entityMap|blocks|key|6phuv|text|尝试下面的语法。我希望它能帮上忙|type|unstyled|depth|inlineStyleRanges|entityRanges|data|9skk6|https://www.google.com/maps/place/?q=place_id:ChIJp4JiUCNP0xQR1JaSjpW_Hms|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

Try below syntax. I hope it helps

<pre><code>https://www.google.com/maps/place/?q=place_id:ChIJp4JiUCNP0xQR1JaSjpW_Hms
</code></pre>

entityMap|blocks|key|7o3on|text|这是一个官方的url，用来搜索|type|unstyled|depth|inlineStyleRanges|entityRanges|data|56vpg|通过回退到|8t403|如果|12g1r|不存在|133dc|​|7cqub|无需令牌，适用于安卓、iOS+(以及iOS+11)和网络^0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|N|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|O|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|P|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|Q|9|@]|A|@]|B|$]]|$2|K|4|L|6|7|8|R|9|@]|A|@]|B|$]]]]

Here is an official url to search for a <code>placeId</code> with a fallback to an <code>address</code> if the <code>placeId</code> does not exist

<code>https://www.google.com/maps/search/?api=1&amp;query=&lt;address&gt;&amp;query_place_id=&lt;placeId&gt;</code>

no token required, works on Android, iOS (as well as iOS 11) and web

entityMap|blocks|key|46bpo|text|有一种方法可以始终获得正确的URL，但它确实需要额外的请求。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|e9sde|你可以用这个地方|c9asi|_|4fkaj|获取地点详细信息的id|e7qbf|从那里得到的响应有一个|84r1k|‘结果’|cdefr|字段，它总是打开正确的位置。例如：^0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|O|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|P|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|Q|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|R|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|S|9|@]|A|@]|B|$]]|$2|K|4|L|6|7|8|T|9|@]|A|@]|B|$]]|$2|M|4|N|6|7|8|U|9|@]|A|@]|B|$]]]]

There is a way of always getting the right URL but it does require an additional request.

You can use the place_id to get the place details with
<code>https://maps.googleapis.com/maps/api/place/details/json?key=YOURAPIKEY&amp;placeid=THEPLACEID</code>

The response from that has a ['result']['url'] field which always opens up the right place. 
eg. <code>https://maps.google.com/?cid=10254754059248426663</code>

entityMap|blocks|key|bm230|text|上没有预期参数的文档|type|unstyled|depth|inlineStyleRanges|entityRanges|data|9nr8m|https://www.google.com/maps/place|91sae|，所以这可能只是一种变通方法(至少目前它对我来说是有效的)。|8ms5m|以下URL格式似乎在大多数情况下都能提供所需的结果：|agiqm|https://www.google.com/maps/place/[place-name]/@[latitude],[longitude],[zoom]z/|code-block|syntax|javascript|24gju|您可以根据地名和地点几何创建此URL|1e2n7|​|fco0a|function+initialize()+{
++var+ac+=+new+google.maps.places.Autocomplete(document.getElementById('pac'));
++google.maps.event.addListener(ac,+'place_changed',+function()+{
++++var+place+=+this.getPlace(),
++++++link+=+document.getElementById('link');
++++if+(!place.geometry)+{
++++++link.textContent+=+'';
++++}+else+{
++++++var+zoom+=+17,
++++++++url+=+'https://www.google.com/maps/place/'+%2B
++++++++encodeURIComponent(place.name)+%2B+'/@'+%2B
++++++++place.geometry.location.toUrlValue()+%2B+','+%2B
++++++++zoom+%2B+'z/';
++++++link.href+=+link.textContent+=+url;

++++}

++});
}
google.maps.event.addDomListener(window,+'load',+initialize);|4fip3^0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|U|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|V|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|W|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|X|9|@]|A|@]|B|$]]|$2|I|4|J|6|K|8|Y|9|@]|A|@]|B|$L|M]]|$2|N|4|O|6|7|8|Z|9|@]|A|@]|B|$]]|$2|P|4|Q|6|7|8|10|9|@]|A|@]|B|$]]|$2|R|4|S|6|K|8|11|9|@]|A|@]|B|$L|M]]|$2|T|4|Q|6|7|8|12|9|@]|A|@]|B|$]]]]

There is no documentation for the expected parameters on <a href="https://www.google.com/maps/place" rel="noreferrer">https://www.google.com/maps/place</a> , so this may only be a workaround(at least it currently works for me).

