Question

我在php中有一个函数：

//simple method with array()
$sensors = array();
$query = "select id, x(transform(wkb_geometry,". $epsg . ")) as lon, y(transform(wkb_geometry,". $epsg . ")) as lat from mytable;";
$result = pg_query($query) or die('Query failed: ' . pg_last_error());
while ($row = pg_fetch_assoc($result)) {
            //var_dump($row);
            $mySensor = new sensor($row['id'],$row['lat'],$row['lon']);
            $sensors[] = $mySensor->geoJSON();
}

echo json_encode($sensors);

输出：

    "features": [{
        "type": "Feature",
        "id": 1579028,
        "x": 4.85310557823,
        "y": 52.7205622103,
        "geometry": {
            "type": "Point",
            "coordinates": [4.85310557823, 52.7205622103],
            "crs": {
                "type": "OGC",
                "properties": {
                    "urn": "urn:ogc:def:crs:OGC:1.3:CRS84"
                }
            }

现在我已经重写了数组，成为这样的对象：

    //advanced method with arrayObject:
    class sensors extends ArrayObject {
        function __construct($epsg){
            $query = "select id, x(transform(wkb_geometry,". $epsg . ")) as lon, y(transform(wkb_geometry,". $epsg . ")) as lat from mytable;";
            $result = pg_query($query) or die('Query failed: ' . pg_last_error());
            while ($row = pg_fetch_assoc($result)) {
                //var_dump($row);
                $mySensor = new sensor($row['id'],$row['lat'],$row['lon']);
                $this[] = $mySensor->geoJSON();
            }
        }
    }
$newsensors = new sensors($epsg);
echo echo json_encode($newsensors);

但是这会将输出更改为：

 "features": {
            "0": {
                "type": "Feature",
                "id": 1579028,
                "x": 4.85310557823,
                "y": 52.7205622103,
                "geometry": {
                    "type": "Point",
                    "coordinates": [4.85310557823, 52.7205622103],
                    "crs": {
                        "type": "OGC",
                        "properties": {
                            "urn": "urn:ogc:def:crs:OGC:1.3:CRS84"
                        }
                    }
                }
            },

这使得它无法用作OpenLayers的geoJSON。为什么json_encode函数会以这种方式运行？我可以关闭索引号的设置吗？这可能是一个小错误吗？

Answer 1

json_encode将显示与任何对象相同的行为，即使是ArrayAccess接口实现ArrayObject接口的行为也是如此;使用公共财产。

要获得您想要的行为，您应该传递一个实际的数组，可以通过调用ArrayObject::getArrayCopy()来检索（或者您可以将对象强制转换为数组）。

echo json_encode($newsensors->getArrayCopy());

Answer 2

不得不转换包含不同级别的ArrayObject的数据树，所以我在输出json之前编写了一段代码来处理和转换每个ArrayObject，希望它有所帮助： https://github.com/nfroidure/Rest4/blob/master/php/class.Json.php

Answer 3

使用JsonSerializable界面（PHP 5＆gt; = 5.4.0，PHP 7），您应该是安全的：

public function jsonSerialize()
{
    return $this->getArrayCopy();
}

json_encode在arrayObject vs array（）上表现不同

3 个答案: