MySQL POINT字段的PHP數據類型

Question

我創建了一個MySQL表，其中的字段為POINT類型，用於存儲經緯度坐標（例如：36.6294654 -93.1725982）。

我在提交表單時遇到錯誤，我認為這是由於數據類型不匹配所致。

SQLSTATE[22003]: Numeric value out of range: 1416 Cannot get geometry object from data you send to the GEOMETRY field

我的理解是POINT應該具有示例的格式（我也嘗試過lat，lon）。 我認為問題在於空格將變量轉換為STRING 。

我在想什么嗎？

這是我的PHP代碼，用於連接到數據庫並插入記錄。

// use google's geocoding service to transform an address into lat/lon coordinates (https://developers.google.com/maps/documentation/geocoding/)
$json = file_get_contents('https://maps.googleapis.com/maps/api/geocode/json?address=' . urlencode($street) . ',+' . urlencode($city) . ',+' . urlencode($state) . ',+' . urlencode($zip) . '&key=YOUR_API_KEY');
$json = json_decode($json, true);
$latlon = $json['results'][0]['geometry']['location']['lat'] . ' ' . $json['results'][0]['geometry']['location']['lng'];
try {
    $dbh = new PDO('mysql:host=someserver;port=someport;dbname=somedatabase', 'someusername', 'somepassword', array(PDO::ATTR_PERSISTENT => true));
    $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $sql = 'INSERT INTO Venues (Name, Street, City, State, Zip, Country, TimeZone, LatLon, Phone, Email, Website, Host, Notes) VALUES (:name, :street, :city, :state, :zip, :country, :timezone, :latlon, :phone, :email, :website, :host, :notes)';
    $stmt = $dbh->prepare($sql);
    $stmt->execute(array(":name" => $name, ":street" => $street, ":city" => $city, ":state" => $state, ":zip" => $zip, ":country" => $country, ":timezone" => $timezone, ":latlon" => $latlon, ":phone" => $phone, ":email" => $email, ":website" => $website, ":host" => $host, ":notes" => $notes));
    if ($stmt->rowCount() == 1) {
        echo '<p>"' . $name . '" creation succeeded.</p>';
    } else {
        echo '<p>"' . $name . '" creation failed.</p>';
    }
    $stmt->setFetchMode(PDO::FETCH_ASSOC);
    $dbh = null;
}
catch(PDOException $e) {
    echo $e->getMessage();
}

另外，我可以有單獨的緯度和經度字段，但是我想練習使用msyql的內置幾何功能。

編輯...

按照這個答案的建議，我用POINT()包圍了$latlon 。 不過，它並沒有改變結果。

編輯2 ...

這是表結構，以防萬一看起來不正確。

VenueKey    int(11)
Name        varchar(255)
Street      varchar(255)
City        varchar(255)
State       varchar(2)
Zip         varchar(10)
Country     varchar(2)
TimeZone    varchar(20)
LatLon      point
Phone       varchar(20)
Email       varchar(255)
Website     varchar(255)
Host        varchar(255)
Notes       text

Answer 1

這應該可以解決您的問題：

$latlng = 'POINT(' . $json['results'][0]['geometry']['location']['lat']. " " . $json['results'][0]['geometry']['location']['lng']. ')';

然后用PointFromText包裝:latlon PointFromText ：

$sql = 'INSERT INTO Venues (Name, Street, City, State, Zip, Country, TimeZone, LatLon, Phone, Email, Website, Host, Notes) VALUES (:name, :street, :city, :state, :zip, :country, :timezone, PointFromText(:latlon), :phone, :email, :website, :host, :notes)';