Php 传递参数查询

我正试图从城市获取lat、lng。当我最近在查询中写下城市“Buffalo”时，它起作用了。我想使用不同的城市

public function getLatLng($data, $city) {
    $count_data = count($data);
    for($i = 0; $i < $count_data; $i++) {
        $latlngs = array();
        $stmt = $this -> conn -> prepare("SELECT g.lat, g.lng
          FROM news_locations AS nl INNER JOIN geolocations AS g
          ON g.geolocation_id = nl.geolocation_id
          WHERE nl.news_id = ". $data[$i]["news_id"] ."
          AND nl.is_deleted = 0 AND g.city = 'Buffalo'");
        if ($stmt -> execute()) {
            $stmt -> bind_result($lat, $lng);
            while($stmt -> fetch()) {
               $row = array();
               $row["lat"] = $lat;
               $row["lng"] = $lng;
               array_push($latlngs, $row);
            }
            $stmt -> close();
        }
        $data[$i]['latlng'] = array_values($latlngs);
    }
    return $data;
}

也试过

AND g.city = :city");
if ($stmt -> execute(array('city' => $city))) {

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两者都不起作用。如何将参数$city传递给查询？

连接字符串和变量：

g.city = '"  . $city . "'"

比如：

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您必须将变量绑定到sql语句。@Jumana您得到的指导很差。如果您在应用程序中使用了接受的答案，那么您的应用程序是不安全的。即使这样做有效，也不是最佳做法，因为它无法利用准备好的语句来保护查询。任何人都不应该使用这个答案。

g.city = '"  . $city . "'"

 $stmt = $this -> conn -> prepare("SELECT g.lat, g.lng
          FROM news_locations AS nl INNER JOIN geolocations AS g
          ON g.geolocation_id = nl.geolocation_id
          WHERE nl.news_id = ". $data[$i]["news_id"] ."
          AND nl.is_deleted = 0 AND g.city = '"  . $city . "'");