The following URL-format seems to give the desired result in most cases:

<pre><code>https://www.google.com/maps/place/[place-name]/@[latitude],[longitude],[zoom]z/
</code></pre>

You may create this URL based on the place-name and the place-geometry

<div class="snippet" data-lang="js" data-hide="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function initialize() {
 var ac = new google.maps.places.Autocomplete(document.getElementById('pac'));
 google.maps.event.addListener(ac, 'place_changed', function() {
 var place = this.getPlace(),
 link = document.getElementById('link');
 if (!place.geometry) {
 link.textContent = '';
 } else {
 var zoom = 17,
 url = 'https://www.google.com/maps/place/' +
 encodeURIComponent(place.name) + '/@' +
 place.geometry.location.toUrlValue() + ',' +
 zoom + 'z/';
 link.href = link.textContent = url;

 }

 });
}
google.maps.event.addDomListener(window, 'load', initialize);</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script type="text/javascript" src="https://maps.googleapis.com/maps/api/js?v=3&amp;libraries=places"&gt;
&lt;/script&gt;
&lt;input id="pac" /&gt;
&lt;br/&gt;
&lt;a id="link" href="" target="_blank"&gt;&lt;/a&gt;</code></pre>
</div>
</div>

entityMap|blocks|key|u7m9|text|您可以使用google+embed+API，并将src作为链接:例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1ggvc|https://www.google.com/maps/embed/v1/place?key=YOUR_API_KEY
++&q=place_id:YOUR_PLACE_ID|code-block|syntax|javascript|94oit|文档：|fc1kn|https://developers.google.com/maps/documentation/embed/guide#place|adirc|_|b0m3r|模式^0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|Q|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|R|9|@]|A|@]|B|$]]|$2|J|4|K|6|7|8|S|9|@]|A|@]|B|$]]|$2|L|4|M|6|7|8|T|9|@]|A|@]|B|$]]|$2|N|4|O|6|7|8|U|9|@]|A|@]|B|$]]]]

You can use google embed API, and take src as a link:
eg: 

<pre><code> https://www.google.com/maps/embed/v1/place?key=YOUR_API_KEY
 &amp;q=place_id:YOUR_PLACE_ID
</code></pre>

Docs:
<a href="https://developers.google.com/maps/documentation/embed/guide#place_mode" rel="nofollow noreferrer">https://developers.google.com/maps/documentation/embed/guide#place_mode</a>

entityMap|blocks|key|7isd9|text|您可以使用google+embed+API，并将src作为链接:例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|bcooq|https://maps.googleapis.com/maps/api/place/details/json?place|ej3su|_|9tt92|id=ChIJN1t|8782o|a7g9m|tDeuEmsRUsoyG83frY4&key=YOUR|75reo|77jl4|API|3vnq|e1hqn|密钥^0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|R|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|S|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|T|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|U|9|@]|A|@]|B|$]]|$2|I|4|F|6|7|8|V|9|@]|A|@]|B|$]]|$2|J|4|K|6|7|8|W|9|@]|A|@]|B|$]]|$2|L|4|F|6|7|8|X|9|@]|A|@]|B|$]]|$2|M|4|N|6|7|8|Y|9|@]|A|@]|B|$]]|$2|O|4|F|6|7|8|Z|9|@]|A|@]|B|$]]|$2|P|4|Q|6|7|8|10|9|@]|A|@]|B|$]]]]

You can use google embed API, and take src as a link: eg:

<a href="https://maps.googleapis.com/maps/api/place/details/json?place_id=ChIJN1t_tDeuEmsRUsoyG83frY4&amp;key=YOUR_API_KEY" rel="nofollow noreferrer">https://maps.googleapis.com/maps/api/place/details/json?place_id=ChIJN1t_tDeuEmsRUsoyG83frY4&amp;key=YOUR_API_KEY</a>

entityMap|blocks|key|525br|text|什么对我有效，没有位置|type|unstyled|depth|inlineStyleRanges|entityRanges|data|85g9q|_|4mphv|id|ch2f6|https://www.google.com/maps/search//@,z|code-block|syntax|javascript|5du0r|示例|f1hrh|https://www.google.com/maps/search/Jouvay%2BNight%2BClub/@40.6961111,-73.8041667,15z^0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|Q|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|R|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|S|9|@]|A|@]|B|$J|K]]|$2|L|4|M|6|7|8|T|9|@]|A|@]|B|$]]|$2|N|4|O|6|I|8|U|9|@]|A|@]|B|$J|K]]]]

What worked for me without place_id
<pre><code>https://www.google.com/maps/search/&lt;search term&gt;/@&lt;coordinates&gt;,&lt;zoom level&gt;z
</code></pre>
Example
<pre><code>https://www.google.com/maps/search/Jouvay+Night+Club/@40.6961111,-73.8041667,15z
</code></pre>

HeIIo, using google map API with search I'm able to find a certain place and then store it's details. However, the array doesn't contain google maps link which would open this place on google maps. 

Via API I receive place_id which, as I feel, should be enough to build a full google maps link; however, I can't find a way to do that. Can anyone advise me on how to do that.

Thank you for your time.

Getting Google Maps link from place_id

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

 HeIIo，使用谷歌地图API进行搜索，我能够找到某个地方，然后存储它的详细信息。然而，该数组不包含google地图链接，该链接将在google地图上打开此位置。 通过API我接收地方 _ 我认为，这应该足以建立一个完整的谷歌地图链接；然而，我找不到一种方法来做到这一点。有人能建议我怎么做吗？ 感谢您的宝贵时间。

问从place_id获取谷歌地图链接
EN

回答 7

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问从place_id获取谷歌地图链接EN

回答 7

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问从place_id获取谷歌地图链接
EN