-097.722382",
"Zipcode" : "78751",
"ZipClass" : "STANDARD",
"County" : "TRAVIS",
"City" : "AUSTIN",
"State" : "TX",
"Latitude" : "+30.310606"
}
]});
You can excess key and value this way...
console.log(response.result[0].County);
console.log(response.result[0].Zipcode);
And also add a key ......
response.result[0].count = 134;
console.log(response);Stack Overflow用户
发布于 2018-08-01 14:05:16
你可以使用lodash库来过滤你的结果。
let _ = require("lodash");
let res = {
result: [
{
"Longitude": "-097.722382",
"Zipcode": "78751",
"ZipClass": "STANDARD",
"County": "TRAVIS",
"City": "AUSTIN",
"State": "TX",
"Latitude": "+30.310606"
},
{
"Longitude": "-097.722382",
"Zipcode": "78751",
"ZipClass": "STANDARD",
"County": "TRAVIS",
"City": "AUSTIN",
"State": "TX",
"Latitude": "+30.310606"
},
{
"Longitude": "-097.722382",
"Zipcode": "78751",
"ZipClass": "STANDARD",
"County": "abc",
"City": "AUSTIN",
"State": "TX",
"Latitude": "+30.310606"
}
]
}
function test() {
let groupedResult = _.groupBy(res.result, 'County')
result = {}
_.forEach(Object.keys(groupedResult), (County)=>{
result[County] = groupedResult[County].length
})
console.log(result)
}
test();Stack Overflow用户
发布于 2015-06-30 01:04:40
如果你使用eval(%yourcode%)，它会返回一个对象，它有一个键'result‘，它指向一个只有一个值的数组--一个包含你的信息对的数组。因此，要访问这些键，您必须迭代此对象- eval(%yourcode%)['result'][0]，如下所示：
initialString = '({"result":[{"Longitude" : "-097.722382","Zipcode" : "78751","ZipClass" : "STANDARD","County" : "TRAVIS","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"}]})';
myObj = eval(initialString)['result'][0];
for (key in myObj) {
document.write(key + ': ' + myObj[key] + '<br>');
}
例如，如果要让result对象包含许多对象，则应遍历(myObj = eval(initialString)['result'])，如下所示：
initialString = '({"result":[{"Longitude" : "1","Zipcode" : "78751","ZipClass" : "STANDARD","County" : "TRAVIS","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"},{"Longitude" : "2","Zipcode" : "78751","ZipClass" : "STANDARD","County" : "TRAVIS","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"},{"Longitude" : "3","Zipcode" : "78751","ZipClass" : "STANDARD","County" : "TRAVIS","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"}]})';
myObj = eval(initialString)['result'];
myObj.forEach(function(infoItem,index) {
document.write('Item #' + (index+1) + "<br>");
for (key in infoItem) {
document.write(' ' + key + ': ' + infoItem[key] + '<br>');
}
});
哦，如果你想创建一个'result‘对象，这个对象将包含county作为键，zipcode作为值，你可以这样做：
// in this example `myObj` has 3 objects inside itself
// we iterate through each and select values that we need
var initialString = '({"result":[{"Longitude" : "1","Zipcode" : "78751","ZipClass" : "STANDARD","County" : "TRAVIS","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"},{"Longitude" : "2","Zipcode" : "37465","ZipClass" : "STANDARD","County" : "SOMECOUNTY","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"},{"Longitude" : "3","Zipcode" : "90210","ZipClass" : "STANDARD","County" : "MYBELOVEDCOUNTY","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"}]})',
result = {};
myObj = eval(initialString)['result'];
myObj.forEach(function(infoItem,index) {
// here we create entries inside 'result' object
// using county values as keys
result[infoItem['County']] = infoItem['Zipcode'];
});
document.write(JSON.stringify(result,null,'<br>'));
对不起，我终于得到你想要的了)看：
// in this example `myObj` has 4 objects inside itself
// we iterate through each and select values that we need
var initialString = '({"result":[{"Longitude" : "1","Zipcode" : "78751","ZipClass" : "STANDARD","County" : "TRAVIS","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"},{"Longitude" : "2","Zipcode" : "37465","ZipClass" : "STANDARD","County" : "SOMECOUNTY","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"},{"Longitude" : "3","Zipcode" : "90210","ZipClass" : "STANDARD","County" : "MYBELOVEDCOUNTY","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"},{"Longitude" : "4","Zipcode" : "90210","ZipClass" : "STANDARD","County" : "MYBELOVEDCOUNTY","City" : "AUSTIN","State" : "TX","Latitude" : "+30.310606"}]})',
result = {};
myObj = eval(initialString)['result'];
myObj.forEach(function(infoItem,index) {
// here we create entries inside 'result' object
// using county value as a keys
// first we check if an entry with current key exists
if (result[infoItem['County']]) {
// if yes, just increment it
result[infoItem['County']]++;
} else {
// if no, make it equal to 1
result[infoItem['County']] = 1;
}
});
document.write(JSON.stringify(result,null,'<br>'));
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原文链接：
https://stackoverflow.com/questions/31121188
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