Question

我想将查询的结果回显到数组中。

这些是原始行：

{
$locations = array();
$locations[0] = array("image"=>"https://www.XX.nl/assets/img/prinsenhof.png","lat"=>"52.012136","lng"=>"4.354596","emp_name"=>"Museum Prinsenhof Delft","emp_id"=>"Museum Prinsenhof Delft");
$locations[1] = array("image"=>"https://www.XX.nl/assets/img/ampelman.png","lat"=>"51.8988595","lng"=>"4.4186571","emp_name"=>"Ampelmann Operations","emp_id"=>"Ampelmann Operations");
$locations[2] = array("image"=>"https://www.XX.nl/assets/img/koekamp.png","lat"=>"52.0821565","lng"=>"4.3202341","emp_name"=>"Koekamp The Hague","emp_id"=>"Koekamp The Hague");
$locations[3] = array("image"=>"","lat"=>"51.9179543","lng"=>"4.3986012","emp_name"=>"Grote of Sint Janskerk Schiedam","emp_id"=>"Grote of Sint Janskerk Schiedam");
$locations[4] = array("image"=>"","lat"=>"52.0596095","lng"=>"4.2219163","emp_name"=>"The International School of The Hague","emp_id"=>"The International School of The Hague");
$locations[5] = array("image"=>"","lat"=>"52.012663","lng"=>"4.3558941","emp_name"=>"Oude Kerk","emp_id"=>"Oude Kerk");
$locations[6] = array("image"=>"","lat"=>"52.0769897","lng"=>"4.3170919","emp_name"=>"Spuiplein","emp_id"=>"Spuiplein");
$locations = json_encode($locations);
}

所以我把这个信息放在数据库中，我想显示它这就是我做的：

    $qry = "SELECT image, lat, lng, name, emp_id FROM googlemaps";
        if(!$result = $connection->query($qry)) {
            echo 'Fout in query: '.$mysqli->error;
        } else {
        $a = 0; 
        while ($location = $result->fetch_assoc()){

    $locations[$a] .= array("image"=>"https://www.XXXX.nl/assets/img/".$location['image'],"lat"=>$location['lat'],"lng"=>$location['lng'],"emp_name"=>$location['name'],"emp_id"=>$location['emp_id']);
    $a++;
            }
        }
        echo $locations;
echo $locations = json_encode($locations);

有更好的主意吗？

错误：

Notice: Undefined offset: 0 in /index.php on line 137
Notice: Array to string conversion in /index.php on line 137
Notice: Array to string conversion in /index.php on line 137
Notice: Array to string conversion in /index.php on line 141

Answer 1

Notice: Undefined offset: 0 in /index.php on line 137
Notice: Array to string conversion in /index.php on line 137
Notice: Array to string conversion in /index.php on line 137

这是因为：

$locations[$a] .= array("image"=>"https://www.XXXX.nl/assets/img/".$location['image'],"lat"=>$location['image'],"lng"=>$location['lng'],"emp_name"=>$location['name'],"emp_id"=>$location['emp_id']);

第一个告诉您$locations[0]不存在。剩下的是因为您试图将字符串追加到没有意义的数组中。

Notice: Array to string conversion in /index.php on line 141

这是因为echo $locations;。您不能只是将数组转换为字符串。您需要使用var_dump或print_r。

有关如何执行此操作。首先，turn on mysqli exceptions so you don't have to constantly check for error。

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

然后使用locations将pushing行的新结果遍历到locations[] = array(...)上。

$result = $connection->query("SELECT image, lat, lng, name, emp_id FROM googlemaps");
while ($location = $result->fetch_assoc()){
    $locations[] = array(
        "image" =>  "https://www.XXXX.nl/assets/img/".$location['image'],
        "lat"   =>  $location['lat'],
        "lng"   =>  $location['lng'],
        "emp_name"  =>  $location['name'],
        "emp_id"    =>  $location['emp_id']
    );
}

您可以通过在查询中完成所有操作来提高效率。

$result = $connection->query(<<<QUERY
    SELECT
        CONCAT("https://www.XXXX.nl/assets/img/", image),
        lat,
        lng,
        name as emp_name,
        emp_id
    FROM googlemaps
QUERY
);
while ($location = $result->fetch_assoc()){
    # now use $location directly
}

这避免了必须复制每一行，并且避免了在内存中建立潜在的巨大行数组。

如果您确实需要构建$locations，现在可以使用fetch_all一步完成，并且速度更快。

$locations = $result->fetch_all;

将结果查询回显到数组

1 个答案: