这可能非常基本,但我正在尝试测试Google Places API。我正在浏览文档并使用它们提供的一些示例。我正在尝试使用JQuery getJSON函数,因为我已经能够成功地使用它来异步访问外部JSON文件,所以我认为这可能是获取Google Places查询的JSON结果的好方法。这是我正在尝试使用的代码:
<body>
<div id="message"></div>
<script type="text/javascript">
var requestURL = 'https://maps.googleapis.com/maps/api/place/search/json?location=-33.8670522,151.1957362&radius=500&types=food&name=harbour&sensor=false&key='my_google_places_key';
$(document).ready(function () {
$.getJSON(requestURL, function (data) {
for (i = 0; i < data.results.length; i++) {
myAddress[i] = data.results[i].formatted_address;
document.getElementById("message").innerHTML += myAddress[i] + "<br>";
console.log(myAddress[i]);
}
});
});
</script>
</body>
根据文档,对查询的结果JSON响应应如下所示:
{
"html_attributions" : [
"Listings by \u003ca href=\"http://www.yellowpages.com.au/\"\u003eYellow Pages\u003c/a\u003e"
],
"results" : [
{
"formatted_address" : "529 Kent Street, Sydney NSW, Australia",
"geometry" : {
"location" : {
"lat" : -33.8750460,
"lng" : 151.2052720
}
},
"icon" : "http://maps.gstatic.com/mapfiles/place_api/icons/restaurant-71.png",
"id" : "827f1ac561d72ec25897df088199315f7cbbc8ed",
"name" : "Tetsuya's",
"rating" : 4.30,
"reference" : "CnRmAAAAmmm3dlSVT3E7rIvwQ0lHBA4sayvxWEc4nZaXSSjRtfKRGoYnfr3d5AvQGk4e0u3oOErXsIJwtd3Wck1Onyw6pCzr8swW4E7dZ6wP4dV6AsXPvodwdVyqHgyGE_K8DqSp5McW_nFcci_-1jXb5Phv-RIQTzv5BjIGS0ufgTslfC6dqBoU7tw8NKUDHg28bPJlL0vGVWVgbTg",
"types" : [ "restaurant", "food", "establishment" ]
},
{
"formatted_address" : "Upper Level, Overseas Passenger Terminal/5 Hickson Road, The Rocks NSW, Australia",
"geometry" : {
"location" : {
"lat" : -33.8583790,
"lng" : 151.2100270
}
},
"icon" : "http://maps.gstatic.com/mapfiles/place_api/icons/cafe-71.png",
"id" : "f181b872b9bc680c8966df3e5770ae9839115440",
"name" : "Quay",
"rating" : 4.10,
"reference" : "CnRiAAAADmPDOkn3znv_fX78Ma6X5_t7caEGNdSWnpwMIdDNZkLpVKPnQJXP1ghlySO-ixqs28UtDmJaOlCHn18pxpj7UQjRzR4Kmye6Gijoqoox9bpkaCAJatbJGZEIIUwRbTNIE_L2jGo5BDqiosqU2F5QdBIQbXKrvfQuo6rmu8285j7bDBoUrGrN4r6XQ-PVm260PFt5kwc3EfY",
"types" : [ "cafe", "bar", "restaurant", "food", "establishment" ]
},
{
"formatted_address" : "107 George Street, The Rocks NSW, Australia",
"geometry" : {
"location" : {
"lat" : -33.8597750,
"lng" : 151.2085920
}
},
"icon" : "http://maps.gstatic.com/mapfiles/place_api/icons/restaurant-71.png",
"id" : "7beacea28938ae42bcac04faf79a607bf84409e6",
"name" : "Rockpool",
"rating" : 4.0,
"reference" : "CnRlAAAAVK4Ek78r9yHV56I-zbaTxo9YiroCbTlel-ZRj2i6yGAkLwNMm_flMhCl3j8ZHN-jJyG1TvKqBBnKQS2z4Tceu-1kZupZ1HSo5JWRBKd7qt2vKgT8VauiEBQL-zJiKVzSy5rFfilKDLSiLusmdi88ThIQqqj6hKHn5awdj6C4f59ifRoUg67KlbpuGuuW7S1tAH_EyBl6KE4",
