从 google map api 获取评论

Question

我必须从 Google 地图 API 获得评论。详细信息在此页面上。

https://developers.google.com/places/documentation/details#PlaceDetailsResults

详细信息将从该页面获取：-

https://maps.googleapis.com/maps/api/place/details/json?reference=CmRYAAAAciqGsTRX1mXRvuXSH2ErwW-jCINE1aLiwP64MCWDN5vkXvXoQGPKldMfmdGyqWSpm7BEYCgDm-iv7Kc2PF7QA7brMAwBbAcqMr5i1f4PwTpaovIZjysCEZTry8Ez30wpEhCNCXpynextCld2EBsDkRKsGhSLayuRyFsex6JA6NPh9dyupoTH3g&sensor=true&key=AddYourOwnKeyHere

我的问题是我在请求中找不到参考。以及如何从我的 Google plus 页面中找到此参数值。

Answer 1

要获取 google 评论，您需要该地点的参考 ID。为了获取此参考密钥，您可以使用 google places search api 请求。

https://maps.googleapis.com/maps/api/place/textsearch/xml?query=restaurants+in+bangalore&sensor=true&key=AddYourOwnKeyHere

它的响应将包含您可以在请求中使用的参考 ID。

    <PlaceSearchResponse>
    <status>OK</status>
    <result>
    <name>Koshy's Restaurant</name>
    <type>bar</type>
    <type>restaurant</type>
    <type>food</type>
    <type>establishment</type>
    <formatted_address>
    39, St Marks Road,Shivajinagar,Bangalore, Karnataka, 560001, India
    </formatted_address>
    <geometry>
    <rating>3.7</rating>
    <icon>
    http://maps.gstatic.com/mapfiles/place_api/icons/bar-71.png
    </icon>
    **<reference>**
    CnRwAAAA1z8aCeII_F2wIVcCnDVPQHQi5zdd-3FsDl6Xhb_16OGrILvvvI4X4M8bFk2U8YvuDCKcFBn_a2rjvYDtvUZJrHykDAntE48L5UX9hUy71Z4n80cO7ve_JXww6zUkoisfFnu6jEHcnKeeTUE42PCA4BIQGhGz0VrXWbADarhKwCQnKhoUOR-Xa9R6Skl0TZmOI4seqt8rO8I
    **</reference>**
    <id>2730db556ca6707ef517e5c165adda05d2395b90</id>
    <opening_hours>
    <open_now>true</open_now>
    </opening_hours>
    <html_attribution>
    <a href="https://plus.google.com/116912222767108657277">Ujaval Gandhi</a>
    </html_attribution>
    </photo>
    </result>

Answer 2

$reqUri = 'https://maps.googleapis.com/maps/api/place/nearbysearch/json?key='.YOURSERVERKEY;
$reqUri .= '&sensor=false&radius=500';
$reqUri .= '&location=38.908310,-104.784035&name='.urlencode(LOCATION NAME);
$reqUri .= '&keyword='.urlencode(WEBSITE PHONE);

我是通过 PHP 实现的，现在像这个 URL 一样调用，你会得到带有引用键的原始结果。

然后像这样解析它：

$data = cURL($reqUri);
$data = json_decode($data);
echo $data ->results[0]->reference;

希望对你有帮助

***注意：location=38.908310,-104.784035 这个变量不是自动的，你必须拥有它。

Answer 3

一种更新的方法：

https://maps.googleapis.com/maps/api/place/details/json?placeid={place_id}&key={api_key}

• place_id：https://developers.google.com/places/place-id
• api_key:https://developers.google.com/places/web-service/get-api-key

回复：

{
  "html_attributions": [],
  "result": {
    ...
    "rating": 4.6,
    "reviews": [
      {
        "author_name": "John Smith",
        "author_url": "https://www.google.com/maps/contrib/106615704148318066456/reviews",
        "language": "en",
        "profile_photo_url": "https://lh4.googleusercontent.com/-2t1b0vo3t-Y/AAAAAAAAAAI/AAAAAAAAAHA/0TUB0z30s-U/s150-c0x00000000-cc-rp-mo/photo.jpg",
        "rating": 5,
        "relative_time_description": "in the last week",
        "text": "Great time! 5 stars!",
        "time": 1508340655
      }
    ]
  }
}

评论仅限于最新的 5 条。

Answer 4

使用 Python 和 Google Places API，您可以检索业务详细信息和评论（最多 5 条评论），如下所示：

api = GooglePlaces("Your API key")

places = api.search_places_by_coordinate("40.819057,-73.914048", "100", "restaurant")

for place in places:
    details = api.get_place_setails(place['place_id'], fields)
    try:
        website = details['result']['website']
    except KeyError:
        website = ""

    try:
        name = details['result']['name']
    except KeyError:
        name = ""

    try:
        address = details['result']['formatted_address']
    except KeyError:
        address = ""

    try:
        phone_number = details['result']['international_phone_number']
    except KeyError:
        phone_number = ""

    try:
        reviews = details['result']['reviews']
    except KeyError:
        reviews = []
    print("===================PLACE===================")
    print("Name:", name)
    print("Website:", website)
    print("Address:", address)
    print("Phone Number", phone_number)
    print("==================REVIEWS==================")
    for review in reviews:
        author_name = review['author_name']
        rating = review['rating']
        text = review['text']
        time = review['relative_time_description']
        profile_photo = review['profile_photo_url']
        print("Author Name:", author_name)
        print("Rating:", rating)
        print("Text:", text)
        print("Time:", time)
        print("Profile photo:", profile_photo)
        print("-----------------------------------------")

有关此代码和 Google Places API 的更多详细信息，您可以查看此教程：https://python.gotrained.com/google-places-api-extracting-location-data-reviews/

Answer 5

我在 localhost 上运行时遇到了问题。我添加了我的 example.com.test 域，但我当然无法验证它，因为无法从外部访问它（ngrok 变体除外）。

我在 GitHub 上发现了一个惊人的肮脏黑客：gaffling/PHP-Grab-Google-Reviews。

对我来说效果很好，除了我必须将 /* CHECK SORT */ 行更改为 if (isset($option['sort_by_reating_best_1']) and $option['sort_by_reating_best_1'] == true) 并且我还通过可选的第二个函数参数将 foreach 限制为仅 5 条评论。

根本不需要 API_KEY