"types" : [ "restaurant", "food", "establishment" ]
},
{
"formatted_address" : "483 George Street, Sydney NSW, Australia",
"events" : [
{
"event_id" : "7lH_gK1GphU",
"summary" : "Google Maps Developer Meetup: Rockin' out with the Places API",
"url" : "https://developers.google.com/places"
}
],
"geometry" : {
"location" : {
"lat" : -33.8731950,
"lng" : 151.2063380
}
},
"icon" : "http://maps.gstatic.com/mapfiles/place_api/icons/civic_building-71.png",
"id" : "017049cb4e82412aaf0efbde890e82b7f2987c16",
"name" : "Chinatown Sydney",
"rating" : 4.0,
"reference" : "CnRuAAAAsLNeRQtKD7TEUXWG6gYD7ByOVKjQE61GSyeGZrX-pOPVps2BaLBlH0zBHlrVU9DKhsuXra075loWmZUCbczKDPdCaP9FVJXB2NsZ1q7188pqRFik58S9Z1lcWjyVoVqvdUUt9bDMLqxVT4ENmolbgBIQ9Wy0sgDy0BgWyg5kfPMHCxoUOvmhfKC-lTefXGgnsRqEQwn8M0I",
"types" : [
"city_hall",
"park",
"restaurant",
"doctor",
"train_station",
"local_government_office",
"food",
"health",
"establishment"
]
}
],
"status" : "OK"
}
如果我复制此JSON脚本并将其保存到文件中,我可以访问它并在浏览器上显示以下结果:
529 Kent Street,Sydney NSW,Australia 上层,海外客运码头/ 5 Hickson Road,The Rocks NSW,澳大利亚 107 George Street,The Rocks NSW,Australia 483 George Street,Sydney NSW,Australia
这意味着它有效。 getJSON函数是否正确解析JSON脚本?
答案 0 :(得分:6)
我建议使用Places Library的Google Maps JavaScript API v3。
您可以找到有关如何使用它的演示和文档here。
答案 1 :(得分:2)
好吧,我或多或少想通了。我想出了你需要做什么,以便getJSON函数返回JSON解析数据。你必须添加“callback =?”到查询字符串。
'https://maps.googleapis.com/maps/api/place/search/json?location=-33.8670522,151.1957362&radius=500&types=food&name=harbour&sensor=false&key="myKey"&callback=?';
但是,现在问题是我现在在我的控制台中出现错误:
SyntaxError: invalid label
[Break On This Error]
"html_attributions" : [
json?l...0080533 (line 2, col 3)
这很奇怪,因为我在JSONLint中检查了响应并且格式有效。此外,如果从本地文件中读取相同的响应。
答案 2 :(得分:0)
请尝试以下代码
<body>
<div id="message"></div>
<script type="text/javascript">
$(document).ready(function () {
$.ajax({
type: 'GET',
url: 'https://maps.googleapis.com/maps/api/place/searc/json?location=-33.8670522,151.1957362&radius=500&types=food&name=harbour&sensor=false&key='my_google_places_key';',
async: false,
jsonpCallback: 'jsonCallback',
contentType: "application/json",
dataType: 'jsonp',
success: function (data) {
for (i = 0; i < data.results.length; i++) {
myAddress[i] = data.results[i].formatted_address;
document.getElementById("message").innerHTML += myAddress[i] + "<br>";
console.log(myAddress[i]);
};
},
error: function (e) {
console.log(e.message);
}
});
});
</script>
</body>
你需要将它作为Json回调,因为它是跨域调用